ABCF->ab-angle angle

Percentage Accurate: 54.6% → 81.9%
Time: 16.5s
Alternatives: 16
Speedup: 3.3×

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 16 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: 54.6% 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: 81.9% accurate, 0.4× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\\ t_1 := \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)\\ \mathbf{if}\;t\_0 \leq -0.5:\\ \;\;\;\;\frac{180}{\frac{\pi}{t\_1}}\\ \mathbf{elif}\;t\_0 \leq 0:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\sqrt[3]{{\pi}^{3}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{1}{\frac{t\_1}{\pi}}}\\ \end{array} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (let* ((t_0
         (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))
        (t_1 (atan (/ (- (- C A) (hypot (- A C) B)) B))))
   (if (<= t_0 -0.5)
     (/ 180.0 (/ PI t_1))
     (if (<= t_0 0.0)
       (*
        180.0
        (/ (atan (* -0.5 (/ (+ B (/ (* B A) C)) C))) (cbrt (pow PI 3.0))))
       (/ 180.0 (/ 1.0 (/ t_1 PI)))))))
double code(double A, double B, double C) {
	double t_0 = (1.0 / B) * ((C - A) - sqrt((pow((A - C), 2.0) + pow(B, 2.0))));
	double t_1 = atan((((C - A) - hypot((A - C), B)) / B));
	double tmp;
	if (t_0 <= -0.5) {
		tmp = 180.0 / (((double) M_PI) / t_1);
	} else if (t_0 <= 0.0) {
		tmp = 180.0 * (atan((-0.5 * ((B + ((B * A) / C)) / C))) / cbrt(pow(((double) M_PI), 3.0)));
	} else {
		tmp = 180.0 / (1.0 / (t_1 / ((double) M_PI)));
	}
	return tmp;
}
public static double code(double A, double B, double C) {
	double t_0 = (1.0 / B) * ((C - A) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0))));
	double t_1 = Math.atan((((C - A) - Math.hypot((A - C), B)) / B));
	double tmp;
	if (t_0 <= -0.5) {
		tmp = 180.0 / (Math.PI / t_1);
	} else if (t_0 <= 0.0) {
		tmp = 180.0 * (Math.atan((-0.5 * ((B + ((B * A) / C)) / C))) / Math.cbrt(Math.pow(Math.PI, 3.0)));
	} else {
		tmp = 180.0 / (1.0 / (t_1 / Math.PI));
	}
	return tmp;
}
function code(A, B, C)
	t_0 = Float64(Float64(1.0 / B) * Float64(Float64(C - A) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0)))))
	t_1 = atan(Float64(Float64(Float64(C - A) - hypot(Float64(A - C), B)) / B))
	tmp = 0.0
	if (t_0 <= -0.5)
		tmp = Float64(180.0 / Float64(pi / t_1));
	elseif (t_0 <= 0.0)
		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(Float64(B + Float64(Float64(B * A) / C)) / C))) / cbrt((pi ^ 3.0))));
	else
		tmp = Float64(180.0 / Float64(1.0 / Float64(t_1 / pi)));
	end
	return tmp
end
code[A_, B_, C_] := Block[{t$95$0 = 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]}, Block[{t$95$1 = N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(A - C), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[t$95$0, -0.5], N[(180.0 / N[(Pi / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, 0.0], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(N[(B + N[(N[(B * A), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Power[N[Power[Pi, 3.0], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(180.0 / N[(1.0 / N[(t$95$1 / Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := \frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\\
t_1 := \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)\\
\mathbf{if}\;t\_0 \leq -0.5:\\
\;\;\;\;\frac{180}{\frac{\pi}{t\_1}}\\

\mathbf{elif}\;t\_0 \leq 0:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\sqrt[3]{{\pi}^{3}}}\\

\mathbf{else}:\\
\;\;\;\;\frac{180}{\frac{1}{\frac{t\_1}{\pi}}}\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))) < -0.5

    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. Add Preprocessing
    3. Applied egg-rr87.9%

