ABCF->ab-angle angle

Percentage Accurate: 53.5% → 81.6%
Time: 22.2s
Alternatives: 19
Speedup: 3.6×

Specification

?
\[\begin{array}{l} \\ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (*
  180.0
  (/
   (atan (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))
   PI)))
double code(double A, double B, double C) {
	return 180.0 * (atan(((1.0 / B) * ((C - A) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / ((double) M_PI));
}
public static double code(double A, double B, double C) {
	return 180.0 * (Math.atan(((1.0 / B) * ((C - A) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / Math.PI);
}
def code(A, B, C):
	return 180.0 * (math.atan(((1.0 / B) * ((C - A) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / math.pi)
function code(A, B, C)
	return Float64(180.0 * Float64(atan(Float64(Float64(1.0 / B) * Float64(Float64(C - A) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0)))))) / pi))
end
function tmp = code(A, B, C)
	tmp = 180.0 * (atan(((1.0 / B) * ((C - A) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / pi);
end
code[A_, B_, C_] := N[(180.0 * N[(N[ArcTan[N[(N[(1.0 / B), $MachinePrecision] * N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 19 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 53.5% accurate, 1.0× speedup?

\[\begin{array}{l} \\ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (*
  180.0
  (/
   (atan (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))
   PI)))
double code(double A, double B, double C) {
	return 180.0 * (atan(((1.0 / B) * ((C - A) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / ((double) M_PI));
}
public static double code(double A, double B, double C) {
	return 180.0 * (Math.atan(((1.0 / B) * ((C - A) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / Math.PI);
}
def code(A, B, C):
	return 180.0 * (math.atan(((1.0 / B) * ((C - A) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / math.pi)
function code(A, B, C)
	return Float64(180.0 * Float64(atan(Float64(Float64(1.0 / B) * Float64(Float64(C - A) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0)))))) / pi))
end
function tmp = code(A, B, C)
	tmp = 180.0 * (atan(((1.0 / B) * ((C - A) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / pi);
end
code[A_, B_, C_] := N[(180.0 * N[(N[ArcTan[N[(N[(1.0 / B), $MachinePrecision] * N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}
\end{array}

Alternative 1: 81.6% accurate, 1.9× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -1.05 \cdot 10^{+136}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}\\ \end{array} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (if (<= A -1.05e+136)
   (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
   (/ 1.0 (/ PI (* 180.0 (atan (/ (- (- C A) (hypot (- A C) B)) B)))))))
double code(double A, double B, double C) {
	double tmp;
	if (A <= -1.05e+136) {
		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
	} else {
		tmp = 1.0 / (((double) M_PI) / (180.0 * atan((((C - A) - hypot((A - C), B)) / B))));
	}
	return tmp;
}
public static double code(double A, double B, double C) {
	double tmp;
	if (A <= -1.05e+136) {
		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
	} else {
		tmp = 1.0 / (Math.PI / (180.0 * Math.atan((((C - A) - Math.hypot((A - C), B)) / B))));
	}
	return tmp;
}
def code(A, B, C):
	tmp = 0
	if A <= -1.05e+136:
		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
	else:
		tmp = 1.0 / (math.pi / (180.0 * math.atan((((C - A) - math.hypot((A - C), B)) / B))))
	return tmp
function code(A, B, C)
	tmp = 0.0
	if (A <= -1.05e+136)
		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
	else
		tmp = Float64(1.0 / Float64(pi / Float64(180.0 * atan(Float64(Float64(Float64(C - A) - hypot(Float64(A - C), B)) / B)))));
	end
	return tmp
end
function tmp_2 = code(A, B, C)
	tmp = 0.0;
	if (A <= -1.05e+136)
		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
	else
		tmp = 1.0 / (pi / (180.0 * atan((((C - A) - hypot((A - C), B)) / B))));
	end
	tmp_2 = tmp;
end
code[A_, B_, C_] := If[LessEqual[A, -1.05e+136], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(1.0 / N[(Pi / N[(180.0 * N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(A - C), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;A \leq -1.05 \cdot 10^{+136}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if A < -1.05e136

    1. Initial program 8.6%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 85.5%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
    4. Step-by-step derivation
      1. mul-1-neg85.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
      2. distribute-neg-frac285.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
      3. distribute-lft-out85.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
      4. associate-/l*89.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
    5. Simplified89.5%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
    6. Taylor expanded in B around 0 89.5%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

    if -1.05e136 < A

    1. Initial program 61.8%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Applied egg-rr83.1%

      \[\leadsto \color{blue}{\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification83.9%

    \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -1.05 \cdot 10^{+136}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 78.9% accurate, 1.9× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -3.5 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 10^{-182}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A\right)\right)}{B}\right)}{\pi}\\ \end{array} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (if (<= A -3.5e+132)
   (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
   (if (<= A 1e-182)
     (/ (* 180.0 (atan (/ (- C (hypot B C)) B))) PI)
     (* 180.0 (/ (atan (/ (- C (+ A (hypot B A))) B)) PI)))))
double code(double A, double B, double C) {
	double tmp;
	if (A <= -3.5e+132) {
		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
	} else if (A <= 1e-182) {
		tmp = (180.0 * atan(((C - hypot(B, C)) / B))) / ((double) M_PI);
	} else {
		tmp = 180.0 * (atan(((C - (A + hypot(B, A))) / B)) / ((double) M_PI));
	}
	return tmp;
}
public static double code(double A, double B, double C) {
	double tmp;
	if (A <= -3.5e+132) {
		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
	} else if (A <= 1e-182) {
		tmp = (180.0 * Math.atan(((C - Math.hypot(B, C)) / B))) / Math.PI;
	} else {
		tmp = 180.0 * (Math.atan(((C - (A + Math.hypot(B, A))) / B)) / Math.PI);
	}
	return tmp;
}
def code(A, B, C):
	tmp = 0
	if A <= -3.5e+132:
		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
	elif A <= 1e-182:
		tmp = (180.0 * math.atan(((C - math.hypot(B, C)) / B))) / math.pi
	else:
		tmp = 180.0 * (math.atan(((C - (A + math.hypot(B, A))) / B)) / math.pi)
	return tmp
function code(A, B, C)
	tmp = 0.0
	if (A <= -3.5e+132)
		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
	elseif (A <= 1e-182)
		tmp = Float64(Float64(180.0 * atan(Float64(Float64(C - hypot(B, C)) / B))) / pi);
	else
		tmp = Float64(180.0 * Float64(atan(Float64(Float64(C - Float64(A + hypot(B, A))) / B)) / pi));
	end
	return tmp
end
function tmp_2 = code(A, B, C)
	tmp = 0.0;
	if (A <= -3.5e+132)
		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
	elseif (A <= 1e-182)
		tmp = (180.0 * atan(((C - hypot(B, C)) / B))) / pi;
	else
		tmp = 180.0 * (atan(((C - (A + hypot(B, A))) / B)) / pi);
	end
	tmp_2 = tmp;
end
code[A_, B_, C_] := If[LessEqual[A, -3.5e+132], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[A, 1e-182], N[(N[(180.0 * N[ArcTan[N[(N[(C - N[Sqrt[B ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / Pi), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(N[(C - N[(A + N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]
\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;A \leq -3.5 \cdot 10^{+132}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\

\mathbf{elif}\;A \leq 10^{-182}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\

\mathbf{else}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A\right)\right)}{B}\right)}{\pi}\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if A < -3.5000000000000002e132

    1. Initial program 8.6%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 85.5%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
    4. Step-by-step derivation
      1. mul-1-neg85.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
      2. distribute-neg-frac285.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
      3. distribute-lft-out85.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
      4. associate-/l*89.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
    5. Simplified89.5%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
    6. Taylor expanded in B around 0 89.5%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

    if -3.5000000000000002e132 < A < 1e-182

    1. Initial program 50.9%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 49.9%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
    4. Step-by-step derivation
      1. unpow249.9%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
      2. unpow249.9%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
      3. hypot-define74.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
    5. Simplified74.6%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
    6. Step-by-step derivation
      1. associate-*r/74.6%

        \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}{\pi}} \]
      2. associate-*l/74.6%

        \[\leadsto \frac{180 \cdot \tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}{B}\right)}}{\pi} \]
      3. *-un-lft-identity74.6%

        \[\leadsto \frac{180 \cdot \tan^{-1} \left(\frac{\color{blue}{C - \mathsf{hypot}\left(B, C\right)}}{B}\right)}{\pi} \]
    7. Applied egg-rr74.6%

      \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}} \]

    if 1e-182 < A

    1. Initial program 73.6%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Step-by-step derivation
      1. Simplified91.9%

        \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}} \]
      2. Add Preprocessing
      3. Taylor expanded in C around 0 73.4%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \color{blue}{\left(A + \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}\right)}{\pi} \]
      4. Step-by-step derivation
        1. +-commutative73.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)}{B}\right)}{\pi} \]
        2. unpow273.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)}{B}\right)}{\pi} \]
        3. unpow273.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)}{B}\right)}{\pi} \]
        4. hypot-define89.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)}{B}\right)}{\pi} \]
      5. Simplified89.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \color{blue}{\left(A + \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)}{\pi} \]
    3. Recombined 3 regimes into one program.
    4. Final simplification82.8%

      \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -3.5 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 10^{-182}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A\right)\right)}{B}\right)}{\pi}\\ \end{array} \]
    5. Add Preprocessing

    Alternative 3: 78.8% accurate, 1.9× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -2 \cdot 10^{+136}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 3 \cdot 10^{-179}:\\ \;\;\;\;\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A\right)\right)}{B}\right)}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (if (<= A -2e+136)
       (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
       (if (<= A 3e-179)
         (/ 1.0 (/ PI (* 180.0 (atan (/ (- C (hypot B C)) B)))))
         (* 180.0 (/ (atan (/ (- C (+ A (hypot B A))) B)) PI)))))
    double code(double A, double B, double C) {
    	double tmp;
    	if (A <= -2e+136) {
    		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
    	} else if (A <= 3e-179) {
    		tmp = 1.0 / (((double) M_PI) / (180.0 * atan(((C - hypot(B, C)) / B))));
    	} else {
    		tmp = 180.0 * (atan(((C - (A + hypot(B, A))) / B)) / ((double) M_PI));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double tmp;
    	if (A <= -2e+136) {
    		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
    	} else if (A <= 3e-179) {
    		tmp = 1.0 / (Math.PI / (180.0 * Math.atan(((C - Math.hypot(B, C)) / B))));
    	} else {
    		tmp = 180.0 * (Math.atan(((C - (A + Math.hypot(B, A))) / B)) / Math.PI);
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	tmp = 0
    	if A <= -2e+136:
    		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
    	elif A <= 3e-179:
    		tmp = 1.0 / (math.pi / (180.0 * math.atan(((C - math.hypot(B, C)) / B))))
    	else:
    		tmp = 180.0 * (math.atan(((C - (A + math.hypot(B, A))) / B)) / math.pi)
    	return tmp
    
    function code(A, B, C)
    	tmp = 0.0
    	if (A <= -2e+136)
    		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
    	elseif (A <= 3e-179)
    		tmp = Float64(1.0 / Float64(pi / Float64(180.0 * atan(Float64(Float64(C - hypot(B, C)) / B)))));
    	else
    		tmp = Float64(180.0 * Float64(atan(Float64(Float64(C - Float64(A + hypot(B, A))) / B)) / pi));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	tmp = 0.0;
    	if (A <= -2e+136)
    		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
    	elseif (A <= 3e-179)
    		tmp = 1.0 / (pi / (180.0 * atan(((C - hypot(B, C)) / B))));
    	else
    		tmp = 180.0 * (atan(((C - (A + hypot(B, A))) / B)) / pi);
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := If[LessEqual[A, -2e+136], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[A, 3e-179], N[(1.0 / N[(Pi / N[(180.0 * N[ArcTan[N[(N[(C - N[Sqrt[B ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(N[(C - N[(A + N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    \mathbf{if}\;A \leq -2 \cdot 10^{+136}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\
    
    \mathbf{elif}\;A \leq 3 \cdot 10^{-179}:\\
    \;\;\;\;\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}}\\
    
    \mathbf{else}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A\right)\right)}{B}\right)}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 3 regimes
    2. if A < -2.00000000000000012e136

      1. Initial program 8.6%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in A around -inf 85.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. mul-1-neg85.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
        2. distribute-neg-frac285.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
        3. distribute-lft-out85.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
        4. associate-/l*89.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
      5. Simplified89.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
      6. Taylor expanded in B around 0 89.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

      if -2.00000000000000012e136 < A < 3.00000000000000006e-179

      1. Initial program 50.9%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr74.9%

        \[\leadsto \color{blue}{\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in A around 0 49.9%

        \[\leadsto \frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{\color{blue}{C - \sqrt{{B}^{2} + {C}^{2}}}}{B}\right)}} \]
      5. Step-by-step derivation
        1. unpow249.9%

          \[\leadsto \frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}}{B}\right)}} \]
        2. unpow249.9%

          \[\leadsto \frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}}{B}\right)}} \]
        3. hypot-undefine74.7%

          \[\leadsto \frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{C - \color{blue}{\mathsf{hypot}\left(B, C\right)}}{B}\right)}} \]
      6. Simplified74.7%

        \[\leadsto \frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{\color{blue}{C - \mathsf{hypot}\left(B, C\right)}}{B}\right)}} \]

      if 3.00000000000000006e-179 < A

      1. Initial program 73.6%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Step-by-step derivation
        1. Simplified91.9%

          \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}} \]
        2. Add Preprocessing
        3. Taylor expanded in C around 0 73.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \color{blue}{\left(A + \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}\right)}{\pi} \]
        4. Step-by-step derivation
          1. +-commutative73.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)}{B}\right)}{\pi} \]
          2. unpow273.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)}{B}\right)}{\pi} \]
          3. unpow273.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)}{B}\right)}{\pi} \]
          4. hypot-define89.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)}{B}\right)}{\pi} \]
        5. Simplified89.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \color{blue}{\left(A + \mathsf{hypot}\left(B, A\right)\right)}}{B}\right)}{\pi} \]
      3. Recombined 3 regimes into one program.
      4. Final simplification82.8%

        \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -2 \cdot 10^{+136}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 3 \cdot 10^{-179}:\\ \;\;\;\;\frac{1}{\frac{\pi}{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A\right)\right)}{B}\right)}{\pi}\\ \end{array} \]
      5. Add Preprocessing

      Alternative 4: 78.4% accurate, 1.9× speedup?

      \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -2.3 \cdot 10^{+135}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 1.22 \cdot 10^{-76}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A + \mathsf{hypot}\left(B, A\right)}{-B}\right)}{\pi}\\ \end{array} \end{array} \]
      (FPCore (A B C)
       :precision binary64
       (if (<= A -2.3e+135)
         (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
         (if (<= A 1.22e-76)
           (/ (* 180.0 (atan (/ (- C (hypot B C)) B))) PI)
           (* 180.0 (/ (atan (/ (+ A (hypot B A)) (- B))) PI)))))
      double code(double A, double B, double C) {
      	double tmp;
      	if (A <= -2.3e+135) {
      		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
      	} else if (A <= 1.22e-76) {
      		tmp = (180.0 * atan(((C - hypot(B, C)) / B))) / ((double) M_PI);
      	} else {
      		tmp = 180.0 * (atan(((A + hypot(B, A)) / -B)) / ((double) M_PI));
      	}
      	return tmp;
      }
      
      public static double code(double A, double B, double C) {
      	double tmp;
      	if (A <= -2.3e+135) {
      		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
      	} else if (A <= 1.22e-76) {
      		tmp = (180.0 * Math.atan(((C - Math.hypot(B, C)) / B))) / Math.PI;
      	} else {
      		tmp = 180.0 * (Math.atan(((A + Math.hypot(B, A)) / -B)) / Math.PI);
      	}
      	return tmp;
      }
      
      def code(A, B, C):
      	tmp = 0
      	if A <= -2.3e+135:
      		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
      	elif A <= 1.22e-76:
      		tmp = (180.0 * math.atan(((C - math.hypot(B, C)) / B))) / math.pi
      	else:
      		tmp = 180.0 * (math.atan(((A + math.hypot(B, A)) / -B)) / math.pi)
      	return tmp
      
      function code(A, B, C)
      	tmp = 0.0
      	if (A <= -2.3e+135)
      		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
      	elseif (A <= 1.22e-76)
      		tmp = Float64(Float64(180.0 * atan(Float64(Float64(C - hypot(B, C)) / B))) / pi);
      	else
      		tmp = Float64(180.0 * Float64(atan(Float64(Float64(A + hypot(B, A)) / Float64(-B))) / pi));
      	end
      	return tmp
      end
      
      function tmp_2 = code(A, B, C)
      	tmp = 0.0;
      	if (A <= -2.3e+135)
      		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
      	elseif (A <= 1.22e-76)
      		tmp = (180.0 * atan(((C - hypot(B, C)) / B))) / pi;
      	else
      		tmp = 180.0 * (atan(((A + hypot(B, A)) / -B)) / pi);
      	end
      	tmp_2 = tmp;
      end
      
      code[A_, B_, C_] := If[LessEqual[A, -2.3e+135], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[A, 1.22e-76], N[(N[(180.0 * N[ArcTan[N[(N[(C - N[Sqrt[B ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / Pi), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(N[(A + N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision] / (-B)), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]
      
      \begin{array}{l}
      
      \\
      \begin{array}{l}
      \mathbf{if}\;A \leq -2.3 \cdot 10^{+135}:\\
      \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\
      