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

    if -0.5 < (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))) < -0.0

    1. Initial program 16.3%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in C around inf 66.9%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B} + \left(-0.5 \cdot \frac{B}{C} + -0.5 \cdot \frac{A \cdot B}{{C}^{2}}\right)\right)}}{\pi} \]
    4. Taylor expanded in C around inf 67.1%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \color{blue}{\frac{-0.5 \cdot B + -0.5 \cdot \frac{A \cdot B}{C}}{C}}\right)}{\pi} \]
    5. Step-by-step derivation
      1. +-commutative67.1%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{\color{blue}{-0.5 \cdot \frac{A \cdot B}{C} + -0.5 \cdot B}}{C}\right)}{\pi} \]
      2. distribute-lft-out67.1%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{\color{blue}{-0.5 \cdot \left(\frac{A \cdot B}{C} + B\right)}}{C}\right)}{\pi} \]
      3. associate-/l*66.7%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{-0.5 \cdot \left(\color{blue}{A \cdot \frac{B}{C}} + B\right)}{C}\right)}{\pi} \]
    6. Simplified66.7%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \color{blue}{\frac{-0.5 \cdot \left(A \cdot \frac{B}{C} + B\right)}{C}}\right)}{\pi} \]
    7. Taylor expanded in A around inf 67.1%

      \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{A \cdot B}{C}}{C}\right)}{\pi}} \]
    8. Step-by-step derivation
      1. rem-cbrt-cube67.1%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{A \cdot B}{C}}{C}\right)}{\color{blue}{\sqrt[3]{{\pi}^{3}}}} \]
    9. Applied egg-rr67.1%

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

    if -0.0 < (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64))))))

    1. Initial program 67.2%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Applied egg-rr91.9%

      \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
    4. Step-by-step derivation
      1. clear-num92.0%

        \[\leadsto \frac{180}{\color{blue}{\frac{1}{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}}}} \]
      2. inv-pow92.0%

        \[\leadsto \frac{180}{\color{blue}{{\left(\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}\right)}^{-1}}} \]
      3. associate--l-88.8%

        \[\leadsto \frac{180}{{\left(\frac{\tan^{-1} \left(\frac{\color{blue}{C - \left(A + \mathsf{hypot}\left(A - C, B\right)\right)}}{B}\right)}{\pi}\right)}^{-1}} \]
    5. Applied egg-rr88.8%

      \[\leadsto \frac{180}{\color{blue}{{\left(\frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(A - C, B\right)\right)}{B}\right)}{\pi}\right)}^{-1}}} \]
    6. Simplified92.0%

      \[\leadsto \frac{180}{\color{blue}{\frac{1}{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}}}} \]
  3. Recombined 3 regimes into one program.
  4. Final simplification86.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right) \leq -0.5:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}\\ \mathbf{elif}\;\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right) \leq 0:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\sqrt[3]{{\pi}^{3}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{1}{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}}}\\ \end{array} \]
  5. Add Preprocessing

Alternative 2: 81.4% accurate, 1.9× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;C \leq 3.1 \cdot 10^{+114}:\\ \;\;\;\;\frac{180}{\frac{1}{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\pi}\\ \end{array} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (if (<= C 3.1e+114)
   (/ 180.0 (/ 1.0 (/ (atan (/ (- (- C A) (hypot (- A C) B)) B)) PI)))
   (* 180.0 (/ (atan (* -0.5 (/ (+ B (/ (* B A) C)) C))) PI))))
double code(double A, double B, double C) {
	double tmp;
	if (C <= 3.1e+114) {
		tmp = 180.0 / (1.0 / (atan((((C - A) - hypot((A - C), B)) / B)) / ((double) M_PI)));
	} else {
		tmp = 180.0 * (atan((-0.5 * ((B + ((B * A) / C)) / C))) / ((double) M_PI));
	}
	return tmp;
}
public static double code(double A, double B, double C) {
	double tmp;
	if (C <= 3.1e+114) {
		tmp = 180.0 / (1.0 / (Math.atan((((C - A) - Math.hypot((A - C), B)) / B)) / Math.PI));
	} else {
		tmp = 180.0 * (Math.atan((-0.5 * ((B + ((B * A) / C)) / C))) / Math.PI);
	}
	return tmp;
}
def code(A, B, C):
	tmp = 0
	if C <= 3.1e+114:
		tmp = 180.0 / (1.0 / (math.atan((((C - A) - math.hypot((A - C), B)) / B)) / math.pi))
	else:
		tmp = 180.0 * (math.atan((-0.5 * ((B + ((B * A) / C)) / C))) / math.pi)
	return tmp
function code(A, B, C)
	tmp = 0.0
	if (C <= 3.1e+114)
		tmp = Float64(180.0 / Float64(1.0 / Float64(atan(Float64(Float64(Float64(C - A) - hypot(Float64(A - C), B)) / B)) / pi)));
	else
		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(Float64(B + Float64(Float64(B * A) / C)) / C))) / pi));
	end
	return tmp
end
function tmp_2 = code(A, B, C)
	tmp = 0.0;
	if (C <= 3.1e+114)
		tmp = 180.0 / (1.0 / (atan((((C - A) - hypot((A - C), B)) / B)) / pi));
	else
		tmp = 180.0 * (atan((-0.5 * ((B + ((B * A) / C)) / C))) / pi);
	end
	tmp_2 = tmp;
end
code[A_, B_, C_] := If[LessEqual[C, 3.1e+114], N[(180.0 / N[(1.0 / N[(N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(A - C), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(N[(B + N[(N[(B * A), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;C \leq 3.1 \cdot 10^{+114}:\\
\;\;\;\;\frac{180}{\frac{1}{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}}}\\