      \mathbf{elif}\;A \leq 1.22 \cdot 10^{-76}:\\
      \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\
      
      \mathbf{else}:\\
      \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A + \mathsf{hypot}\left(B, A\right)}{-B}\right)}{\pi}\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 3 regimes
      2. if A < -2.3000000000000001e135

        1. Initial program 8.6%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in A around -inf 85.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
        4. Step-by-step derivation
          1. mul-1-neg85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
          2. distribute-neg-frac285.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
          3. distribute-lft-out85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
          4. associate-/l*89.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
        5. Simplified89.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
        6. Taylor expanded in B around 0 89.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

        if -2.3000000000000001e135 < A < 1.22e-76

        1. Initial program 52.5%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in A around 0 49.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
        4. Step-by-step derivation
          1. unpow249.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
          2. unpow249.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
          3. hypot-define74.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
        5. Simplified74.8%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
        6. Step-by-step derivation
          1. associate-*r/74.9%

            \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}{\pi}} \]
          2. associate-*l/74.9%

            \[\leadsto \frac{180 \cdot \tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}{B}\right)}}{\pi} \]
          3. *-un-lft-identity74.9%

            \[\leadsto \frac{180 \cdot \tan^{-1} \left(\frac{\color{blue}{C - \mathsf{hypot}\left(B, C\right)}}{B}\right)}{\pi} \]
        7. Applied egg-rr74.9%

          \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}} \]

        if 1.22e-76 < A

        1. Initial program 79.1%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in C around 0 78.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + \sqrt{{A}^{2} + {B}^{2}}}{B}\right)}}{\pi} \]
        4. Step-by-step derivation
          1. mul-1-neg78.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{A + \sqrt{{A}^{2} + {B}^{2}}}{B}\right)}}{\pi} \]
          2. distribute-neg-frac278.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{A + \sqrt{{A}^{2} + {B}^{2}}}{-B}\right)}}{\pi} \]
          3. +-commutative78.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{A + \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}}{-B}\right)}{\pi} \]
          4. unpow278.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{A + \sqrt{\color{blue}{B \cdot B} + {A}^{2}}}{-B}\right)}{\pi} \]
          5. unpow278.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{A + \sqrt{B \cdot B + \color{blue}{A \cdot A}}}{-B}\right)}{\pi} \]
          6. hypot-define91.3%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{A + \color{blue}{\mathsf{hypot}\left(B, A\right)}}{-B}\right)}{\pi} \]
        5. Simplified91.3%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{A + \mathsf{hypot}\left(B, A\right)}{-B}\right)}}{\pi} \]
      3. Recombined 3 regimes into one program.
      4. Final simplification81.7%

        \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -2.3 \cdot 10^{+135}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 1.22 \cdot 10^{-76}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A + \mathsf{hypot}\left(B, A\right)}{-B}\right)}{\pi}\\ \end{array} \]
      5. Add Preprocessing

      Alternative 5: 75.9% accurate, 1.9× speedup?

      \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -2.2 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 1.8 \cdot 10^{+39}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \end{array} \end{array} \]
      (FPCore (A B C)
       :precision binary64
       (if (<= A -2.2e+132)
         (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
         (if (<= A 1.8e+39)
           (* 180.0 (/ (atan (/ (- C (hypot B C)) B)) PI))
           (/ 180.0 (/ PI (atan (+ 1.0 (/ (- C A) B))))))))
      double code(double A, double B, double C) {
      	double tmp;
      	if (A <= -2.2e+132) {
      		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
      	} else if (A <= 1.8e+39) {
      		tmp = 180.0 * (atan(((C - hypot(B, C)) / B)) / ((double) M_PI));
      	} else {
      		tmp = 180.0 / (((double) M_PI) / atan((1.0 + ((C - A) / B))));
      	}
      	return tmp;
      }
      
      public static double code(double A, double B, double C) {
      	double tmp;
      	if (A <= -2.2e+132) {
      		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
      	} else if (A <= 1.8e+39) {
      		tmp = 180.0 * (Math.atan(((C - Math.hypot(B, C)) / B)) / Math.PI);
      	} else {
      		tmp = 180.0 / (Math.PI / Math.atan((1.0 + ((C - A) / B))));
      	}
      	return tmp;
      }
      
      def code(A, B, C):
      	tmp = 0
      	if A <= -2.2e+132:
      		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
      	elif A <= 1.8e+39:
      		tmp = 180.0 * (math.atan(((C - math.hypot(B, C)) / B)) / math.pi)
      	else:
      		tmp = 180.0 / (math.pi / math.atan((1.0 + ((C - A) / B))))
      	return tmp
      
      function code(A, B, C)
      	tmp = 0.0
      	if (A <= -2.2e+132)
      		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
      	elseif (A <= 1.8e+39)
      		tmp = Float64(180.0 * Float64(atan(Float64(Float64(C - hypot(B, C)) / B)) / pi));
      	else
      		tmp = Float64(180.0 / Float64(pi / atan(Float64(1.0 + Float64(Float64(C - A) / B)))));
      	end
      	return tmp
      end
      
      function tmp_2 = code(A, B, C)
      	tmp = 0.0;
      	if (A <= -2.2e+132)
      		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
      	elseif (A <= 1.8e+39)
      		tmp = 180.0 * (atan(((C - hypot(B, C)) / B)) / pi);
      	else
      		tmp = 180.0 / (pi / atan((1.0 + ((C - A) / B))));
      	end
      	tmp_2 = tmp;
      end
      
      code[A_, B_, C_] := If[LessEqual[A, -2.2e+132], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[A, 1.8e+39], N[(180.0 * N[(N[ArcTan[N[(N[(C - N[Sqrt[B ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 / N[(Pi / N[ArcTan[N[(1.0 + N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]
      
      \begin{array}{l}
      
      \\
      \begin{array}{l}
      \mathbf{if}\;A \leq -2.2 \cdot 10^{+132}:\\
      \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\
      
      \mathbf{elif}\;A \leq 1.8 \cdot 10^{+39}:\\
      \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\
      
      \mathbf{else}:\\
      \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 3 regimes
      2. if A < -2.19999999999999989e132

        1. Initial program 8.6%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in A around -inf 85.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
        4. Step-by-step derivation
          1. mul-1-neg85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
          2. distribute-neg-frac285.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
          3. distribute-lft-out85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
          4. associate-/l*89.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
        5. Simplified89.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
        6. Taylor expanded in B around 0 89.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

        if -2.19999999999999989e132 < A < 1.79999999999999992e39

        1. Initial program 54.6%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in A around 0 49.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{C - \sqrt{{B}^{2} + {C}^{2}}}{B}\right)}}{\pi} \]
        4. Step-by-step derivation
          1. unpow249.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}}{B}\right)}{\pi} \]
          2. unpow249.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}}{B}\right)}{\pi} \]
          3. hypot-define75.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \color{blue}{\mathsf{hypot}\left(B, C\right)}}{B}\right)}{\pi} \]
        5. Simplified75.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}}{\pi} \]

        if 1.79999999999999992e39 < A

        1. Initial program 85.1%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Applied egg-rr94.6%

          \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
        4. Taylor expanded in B around -inf 88.9%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
        5. Step-by-step derivation
          1. associate--l+88.9%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
          2. div-sub88.9%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
        6. Simplified88.9%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \frac{C - A}{B}\right)}}} \]
      3. Recombined 3 regimes into one program.
      4. Final simplification79.7%

        \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -2.2 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 1.8 \cdot 10^{+39}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \end{array} \]
      5. Add Preprocessing

      Alternative 6: 75.9% accurate, 1.9× speedup?