\mathbf{else}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\pi}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if C < 3.1e114

    1. Initial program 66.9%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Applied egg-rr85.4%

      \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
    4. Step-by-step derivation
      1. clear-num85.5%

        \[\leadsto \frac{180}{\color{blue}{\frac{1}{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}}}} \]
      2. inv-pow85.5%

        \[\leadsto \frac{180}{\color{blue}{{\left(\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}\right)}^{-1}}} \]
      3. associate--l-81.3%

        \[\leadsto \frac{180}{{\left(\frac{\tan^{-1} \left(\frac{\color{blue}{C - \left(A + \mathsf{hypot}\left(A - C, B\right)\right)}}{B}\right)}{\pi}\right)}^{-1}} \]
    5. Applied egg-rr81.3%

      \[\leadsto \frac{180}{\color{blue}{{\left(\frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(A - C, B\right)\right)}{B}\right)}{\pi}\right)}^{-1}}} \]
    6. Simplified85.5%

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

    if 3.1e114 < C

    1. Initial program 14.1%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in C around inf 77.3%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B} + \left(-0.5 \cdot \frac{B}{C} + -0.5 \cdot \frac{A \cdot B}{{C}^{2}}\right)\right)}}{\pi} \]
    4. Taylor expanded in C around inf 79.6%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \color{blue}{\frac{-0.5 \cdot B + -0.5 \cdot \frac{A \cdot B}{C}}{C}}\right)}{\pi} \]
    5. Step-by-step derivation
      1. +-commutative79.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{\color{blue}{-0.5 \cdot \frac{A \cdot B}{C} + -0.5 \cdot B}}{C}\right)}{\pi} \]
      2. distribute-lft-out79.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{\color{blue}{-0.5 \cdot \left(\frac{A \cdot B}{C} + B\right)}}{C}\right)}{\pi} \]
      3. associate-/l*77.4%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{-0.5 \cdot \left(\color{blue}{A \cdot \frac{B}{C}} + B\right)}{C}\right)}{\pi} \]
    6. Simplified77.4%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \color{blue}{\frac{-0.5 \cdot \left(A \cdot \frac{B}{C} + B\right)}{C}}\right)}{\pi} \]
    7. Taylor expanded in A around inf 79.6%

      \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{A \cdot B}{C}}{C}\right)}{\pi}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification84.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 3.1 \cdot 10^{+114}:\\ \;\;\;\;\frac{180}{\frac{1}{\frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\pi}\\ \end{array} \]
  5. Add Preprocessing

Alternative 3: 75.9% accurate, 1.9× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -1.02 \cdot 10^{+116}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;A \leq 1.15 \cdot 10^{+68}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \end{array} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (if (<= A -1.02e+116)
   (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
   (if (<= A 1.15e+68)
     (* 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 <= -1.02e+116) {
		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
	} else if (A <= 1.15e+68) {
		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 <= -1.02e+116) {
		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
	} else if (A <= 1.15e+68) {
		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 <= -1.02e+116:
		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
	elif A <= 1.15e+68:
		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 <= -1.02e+116)
		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
	elseif (A <= 1.15e+68)
		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 <= -1.02e+116)
		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
	elseif (A <= 1.15e+68)
		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, -1.02e+116], 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, 1.15e+68], 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 -1.02 \cdot 10^{+116}:\\
\;\;\;\;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 1.15 \cdot 10^{+68}:\\
\;\;\;\;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 < -1.0199999999999999e116