      \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -2.3 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 5.2 \cdot 10^{+37}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \end{array} \end{array} \]
      (FPCore (A B C)
       :precision binary64
       (if (<= A -2.3e+132)
         (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
         (if (<= A 5.2e+37)
           (/ (* 180.0 (atan (/ (- C (hypot B C)) B))) PI)
           (/ 180.0 (/ PI (atan (+ 1.0 (/ (- C A) B))))))))
      double code(double A, double B, double C) {
      	double tmp;
      	if (A <= -2.3e+132) {
      		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
      	} else if (A <= 5.2e+37) {
      		tmp = (180.0 * atan(((C - hypot(B, C)) / B))) / ((double) M_PI);
      	} else {
      		tmp = 180.0 / (((double) M_PI) / atan((1.0 + ((C - A) / B))));
      	}
      	return tmp;
      }
      
      public static double code(double A, double B, double C) {
      	double tmp;
      	if (A <= -2.3e+132) {
      		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
      	} else if (A <= 5.2e+37) {
      		tmp = (180.0 * Math.atan(((C - Math.hypot(B, C)) / B))) / Math.PI;
      	} else {
      		tmp = 180.0 / (Math.PI / Math.atan((1.0 + ((C - A) / B))));
      	}
      	return tmp;
      }
      
      def code(A, B, C):
      	tmp = 0
      	if A <= -2.3e+132:
      		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
      	elif A <= 5.2e+37:
      		tmp = (180.0 * math.atan(((C - math.hypot(B, C)) / B))) / math.pi
      	else:
      		tmp = 180.0 / (math.pi / math.atan((1.0 + ((C - A) / B))))
      	return tmp
      
      function code(A, B, C)
      	tmp = 0.0
      	if (A <= -2.3e+132)
      		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
      	elseif (A <= 5.2e+37)
      		tmp = Float64(Float64(180.0 * atan(Float64(Float64(C - hypot(B, C)) / B))) / pi);
      	else
      		tmp = Float64(180.0 / Float64(pi / atan(Float64(1.0 + Float64(Float64(C - A) / B)))));
      	end
      	return tmp
      end
      
      function tmp_2 = code(A, B, C)
      	tmp = 0.0;
      	if (A <= -2.3e+132)
      		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
      	elseif (A <= 5.2e+37)
      		tmp = (180.0 * atan(((C - hypot(B, C)) / B))) / pi;
      	else
      		tmp = 180.0 / (pi / atan((1.0 + ((C - A) / B))));
      	end
      	tmp_2 = tmp;
      end
      
      code[A_, B_, C_] := If[LessEqual[A, -2.3e+132], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[A, 5.2e+37], N[(N[(180.0 * N[ArcTan[N[(N[(C - N[Sqrt[B ^ 2 + C ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / Pi), $MachinePrecision], N[(180.0 / N[(Pi / N[ArcTan[N[(1.0 + N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]
      
      \begin{array}{l}
      
      \\
      \begin{array}{l}
      \mathbf{if}\;A \leq -2.3 \cdot 10^{+132}:\\
      \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\
      
      \mathbf{elif}\;A \leq 5.2 \cdot 10^{+37}:\\
      \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\
      
      \mathbf{else}:\\
      \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 3 regimes
      2. if A < -2.3000000000000002e132

        1. Initial program 8.6%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in A around -inf 85.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
        4. Step-by-step derivation
          1. mul-1-neg85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
          2. distribute-neg-frac285.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
          3. distribute-lft-out85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
          4. associate-/l*89.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
        5. Simplified89.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
        6. Taylor expanded in B around 0 89.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

        if -2.3000000000000002e132 < A < 5.1999999999999998e37

        1. Initial program 54.6%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in A around 0 49.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
        4. Step-by-step derivation
          1. unpow249.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
          2. unpow249.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
          3. hypot-define75.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
        5. Simplified75.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
        6. Step-by-step derivation
          1. associate-*r/75.0%

            \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)\right)}{\pi}} \]
          2. associate-*l/75.0%

            \[\leadsto \frac{180 \cdot \tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(C - \mathsf{hypot}\left(B, C\right)\right)}{B}\right)}}{\pi} \]
          3. *-un-lft-identity75.0%

            \[\leadsto \frac{180 \cdot \tan^{-1} \left(\frac{\color{blue}{C - \mathsf{hypot}\left(B, C\right)}}{B}\right)}{\pi} \]
        7. Applied egg-rr75.0%

          \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}} \]

        if 5.1999999999999998e37 < A

        1. Initial program 85.1%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Applied egg-rr94.6%

          \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
        4. Taylor expanded in B around -inf 88.9%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
        5. Step-by-step derivation
          1. associate--l+88.9%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
          2. div-sub88.9%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
        6. Simplified88.9%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \frac{C - A}{B}\right)}}} \]
      3. Recombined 3 regimes into one program.
      4. Final simplification79.7%

        \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -2.3 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 5.2 \cdot 10^{+37}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \end{array} \]
      5. Add Preprocessing

      Alternative 7: 81.1% accurate, 1.9× speedup?

      \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -3.3 \cdot 10^{+115}:\\ \;\;\;\;\frac{-180 \cdot \tan^{-1} \left(-0.5 \cdot \frac{\mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}\\ \end{array} \end{array} \]
      (FPCore (A B C)
       :precision binary64
       (if (<= A -3.3e+115)
         (/ (* -180.0 (atan (* -0.5 (/ (fma B (/ C A) B) A)))) PI)
         (* 180.0 (/ (atan (/ (- C (+ A (hypot B (- A C)))) B)) PI))))
      double code(double A, double B, double C) {
      	double tmp;
      	if (A <= -3.3e+115) {
      		tmp = (-180.0 * atan((-0.5 * (fma(B, (C / A), B) / A)))) / ((double) M_PI);
      	} else {
      		tmp = 180.0 * (atan(((C - (A + hypot(B, (A - C)))) / B)) / ((double) M_PI));
      	}
      	return tmp;
      }
      
      function code(A, B, C)
      	tmp = 0.0
      	if (A <= -3.3e+115)
      		tmp = Float64(Float64(-180.0 * atan(Float64(-0.5 * Float64(fma(B, Float64(C / A), B) / A)))) / pi);
      	else
      		tmp = Float64(180.0 * Float64(atan(Float64(Float64(C - Float64(A + hypot(B, Float64(A - C)))) / B)) / pi));
      	end
      	return tmp
      end
      
      code[A_, B_, C_] := If[LessEqual[A, -3.3e+115], N[(N[(-180.0 * N[ArcTan[N[(-0.5 * N[(N[(B * N[(C / A), $MachinePrecision] + B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / Pi), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(N[(C - N[(A + N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]
      
      \begin{array}{l}
      
      \\
      \begin{array}{l}
      \mathbf{if}\;A \leq -3.3 \cdot 10^{+115}:\\
      \;\;\;\;\frac{-180 \cdot \tan^{-1} \left(-0.5 \cdot \frac{\mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}{\pi}\\
      
      \mathbf{else}:\\
      \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 2 regimes
      2. if A < -3.30000000000000005e115

        1. Initial program 13.0%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in A around -inf 78.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
        4. Step-by-step derivation
          1. mul-1-neg78.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
          2. distribute-neg-frac278.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
          3. distribute-lft-out78.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
          4. associate-/l*81.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
        5. Simplified81.8%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
        6. Step-by-step derivation
          1. associate-*r/81.8%

            \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}{\pi}} \]
          2. distribute-frac-neg281.8%

            \[\leadsto \frac{180 \cdot \tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{A}\right)}}{\pi} \]
          3. atan-neg81.8%

            \[\leadsto \frac{180 \cdot \color{blue}{\left(-\tan^{-1} \left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{A}\right)\right)}}{\pi} \]
          4. +-commutative81.8%

            \[\leadsto \frac{180 \cdot \left(-\tan^{-1} \left(\frac{-0.5 \cdot \color{blue}{\left(B \cdot \frac{C}{A} + B\right)}}{A}\right)\right)}{\pi} \]
          5. fma-define81.8%

            \[\leadsto \frac{180 \cdot \left(-\tan^{-1} \left(\frac{-0.5 \cdot \color{blue}{\mathsf{fma}\left(B, \frac{C}{A}, B\right)}}{A}\right)\right)}{\pi} \]
        7. Applied egg-rr81.8%

          \[\leadsto \color{blue}{\frac{180 \cdot \left(-\tan^{-1} \left(\frac{-0.5 \cdot \mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)\right)}{\pi}} \]
        8. Step-by-step derivation
          1. distribute-rgt-neg-out81.8%

            \[\leadsto \frac{\color{blue}{-180 \cdot \tan^{-1} \left(\frac{-0.5 \cdot \mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}}{\pi} \]
          2. distribute-lft-neg-in81.8%

            \[\leadsto \frac{\color{blue}{\left(-180\right) \cdot \tan^{-1} \left(\frac{-0.5 \cdot \mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}}{\pi} \]
          3. metadata-eval81.8%

            \[\leadsto \frac{\color{blue}{-180} \cdot \tan^{-1} \left(\frac{-0.5 \cdot \mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}{\pi} \]
          4. associate-/l*81.8%

            \[\leadsto \frac{-180 \cdot \tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{\mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}}{\pi} \]
        9. Simplified81.8%

          \[\leadsto \color{blue}{\frac{-180 \cdot \tan^{-1} \left(-0.5 \cdot \frac{\mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}{\pi}} \]

        if -3.30000000000000005e115 < A

        1. Initial program 62.7%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Step-by-step derivation
          1. Simplified83.5%

            \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}} \]
          2. Add Preprocessing
        3. Recombined 2 regimes into one program.
        4. Final simplification83.3%

          \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -3.3 \cdot 10^{+115}:\\ \;\;\;\;\frac{-180 \cdot \tan^{-1} \left(-0.5 \cdot \frac{\mathsf{fma}\left(B, \frac{C}{A}, B\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 8: 81.7% accurate, 1.9× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -3.9 \cdot 10^{+143}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= A -3.9e+143)
           (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
           (* 180.0 (/ (atan (/ (- (- C A) (hypot B (- A C))) B)) PI))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (A <= -3.9e+143) {
        		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan((((C - A) - hypot(B, (A - C))) / B)) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (A <= -3.9e+143) {
        		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan((((C - A) - Math.hypot(B, (A - C))) / B)) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if A <= -3.9e+143:
        		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan((((C - A) - math.hypot(B, (A - C))) / B)) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (A <= -3.9e+143)
        		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(Float64(Float64(Float64(C - A) - hypot(B, Float64(A - C))) / B)) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (A <= -3.9e+143)
        		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
        	else
        		tmp = 180.0 * (atan((((C - A) - hypot(B, (A - C))) / B)) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[A, -3.9e+143], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;A \leq -3.9 \cdot 10^{+143}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if A < -3.8999999999999998e143

          1. Initial program 8.6%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around -inf 85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
          4. Step-by-step derivation
            1. mul-1-neg85.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
            2. distribute-neg-frac285.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
            3. distribute-lft-out85.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
            4. associate-/l*89.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
          5. Simplified89.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
          6. Taylor expanded in B around 0 89.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

          if -3.8999999999999998e143 < A

          1. Initial program 61.8%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Step-by-step derivation
            1. associate-*l/61.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}{B}\right)}}{\pi} \]
            2. *-lft-identity61.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}}}{B}\right)}{\pi} \]
            3. +-commutative61.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}}{B}\right)}{\pi} \]
            4. unpow261.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}}{B}\right)}{\pi} \]
            5. unpow261.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}}{B}\right)}{\pi} \]
            6. hypot-define83.0%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}}{B}\right)}{\pi} \]
          3. Simplified83.0%

            \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}} \]
          4. Add Preprocessing
        3. Recombined 2 regimes into one program.
        4. Final simplification83.9%

          \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -3.9 \cdot 10^{+143}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 9: 81.6% accurate, 1.9× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -5.4 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= A -5.4e+132)
           (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
           (/ 180.0 (/ PI (atan (/ (- (- C A) (hypot (- A C) B)) B))))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (A <= -5.4e+132) {
        		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
        	} else {
        		tmp = 180.0 / (((double) M_PI) / atan((((C - A) - hypot((A - C), B)) / B)));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (A <= -5.4e+132) {
        		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
        	} else {
        		tmp = 180.0 / (Math.PI / Math.atan((((C - A) - Math.hypot((A - C), B)) / B)));
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if A <= -5.4e+132:
        		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
        	else:
        		tmp = 180.0 / (math.pi / math.atan((((C - A) - math.hypot((A - C), B)) / B)))
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (A <= -5.4e+132)
        		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
        	else
        		tmp = Float64(180.0 / Float64(pi / atan(Float64(Float64(Float64(C - A) - hypot(Float64(A - C), B)) / B))));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (A <= -5.4e+132)
        		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
        	else
        		tmp = 180.0 / (pi / atan((((C - A) - hypot((A - C), B)) / B)));
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[A, -5.4e+132], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B * N[(1.0 + N[(C / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 / N[(Pi / N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(A - C), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;A \leq -5.4 \cdot 10^{+132}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if A < -5.3999999999999999e132

          1. Initial program 8.6%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around -inf 85.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
          4. Step-by-step derivation
            1. mul-1-neg85.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
            2. distribute-neg-frac285.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
            3. distribute-lft-out85.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
            4. associate-/l*89.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
          5. Simplified89.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
          6. Taylor expanded in B around 0 89.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}}{\pi} \]

          if -5.3999999999999999e132 < A

          1. Initial program 61.8%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Applied egg-rr83.1%

            \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
        3. Recombined 2 regimes into one program.
        4. Final simplification83.9%

          \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -5.4 \cdot 10^{+132}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 10: 46.8% accurate, 3.2× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;B \leq -5.2 \cdot 10^{-5}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.65 \cdot 10^{-203}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq -1.75 \cdot 10^{-250}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C}{B} \cdot 2\right)}{\pi}\\ \mathbf{elif}\;B \leq 2.8 \cdot 10^{+52}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= B -5.2e-5)
           (* 180.0 (/ (atan 1.0) PI))
           (if (<= B -1.65e-203)
             (* 180.0 (/ (atan (* -2.0 (/ A B))) PI))
             (if (<= B -1.75e-250)
               (* 180.0 (/ (atan (* (/ C B) 2.0)) PI))
               (if (<= B 2.8e+52)
                 (* 180.0 (/ (atan (* -0.5 (/ B C))) PI))
                 (* 180.0 (/ (atan -1.0) PI)))))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -5.2e-5) {
        		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
        	} else if (B <= -1.65e-203) {
        		tmp = 180.0 * (atan((-2.0 * (A / B))) / ((double) M_PI));
        	} else if (B <= -1.75e-250) {
        		tmp = 180.0 * (atan(((C / B) * 2.0)) / ((double) M_PI));
        	} else if (B <= 2.8e+52) {
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -5.2e-5) {
        		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
        	} else if (B <= -1.65e-203) {
        		tmp = 180.0 * (Math.atan((-2.0 * (A / B))) / Math.PI);
        	} else if (B <= -1.75e-250) {
        		tmp = 180.0 * (Math.atan(((C / B) * 2.0)) / Math.PI);
        	} else if (B <= 2.8e+52) {
        		tmp = 180.0 * (Math.atan((-0.5 * (B / C))) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if B <= -5.2e-5:
        		tmp = 180.0 * (math.atan(1.0) / math.pi)
        	elif B <= -1.65e-203:
        		tmp = 180.0 * (math.atan((-2.0 * (A / B))) / math.pi)
        	elif B <= -1.75e-250:
        		tmp = 180.0 * (math.atan(((C / B) * 2.0)) / math.pi)
        	elif B <= 2.8e+52:
        		tmp = 180.0 * (math.atan((-0.5 * (B / C))) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan(-1.0) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (B <= -5.2e-5)
        		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
        	elseif (B <= -1.65e-203)
        		tmp = Float64(180.0 * Float64(atan(Float64(-2.0 * Float64(A / B))) / pi));
        	elseif (B <= -1.75e-250)
        		tmp = Float64(180.0 * Float64(atan(Float64(Float64(C / B) * 2.0)) / pi));
        	elseif (B <= 2.8e+52)
        		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(B / C))) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (B <= -5.2e-5)
        		tmp = 180.0 * (atan(1.0) / pi);
        	elseif (B <= -1.65e-203)
        		tmp = 180.0 * (atan((-2.0 * (A / B))) / pi);
        	elseif (B <= -1.75e-250)
        		tmp = 180.0 * (atan(((C / B) * 2.0)) / pi);
        	elseif (B <= 2.8e+52)
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / pi);
        	else
        		tmp = 180.0 * (atan(-1.0) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[B, -5.2e-5], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -1.65e-203], N[(180.0 * N[(N[ArcTan[N[(-2.0 * N[(A / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -1.75e-250], N[(180.0 * N[(N[ArcTan[N[(N[(C / B), $MachinePrecision] * 2.0), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 2.8e+52], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;B \leq -5.2 \cdot 10^{-5}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
        
        \mathbf{elif}\;B \leq -1.65 \cdot 10^{-203}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\
        
        \mathbf{elif}\;B \leq -1.75 \cdot 10^{-250}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C}{B} \cdot 2\right)}{\pi}\\
        
        \mathbf{elif}\;B \leq 2.8 \cdot 10^{+52}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 5 regimes
        2. if B < -5.19999999999999968e-5

          1. Initial program 53.2%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around -inf 63.7%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{1}}{\pi} \]

          if -5.19999999999999968e-5 < B < -1.65000000000000012e-203

          1. Initial program 53.9%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around inf 42.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-2 \cdot \frac{A}{B}\right)}}{\pi} \]

          if -1.65000000000000012e-203 < B < -1.7499999999999999e-250

          1. Initial program 78.3%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in C around -inf 67.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(2 \cdot \frac{C}{B}\right)}}{\pi} \]

          if -1.7499999999999999e-250 < B < 2.8e52

          1. Initial program 54.2%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around 0 39.1%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
          4. Step-by-step derivation
            1. unpow239.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
            2. unpow239.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
            3. hypot-define50.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
          5. Simplified50.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
          6. Taylor expanded in B around 0 49.2%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}{\pi} \]

          if 2.8e52 < B

          1. Initial program 52.6%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around inf 66.2%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{-1}}{\pi} \]
        3. Recombined 5 regimes into one program.
        4. Final simplification56.5%