    1. Initial program 18.3%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 76.0%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
    4. Step-by-step derivation
      1. mul-1-neg76.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
      2. distribute-neg-frac276.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
      3. distribute-lft-out76.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
      4. associate-/l*77.8%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
    5. Simplified77.8%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
    6. Taylor expanded in B around 0 77.8%

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

    if -1.0199999999999999e116 < A < 1.15e68

    1. Initial program 57.4%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in A around 0 55.6%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{C - \sqrt{{B}^{2} + {C}^{2}}}{B}\right)}}{\pi} \]
    4. Step-by-step derivation
      1. unpow255.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \sqrt{\color{blue}{B \cdot B} + {C}^{2}}}{B}\right)}{\pi} \]
      2. unpow255.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \sqrt{B \cdot B + \color{blue}{C \cdot C}}}{B}\right)}{\pi} \]
      3. hypot-define76.7%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - \color{blue}{\mathsf{hypot}\left(B, C\right)}}{B}\right)}{\pi} \]
    5. Simplified76.7%

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

    if 1.15e68 < A

    1. Initial program 87.4%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Applied egg-rr97.6%

      \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
    4. Taylor expanded in B around -inf 87.6%

      \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
    5. Step-by-step derivation
      1. associate--l+87.6%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
      2. div-sub90.0%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
    6. Simplified90.0%

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

Alternative 4: 81.4% accurate, 1.9× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;C \leq 1.6 \cdot 10^{+116}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\pi}\\ \end{array} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (if (<= C 1.6e+116)
   (* 180.0 (/ (atan (/ (- (- C A) (hypot B (- A C))) B)) PI))
   (* 180.0 (/ (atan (* -0.5 (/ (+ B (/ (* B A) C)) C))) PI))))
double code(double A, double B, double C) {
	double tmp;
	if (C <= 1.6e+116) {
		tmp = 180.0 * (atan((((C - A) - hypot(B, (A - C))) / B)) / ((double) M_PI));
	} else {
		tmp = 180.0 * (atan((-0.5 * ((B + ((B * A) / C)) / C))) / ((double) M_PI));
	}
	return tmp;
}
public static double code(double A, double B, double C) {
	double tmp;
	if (C <= 1.6e+116) {
		tmp = 180.0 * (Math.atan((((C - A) - Math.hypot(B, (A - C))) / B)) / Math.PI);
	} else {
		tmp = 180.0 * (Math.atan((-0.5 * ((B + ((B * A) / C)) / C))) / Math.PI);
	}
	return tmp;
}
def code(A, B, C):
	tmp = 0
	if C <= 1.6e+116:
		tmp = 180.0 * (math.atan((((C - A) - math.hypot(B, (A - C))) / B)) / math.pi)
	else:
		tmp = 180.0 * (math.atan((-0.5 * ((B + ((B * A) / C)) / C))) / math.pi)
	return tmp
function code(A, B, C)
	tmp = 0.0
	if (C <= 1.6e+116)
		tmp = Float64(180.0 * Float64(atan(Float64(Float64(Float64(C - A) - hypot(B, Float64(A - C))) / B)) / pi));
	else
		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(Float64(B + Float64(Float64(B * A) / C)) / C))) / pi));
	end
	return tmp
end
function tmp_2 = code(A, B, C)
	tmp = 0.0;
	if (C <= 1.6e+116)
		tmp = 180.0 * (atan((((C - A) - hypot(B, (A - C))) / B)) / pi);
	else
		tmp = 180.0 * (atan((-0.5 * ((B + ((B * A) / C)) / C))) / pi);
	end
	tmp_2 = tmp;
end
code[A_, B_, C_] := If[LessEqual[C, 1.6e+116], 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], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(N[(B + N[(N[(B * A), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;C \leq 1.6 \cdot 10^{+116}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\

\mathbf{else}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\pi}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if C < 1.6e116

    1. Initial program 66.9%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Step-by-step derivation
      1. associate-*l/66.9%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}{B}\right)}}{\pi} \]
      2. *-lft-identity66.9%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}}}{B}\right)}{\pi} \]
      3. +-commutative66.9%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}}{B}\right)}{\pi} \]
      4. unpow266.9%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}}{B}\right)}{\pi} \]
      5. unpow266.9%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}}{B}\right)}{\pi} \]
      6. hypot-define85.4%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}}{B}\right)}{\pi} \]
    3. Simplified85.4%