          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -5.2 \cdot 10^{-5}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.65 \cdot 10^{-203}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq -1.75 \cdot 10^{-250}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C}{B} \cdot 2\right)}{\pi}\\ \mathbf{elif}\;B \leq 2.8 \cdot 10^{+52}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 11: 52.8% accurate, 3.4× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;C \leq 1.65 \cdot 10^{-307}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(1 + \frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq 3.7 \cdot 10^{-111} \lor \neg \left(C \leq 5.4 \cdot 10^{-31}\right):\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= C 1.65e-307)
           (* 180.0 (/ (atan (+ 1.0 (/ C B))) PI))
           (if (or (<= C 3.7e-111) (not (<= C 5.4e-31)))
             (* 180.0 (/ (atan (* -0.5 (/ B C))) PI))
             (* 180.0 (/ (atan 1.0) PI)))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (C <= 1.65e-307) {
        		tmp = 180.0 * (atan((1.0 + (C / B))) / ((double) M_PI));
        	} else if ((C <= 3.7e-111) || !(C <= 5.4e-31)) {
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (C <= 1.65e-307) {
        		tmp = 180.0 * (Math.atan((1.0 + (C / B))) / Math.PI);
        	} else if ((C <= 3.7e-111) || !(C <= 5.4e-31)) {
        		tmp = 180.0 * (Math.atan((-0.5 * (B / C))) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if C <= 1.65e-307:
        		tmp = 180.0 * (math.atan((1.0 + (C / B))) / math.pi)
        	elif (C <= 3.7e-111) or not (C <= 5.4e-31):
        		tmp = 180.0 * (math.atan((-0.5 * (B / C))) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan(1.0) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (C <= 1.65e-307)
        		tmp = Float64(180.0 * Float64(atan(Float64(1.0 + Float64(C / B))) / pi));
        	elseif ((C <= 3.7e-111) || !(C <= 5.4e-31))
        		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(B / C))) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (C <= 1.65e-307)
        		tmp = 180.0 * (atan((1.0 + (C / B))) / pi);
        	elseif ((C <= 3.7e-111) || ~((C <= 5.4e-31)))
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / pi);
        	else
        		tmp = 180.0 * (atan(1.0) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[C, 1.65e-307], N[(180.0 * N[(N[ArcTan[N[(1.0 + N[(C / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[Or[LessEqual[C, 3.7e-111], N[Not[LessEqual[C, 5.4e-31]], $MachinePrecision]], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;C \leq 1.65 \cdot 10^{-307}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(1 + \frac{C}{B}\right)}{\pi}\\
        
        \mathbf{elif}\;C \leq 3.7 \cdot 10^{-111} \lor \neg \left(C \leq 5.4 \cdot 10^{-31}\right):\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 3 regimes
        2. if C < 1.65e-307

          1. Initial program 65.5%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around 0 59.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
          4. Step-by-step derivation
            1. unpow259.6%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
            2. unpow259.6%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
            3. hypot-define73.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
          5. Simplified73.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
          6. Taylor expanded in B around -inf 59.3%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(1 + \frac{C}{B}\right)}}{\pi} \]

          if 1.65e-307 < C < 3.7000000000000002e-111 or 5.40000000000000027e-31 < C

          1. Initial program 43.7%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around 0 26.3%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
          4. Step-by-step derivation
            1. unpow226.3%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
            2. unpow226.3%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
            3. hypot-define49.2%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
          5. Simplified49.2%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
          6. Taylor expanded in B around 0 61.9%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}{\pi} \]

          if 3.7000000000000002e-111 < C < 5.40000000000000027e-31

          1. Initial program 49.0%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around -inf 47.7%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{1}}{\pi} \]
        3. Recombined 3 regimes into one program.
        4. Final simplification59.9%

          \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 1.65 \cdot 10^{-307}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(1 + \frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq 3.7 \cdot 10^{-111} \lor \neg \left(C \leq 5.4 \cdot 10^{-31}\right):\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 12: 46.9% accurate, 3.4× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;B \leq -9 \cdot 10^{-6}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.7 \cdot 10^{-249}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 7.6 \cdot 10^{+53}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= B -9e-6)
           (* 180.0 (/ (atan 1.0) PI))
           (if (<= B -1.7e-249)
             (* 180.0 (/ (atan (* -2.0 (/ A B))) PI))
             (if (<= B 7.6e+53)
               (* 180.0 (/ (atan (* -0.5 (/ B C))) PI))
               (* 180.0 (/ (atan -1.0) PI))))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -9e-6) {
        		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
        	} else if (B <= -1.7e-249) {
        		tmp = 180.0 * (atan((-2.0 * (A / B))) / ((double) M_PI));
        	} else if (B <= 7.6e+53) {
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -9e-6) {
        		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
        	} else if (B <= -1.7e-249) {
        		tmp = 180.0 * (Math.atan((-2.0 * (A / B))) / Math.PI);
        	} else if (B <= 7.6e+53) {
        		tmp = 180.0 * (Math.atan((-0.5 * (B / C))) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if B <= -9e-6:
        		tmp = 180.0 * (math.atan(1.0) / math.pi)
        	elif B <= -1.7e-249:
        		tmp = 180.0 * (math.atan((-2.0 * (A / B))) / math.pi)
        	elif B <= 7.6e+53:
        		tmp = 180.0 * (math.atan((-0.5 * (B / C))) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan(-1.0) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (B <= -9e-6)
        		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
        	elseif (B <= -1.7e-249)
        		tmp = Float64(180.0 * Float64(atan(Float64(-2.0 * Float64(A / B))) / pi));
        	elseif (B <= 7.6e+53)
        		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(B / C))) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (B <= -9e-6)
        		tmp = 180.0 * (atan(1.0) / pi);
        	elseif (B <= -1.7e-249)
        		tmp = 180.0 * (atan((-2.0 * (A / B))) / pi);
        	elseif (B <= 7.6e+53)
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / pi);
        	else
        		tmp = 180.0 * (atan(-1.0) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[B, -9e-6], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -1.7e-249], N[(180.0 * N[(N[ArcTan[N[(-2.0 * N[(A / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 7.6e+53], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;B \leq -9 \cdot 10^{-6}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
        
        \mathbf{elif}\;B \leq -1.7 \cdot 10^{-249}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\
        
        \mathbf{elif}\;B \leq 7.6 \cdot 10^{+53}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 4 regimes
        2. if B < -9.00000000000000023e-6

          1. Initial program 53.2%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around -inf 63.7%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{1}}{\pi} \]

          if -9.00000000000000023e-6 < B < -1.6999999999999999e-249

          1. Initial program 60.0%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around inf 41.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-2 \cdot \frac{A}{B}\right)}}{\pi} \]

          if -1.6999999999999999e-249 < B < 7.59999999999999995e53

          1. Initial program 54.2%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around 0 39.1%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
          4. Step-by-step derivation
            1. unpow239.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
            2. unpow239.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
            3. hypot-define50.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
          5. Simplified50.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
          6. Taylor expanded in B around 0 49.2%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}{\pi} \]

          if 7.59999999999999995e53 < B

          1. Initial program 52.6%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around inf 66.2%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{-1}}{\pi} \]
        3. Recombined 4 regimes into one program.
        4. Final simplification54.9%

          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -9 \cdot 10^{-6}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.7 \cdot 10^{-249}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 7.6 \cdot 10^{+53}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 13: 46.6% accurate, 3.4× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;B \leq -3.45 \cdot 10^{-6}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -2.9 \cdot 10^{-248}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 5.5 \cdot 10^{-119}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= B -3.45e-6)
           (* 180.0 (/ (atan 1.0) PI))
           (if (<= B -2.9e-248)
             (* 180.0 (/ (atan (* -2.0 (/ A B))) PI))
             (if (<= B 5.5e-119)
               (* 180.0 (/ (atan (/ 0.0 B)) PI))
               (* 180.0 (/ (atan -1.0) PI))))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -3.45e-6) {
        		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
        	} else if (B <= -2.9e-248) {
        		tmp = 180.0 * (atan((-2.0 * (A / B))) / ((double) M_PI));
        	} else if (B <= 5.5e-119) {
        		tmp = 180.0 * (atan((0.0 / B)) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -3.45e-6) {
        		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
        	} else if (B <= -2.9e-248) {
        		tmp = 180.0 * (Math.atan((-2.0 * (A / B))) / Math.PI);
        	} else if (B <= 5.5e-119) {
        		tmp = 180.0 * (Math.atan((0.0 / B)) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if B <= -3.45e-6:
        		tmp = 180.0 * (math.atan(1.0) / math.pi)
        	elif B <= -2.9e-248:
        		tmp = 180.0 * (math.atan((-2.0 * (A / B))) / math.pi)
        	elif B <= 5.5e-119:
        		tmp = 180.0 * (math.atan((0.0 / B)) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan(-1.0) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (B <= -3.45e-6)
        		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
        	elseif (B <= -2.9e-248)
        		tmp = Float64(180.0 * Float64(atan(Float64(-2.0 * Float64(A / B))) / pi));
        	elseif (B <= 5.5e-119)
        		tmp = Float64(180.0 * Float64(atan(Float64(0.0 / B)) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (B <= -3.45e-6)
        		tmp = 180.0 * (atan(1.0) / pi);
        	elseif (B <= -2.9e-248)
        		tmp = 180.0 * (atan((-2.0 * (A / B))) / pi);
        	elseif (B <= 5.5e-119)
        		tmp = 180.0 * (atan((0.0 / B)) / pi);
        	else
        		tmp = 180.0 * (atan(-1.0) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[B, -3.45e-6], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -2.9e-248], N[(180.0 * N[(N[ArcTan[N[(-2.0 * N[(A / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 5.5e-119], N[(180.0 * N[(N[ArcTan[N[(0.0 / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;B \leq -3.45 \cdot 10^{-6}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
        