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

    if 1.6e116 < C

    1. Initial program 14.1%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in C around inf 77.3%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B} + \left(-0.5 \cdot \frac{B}{C} + -0.5 \cdot \frac{A \cdot B}{{C}^{2}}\right)\right)}}{\pi} \]
    4. Taylor expanded in C around inf 79.6%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \color{blue}{\frac{-0.5 \cdot B + -0.5 \cdot \frac{A \cdot B}{C}}{C}}\right)}{\pi} \]
    5. Step-by-step derivation
      1. +-commutative79.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{\color{blue}{-0.5 \cdot \frac{A \cdot B}{C} + -0.5 \cdot B}}{C}\right)}{\pi} \]
      2. distribute-lft-out79.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{\color{blue}{-0.5 \cdot \left(\frac{A \cdot B}{C} + B\right)}}{C}\right)}{\pi} \]
      3. associate-/l*77.4%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \frac{-0.5 \cdot \left(\color{blue}{A \cdot \frac{B}{C}} + B\right)}{C}\right)}{\pi} \]
    6. Simplified77.4%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-1 \cdot \frac{A + -1 \cdot A}{B} + \color{blue}{\frac{-0.5 \cdot \left(A \cdot \frac{B}{C} + B\right)}{C}}\right)}{\pi} \]
    7. Taylor expanded in A around inf 79.6%

      \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{A \cdot B}{C}}{C}\right)}{\pi}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification84.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 1.6 \cdot 10^{+116}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B + \frac{B \cdot A}{C}}{C}\right)}{\pi}\\ \end{array} \]
  5. Add Preprocessing

Alternative 5: 80.9% accurate, 1.9× speedup?

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;A \leq -2.5 \cdot 10^{+118}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\pi}\\ \end{array} \end{array} \]
(FPCore (A B C)
 :precision binary64
 (if (<= A -2.5e+118)
   (* 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 <= -2.5e+118) {
		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 <= -2.5e+118) {
		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 <= -2.5e+118:
		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 <= -2.5e+118)
		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(C - Float64(A + hypot(B, Float64(A - C)))) / B)) / pi));
	end
	return tmp
end
function tmp_2 = code(A, B, C)
	tmp = 0.0;
	if (A <= -2.5e+118)
		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, -2.5e+118], 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[(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 -2.5 \cdot 10^{+118}:\\
\;\;\;\;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{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 < -2.49999999999999986e118

    1. Initial program 18.3%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in A around -inf 76.0%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
    4. Step-by-step derivation
      1. mul-1-neg76.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
      2. distribute-neg-frac276.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
      3. distribute-lft-out76.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
      4. associate-/l*77.8%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
    5. Simplified77.8%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
    6. Taylor expanded in B around 0 77.8%

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

    if -2.49999999999999986e118 < 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. Simplified81.8%

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

    Alternative 6: 45.4% accurate, 3.1× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\ \mathbf{if}\;B \leq -4.2 \cdot 10^{-91}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -2.1 \cdot 10^{-210}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;B \leq 3.2 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 2500000000:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;B \leq 3.5 \cdot 10^{+67}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (* 180.0 (/ (atan (* 2.0 (/ C B))) PI))))
       (if (<= B -4.2e-91)
         (* 180.0 (/ (atan 1.0) PI))
         (if (<= B -2.1e-210)
           t_0
           (if (<= B 3.2e-202)
             (* 180.0 (/ (atan (/ (* B 0.5) A)) PI))
             (if (<= B 2500000000.0)
               t_0
               (if (<= B 3.5e+67)
                 (/ 180.0 (/ PI (atan (* -0.5 (/ B C)))))
                 (* 180.0 (/ (atan -1.0) PI)))))))))
    double code(double A, double B, double C) {
    	double t_0 = 180.0 * (atan((2.0 * (C / B))) / ((double) M_PI));
    	double tmp;
    	if (B <= -4.2e-91) {
    		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
    	} else if (B <= -2.1e-210) {
    		tmp = t_0;
    	} else if (B <= 3.2e-202) {
    		tmp = 180.0 * (atan(((B * 0.5) / A)) / ((double) M_PI));
    	} else if (B <= 2500000000.0) {
    		tmp = t_0;
    	} else if (B <= 3.5e+67) {
    		tmp = 180.0 / (((double) M_PI) / atan((-0.5 * (B / C))));
    	} else {
    		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = 180.0 * (Math.atan((2.0 * (C / B))) / Math.PI);
    	double tmp;
    	if (B <= -4.2e-91) {
    		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
    	} else if (B <= -2.1e-210) {
    		tmp = t_0;
    	} else if (B <= 3.2e-202) {
    		tmp = 180.0 * (Math.atan(((B * 0.5) / A)) / Math.PI);
    	} else if (B <= 2500000000.0) {
    		tmp = t_0;
    	} else if (B <= 3.5e+67) {
    		tmp = 180.0 / (Math.PI / Math.atan((-0.5 * (B / C))));
    	} else {
    		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = 180.0 * (math.atan((2.0 * (C / B))) / math.pi)
    	tmp = 0
    	if B <= -4.2e-91:
    		tmp = 180.0 * (math.atan(1.0) / math.pi)
    	elif B <= -2.1e-210:
    		tmp = t_0
    	elif B <= 3.2e-202:
    		tmp = 180.0 * (math.atan(((B * 0.5) / A)) / math.pi)
    	elif B <= 2500000000.0:
    		tmp = t_0
    	elif B <= 3.5e+67:
    		tmp = 180.0 / (math.pi / math.atan((-0.5 * (B / C))))
    	else:
    		tmp = 180.0 * (math.atan(-1.0) / math.pi)
    	return tmp
    