        \mathbf{elif}\;B \leq -2.9 \cdot 10^{-248}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\
        
        \mathbf{elif}\;B \leq 5.5 \cdot 10^{-119}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 4 regimes
        2. if B < -3.45e-6

          1. Initial program 53.2%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around -inf 63.7%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{1}}{\pi} \]

          if -3.45e-6 < B < -2.9000000000000001e-248

          1. Initial program 60.0%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around inf 41.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-2 \cdot \frac{A}{B}\right)}}{\pi} \]

          if -2.9000000000000001e-248 < B < 5.49999999999999959e-119

          1. Initial program 55.2%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in C around inf 37.1%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
          4. Step-by-step derivation
            1. associate-*r/37.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
            2. distribute-rgt1-in37.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
            3. metadata-eval37.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
            4. mul0-lft37.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
            5. metadata-eval37.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
          5. Simplified37.1%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0}{B}\right)}}{\pi} \]

          if 5.49999999999999959e-119 < B

          1. Initial program 52.4%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around inf 53.3%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{-1}}{\pi} \]
        3. Recombined 4 regimes into one program.
        4. Final simplification50.7%

          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -3.45 \cdot 10^{-6}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -2.9 \cdot 10^{-248}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 5.5 \cdot 10^{-119}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 14: 64.8% accurate, 3.5× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{C - A}{B}\\ \mathbf{if}\;B \leq -1.02 \cdot 10^{-250}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + t\_0\right)}}\\ \mathbf{elif}\;B \leq 1.3 \cdot 10^{-248}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(t\_0 + -1\right)}}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (let* ((t_0 (/ (- C A) B)))
           (if (<= B -1.02e-250)
             (/ 180.0 (/ PI (atan (+ 1.0 t_0))))
             (if (<= B 1.3e-248)
               (* 180.0 (/ (atan (/ 0.0 B)) PI))
               (/ 180.0 (/ PI (atan (+ t_0 -1.0))))))))
        double code(double A, double B, double C) {
        	double t_0 = (C - A) / B;
        	double tmp;
        	if (B <= -1.02e-250) {
        		tmp = 180.0 / (((double) M_PI) / atan((1.0 + t_0)));
        	} else if (B <= 1.3e-248) {
        		tmp = 180.0 * (atan((0.0 / B)) / ((double) M_PI));
        	} else {
        		tmp = 180.0 / (((double) M_PI) / atan((t_0 + -1.0)));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double t_0 = (C - A) / B;
        	double tmp;
        	if (B <= -1.02e-250) {
        		tmp = 180.0 / (Math.PI / Math.atan((1.0 + t_0)));
        	} else if (B <= 1.3e-248) {
        		tmp = 180.0 * (Math.atan((0.0 / B)) / Math.PI);
        	} else {
        		tmp = 180.0 / (Math.PI / Math.atan((t_0 + -1.0)));
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	t_0 = (C - A) / B
        	tmp = 0
        	if B <= -1.02e-250:
        		tmp = 180.0 / (math.pi / math.atan((1.0 + t_0)))
        	elif B <= 1.3e-248:
        		tmp = 180.0 * (math.atan((0.0 / B)) / math.pi)
        	else:
        		tmp = 180.0 / (math.pi / math.atan((t_0 + -1.0)))
        	return tmp
        
        function code(A, B, C)
        	t_0 = Float64(Float64(C - A) / B)
        	tmp = 0.0
        	if (B <= -1.02e-250)
        		tmp = Float64(180.0 / Float64(pi / atan(Float64(1.0 + t_0))));
        	elseif (B <= 1.3e-248)
        		tmp = Float64(180.0 * Float64(atan(Float64(0.0 / B)) / pi));
        	else
        		tmp = Float64(180.0 / Float64(pi / atan(Float64(t_0 + -1.0))));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	t_0 = (C - A) / B;
        	tmp = 0.0;
        	if (B <= -1.02e-250)
        		tmp = 180.0 / (pi / atan((1.0 + t_0)));
        	elseif (B <= 1.3e-248)
        		tmp = 180.0 * (atan((0.0 / B)) / pi);
        	else
        		tmp = 180.0 / (pi / atan((t_0 + -1.0)));
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := Block[{t$95$0 = N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]}, If[LessEqual[B, -1.02e-250], N[(180.0 / N[(Pi / N[ArcTan[N[(1.0 + t$95$0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 1.3e-248], N[(180.0 * N[(N[ArcTan[N[(0.0 / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 / N[(Pi / N[ArcTan[N[(t$95$0 + -1.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        t_0 := \frac{C - A}{B}\\
        \mathbf{if}\;B \leq -1.02 \cdot 10^{-250}:\\
        \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + t\_0\right)}}\\
        
        \mathbf{elif}\;B \leq 1.3 \cdot 10^{-248}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(t\_0 + -1\right)}}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 3 regimes
        2. if B < -1.02000000000000001e-250

          1. Initial program 56.2%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Applied egg-rr78.4%

            \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
          4. Taylor expanded in B around -inf 70.5%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
          5. Step-by-step derivation
            1. associate--l+70.5%

              \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
            2. div-sub70.5%

              \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
          6. Simplified70.5%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \frac{C - A}{B}\right)}}} \]

          if -1.02000000000000001e-250 < B < 1.30000000000000003e-248

          1. Initial program 49.3%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in C around inf 54.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
          4. Step-by-step derivation
            1. associate-*r/54.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
            2. distribute-rgt1-in54.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
            3. metadata-eval54.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
            4. mul0-lft54.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
            5. metadata-eval54.8%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
          5. Simplified54.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0}{B}\right)}}{\pi} \]

          if 1.30000000000000003e-248 < B

          1. Initial program 54.3%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Applied egg-rr79.3%

            \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
          4. Taylor expanded in B around inf 69.0%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\frac{C}{B} - \left(1 + \frac{A}{B}\right)\right)}}} \]
          5. Step-by-step derivation
            1. +-commutative69.0%

              \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C}{B} - \color{blue}{\left(\frac{A}{B} + 1\right)}\right)}} \]
            2. associate--r+69.0%

              \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(\frac{C}{B} - \frac{A}{B}\right) - 1\right)}}} \]
            3. div-sub69.0%

              \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{\frac{C - A}{B}} - 1\right)}} \]
          6. Simplified69.0%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\frac{C - A}{B} - 1\right)}}} \]
        3. Recombined 3 regimes into one program.
        4. Final simplification68.8%

          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -1.02 \cdot 10^{-250}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \mathbf{elif}\;B \leq 1.3 \cdot 10^{-248}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 15: 45.4% accurate, 3.6× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;B \leq -2 \cdot 10^{-170}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 1.8 \cdot 10^{-121}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= B -2e-170)
           (* 180.0 (/ (atan 1.0) PI))
           (if (<= B 1.8e-121)
             (* 180.0 (/ (atan (/ 0.0 B)) PI))
             (* 180.0 (/ (atan -1.0) PI)))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -2e-170) {
        		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
        	} else if (B <= 1.8e-121) {
        		tmp = 180.0 * (atan((0.0 / B)) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -2e-170) {
        		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
        	} else if (B <= 1.8e-121) {
        		tmp = 180.0 * (Math.atan((0.0 / B)) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if B <= -2e-170:
        		tmp = 180.0 * (math.atan(1.0) / math.pi)
        	elif B <= 1.8e-121:
        		tmp = 180.0 * (math.atan((0.0 / B)) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan(-1.0) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (B <= -2e-170)
        		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
        	elseif (B <= 1.8e-121)
        		tmp = Float64(180.0 * Float64(atan(Float64(0.0 / B)) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (B <= -2e-170)
        		tmp = 180.0 * (atan(1.0) / pi);
        	elseif (B <= 1.8e-121)
        		tmp = 180.0 * (atan((0.0 / B)) / pi);
        	else
        		tmp = 180.0 * (atan(-1.0) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[B, -2e-170], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 1.8e-121], N[(180.0 * N[(N[ArcTan[N[(0.0 / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;B \leq -2 \cdot 10^{-170}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
        
        \mathbf{elif}\;B \leq 1.8 \cdot 10^{-121}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 3 regimes
        2. if B < -1.99999999999999997e-170

          1. Initial program 54.8%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around -inf 50.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{1}}{\pi} \]

          if -1.99999999999999997e-170 < B < 1.79999999999999992e-121

          1. Initial program 57.7%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in C around inf 32.1%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
          4. Step-by-step derivation
            1. associate-*r/32.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
            2. distribute-rgt1-in32.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
            3. metadata-eval32.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
            4. mul0-lft32.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
            5. metadata-eval32.1%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
          5. Simplified32.1%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0}{B}\right)}}{\pi} \]

          if 1.79999999999999992e-121 < B

          1. Initial program 52.4%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around inf 53.3%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{-1}}{\pi} \]
        3. Recombined 3 regimes into one program.
        4. Final simplification46.1%

          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -2 \cdot 10^{-170}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 1.8 \cdot 10^{-121}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 16: 60.5% accurate, 3.6× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;C \leq 1.55 \cdot 10^{-30}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= C 1.55e-30)
           (* 180.0 (/ (atan (+ 1.0 (/ (- C A) B))) PI))
           (* 180.0 (/ (atan (* -0.5 (/ B C))) PI))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (C <= 1.55e-30) {
        		tmp = 180.0 * (atan((1.0 + ((C - A) / B))) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (C <= 1.55e-30) {
        		tmp = 180.0 * (Math.atan((1.0 + ((C - A) / B))) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan((-0.5 * (B / C))) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if C <= 1.55e-30:
        		tmp = 180.0 * (math.atan((1.0 + ((C - A) / B))) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan((-0.5 * (B / C))) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (C <= 1.55e-30)
        		tmp = Float64(180.0 * Float64(atan(Float64(1.0 + Float64(Float64(C - A) / B))) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(B / C))) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (C <= 1.55e-30)
        		tmp = 180.0 * (atan((1.0 + ((C - A) / B))) / pi);
        	else
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[C, 1.55e-30], N[(180.0 * N[(N[ArcTan[N[(1.0 + N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;C \leq 1.55 \cdot 10^{-30}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if C < 1.54999999999999995e-30

          1. Initial program 66.4%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around -inf 67.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}{\pi} \]
          4. Step-by-step derivation
            1. associate--l+67.0%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}{\pi} \]
            2. div-sub67.0%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}{\pi} \]
          5. Simplified67.0%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(1 + \frac{C - A}{B}\right)}}{\pi} \]

          if 1.54999999999999995e-30 < C

          1. Initial program 28.3%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around 0 21.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
          4. Step-by-step derivation
            1. unpow221.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
            2. unpow221.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
            3. hypot-define49.4%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
          5. Simplified49.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
          6. Taylor expanded in B around 0 64.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}{\pi} \]
        3. Recombined 2 regimes into one program.
        4. Final simplification66.3%

          \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 1.55 \cdot 10^{-30}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 17: 60.5% accurate, 3.6× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;C \leq 1.05 \cdot 10^{-29}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= C 1.05e-29)
           (/ 180.0 (/ PI (atan (+ 1.0 (/ (- C A) B)))))
           (* 180.0 (/ (atan (* -0.5 (/ B C))) PI))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (C <= 1.05e-29) {
        		tmp = 180.0 / (((double) M_PI) / atan((1.0 + ((C - A) / B))));
        	} else {
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (C <= 1.05e-29) {
        		tmp = 180.0 / (Math.PI / Math.atan((1.0 + ((C - A) / B))));
        	} else {
        		tmp = 180.0 * (Math.atan((-0.5 * (B / C))) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if C <= 1.05e-29:
        		tmp = 180.0 / (math.pi / math.atan((1.0 + ((C - A) / B))))
        	else:
        		tmp = 180.0 * (math.atan((-0.5 * (B / C))) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (C <= 1.05e-29)
        		tmp = Float64(180.0 / Float64(pi / atan(Float64(1.0 + Float64(Float64(C - A) / B)))));
        	else
        		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(B / C))) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (C <= 1.05e-29)
        		tmp = 180.0 / (pi / atan((1.0 + ((C - A) / B))));
        	else
        		tmp = 180.0 * (atan((-0.5 * (B / C))) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[C, 1.05e-29], N[(180.0 / N[(Pi / N[ArcTan[N[(1.0 + N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;C \leq 1.05 \cdot 10^{-29}:\\
        \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if C < 1.04999999999999995e-29

          1. Initial program 66.4%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Applied egg-rr87.0%

            \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
          4. Taylor expanded in B around -inf 67.0%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
          5. Step-by-step derivation
            1. associate--l+67.0%

              \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
            2. div-sub67.0%

              \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
          6. Simplified67.0%

            \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \frac{C - A}{B}\right)}}} \]

          if 1.04999999999999995e-29 < C

          1. Initial program 28.3%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in A around 0 21.5%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}\right)}{\pi} \]
          4. Step-by-step derivation
            1. unpow221.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}\right)\right)}{\pi} \]
            2. unpow221.5%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}\right)\right)}{\pi} \]
            3. hypot-define49.4%

              \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(C - \color{blue}{\mathsf{hypot}\left(B, C\right)}\right)\right)}{\pi} \]
          5. Simplified49.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(C - \mathsf{hypot}\left(B, C\right)\right)}\right)}{\pi} \]
          6. Taylor expanded in B around 0 64.8%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}{\pi} \]
        3. Recombined 2 regimes into one program.
        4. Final simplification66.3%

          \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 1.05 \cdot 10^{-29}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 18: 40.5% accurate, 3.8× speedup?

        \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;B \leq -8.5 \cdot 10^{-302}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
        (FPCore (A B C)
         :precision binary64
         (if (<= B -8.5e-302)
           (* 180.0 (/ (atan 1.0) PI))
           (* 180.0 (/ (atan -1.0) PI))))
        double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -8.5e-302) {
        		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
        	} else {
        		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
        	}
        	return tmp;
        }
        
        public static double code(double A, double B, double C) {
        	double tmp;
        	if (B <= -8.5e-302) {
        		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
        	} else {
        		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
        	}
        	return tmp;
        }
        
        def code(A, B, C):
        	tmp = 0
        	if B <= -8.5e-302:
        		tmp = 180.0 * (math.atan(1.0) / math.pi)
        	else:
        		tmp = 180.0 * (math.atan(-1.0) / math.pi)
        	return tmp
        
        function code(A, B, C)
        	tmp = 0.0
        	if (B <= -8.5e-302)
        		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
        	else
        		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
        	end
        	return tmp
        end
        
        function tmp_2 = code(A, B, C)
        	tmp = 0.0;
        	if (B <= -8.5e-302)
        		tmp = 180.0 * (atan(1.0) / pi);
        	else
        		tmp = 180.0 * (atan(-1.0) / pi);
        	end
        	tmp_2 = tmp;
        end
        
        code[A_, B_, C_] := If[LessEqual[B, -8.5e-302], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]
        
        \begin{array}{l}
        
        \\
        \begin{array}{l}
        \mathbf{if}\;B \leq -8.5 \cdot 10^{-302}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
        
        \mathbf{else}:\\
        \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if B < -8.5000000000000005e-302

          1. Initial program 55.0%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around -inf 42.4%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{1}}{\pi} \]

          if -8.5000000000000005e-302 < B

          1. Initial program 54.9%

            \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
          2. Add Preprocessing
          3. Taylor expanded in B around inf 37.6%

            \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{-1}}{\pi} \]
        3. Recombined 2 regimes into one program.
        4. Final simplification40.4%

          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -8.5 \cdot 10^{-302}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \]
        5. Add Preprocessing

        Alternative 19: 21.7% accurate, 4.0× speedup?

        \[\begin{array}{l} \\ 180 \cdot \frac{\tan^{-1} -1}{\pi} \end{array} \]
        (FPCore (A B C) :precision binary64 (* 180.0 (/ (atan -1.0) PI)))
        double code(double A, double B, double C) {
        	return 180.0 * (atan(-1.0) / ((double) M_PI));
        }
        
        public static double code(double A, double B, double C) {
        	return 180.0 * (Math.atan(-1.0) / Math.PI);
        }
        
        def code(A, B, C):
        	return 180.0 * (math.atan(-1.0) / math.pi)
        
        function code(A, B, C)
        	return Float64(180.0 * Float64(atan(-1.0) / pi))
        end
        
        function tmp = code(A, B, C)
        	tmp = 180.0 * (atan(-1.0) / pi);
        end
        
        code[A_, B_, C_] := N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]
        
        \begin{array}{l}
        
        \\
        180 \cdot \frac{\tan^{-1} -1}{\pi}
        \end{array}
        
        Derivation
        1. Initial program 55.0%

          \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
        2. Add Preprocessing
        3. Taylor expanded in B around inf 17.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{-1}}{\pi} \]
        4. Final simplification17.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} -1}{\pi} \]
        5. Add Preprocessing

        Reproduce

        ?
        herbie shell --seed 2024071 
        (FPCore (A B C)
          :name "ABCF->ab-angle angle"
          :precision binary64
          (* 180.0 (/ (atan (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))) PI)))