    function code(A, B, C)
    	t_0 = Float64(180.0 * Float64(atan(Float64(2.0 * Float64(C / B))) / pi))
    	tmp = 0.0
    	if (B <= -4.2e-91)
    		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
    	elseif (B <= -2.1e-210)
    		tmp = t_0;
    	elseif (B <= 3.2e-202)
    		tmp = Float64(180.0 * Float64(atan(Float64(Float64(B * 0.5) / A)) / pi));
    	elseif (B <= 2500000000.0)
    		tmp = t_0;
    	elseif (B <= 3.5e+67)
    		tmp = Float64(180.0 / Float64(pi / atan(Float64(-0.5 * Float64(B / C)))));
    	else
    		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = 180.0 * (atan((2.0 * (C / B))) / pi);
    	tmp = 0.0;
    	if (B <= -4.2e-91)
    		tmp = 180.0 * (atan(1.0) / pi);
    	elseif (B <= -2.1e-210)
    		tmp = t_0;
    	elseif (B <= 3.2e-202)
    		tmp = 180.0 * (atan(((B * 0.5) / A)) / pi);
    	elseif (B <= 2500000000.0)
    		tmp = t_0;
    	elseif (B <= 3.5e+67)
    		tmp = 180.0 / (pi / atan((-0.5 * (B / C))));
    	else
    		tmp = 180.0 * (atan(-1.0) / pi);
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[(180.0 * N[(N[ArcTan[N[(2.0 * N[(C / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, -4.2e-91], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -2.1e-210], t$95$0, If[LessEqual[B, 3.2e-202], N[(180.0 * N[(N[ArcTan[N[(N[(B * 0.5), $MachinePrecision] / A), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 2500000000.0], t$95$0, If[LessEqual[B, 3.5e+67], N[(180.0 / N[(Pi / N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := 180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\
    \mathbf{if}\;B \leq -4.2 \cdot 10^{-91}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
    
    \mathbf{elif}\;B \leq -2.1 \cdot 10^{-210}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{elif}\;B \leq 3.2 \cdot 10^{-202}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\
    
    \mathbf{elif}\;B \leq 2500000000:\\
    \;\;\;\;t\_0\\
    
    \mathbf{elif}\;B \leq 3.5 \cdot 10^{+67}:\\
    \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}}\\
    
    \mathbf{else}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 5 regimes
    2. if B < -4.1999999999999998e-91

      1. Initial program 62.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around -inf 66.3%

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

      if -4.1999999999999998e-91 < B < -2.10000000000000016e-210 or 3.2000000000000001e-202 < B < 2.5e9

      1. Initial program 62.6%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in C around -inf 43.1%

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

      if -2.10000000000000016e-210 < B < 3.2000000000000001e-202

      1. Initial program 45.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in A around -inf 44.3%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B}{A}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/44.3%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0.5 \cdot B}{A}\right)}}{\pi} \]
      5. Simplified44.3%

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

      if 2.5e9 < B < 3.5e67

      1. Initial program 68.1%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr68.1%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in C around inf 57.0%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B} + -0.5 \cdot \frac{B}{C}\right)}}} \]
      5. Step-by-step derivation
        1. distribute-rgt1-in57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{\left(-1 + 1\right) \cdot A}}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        2. metadata-eval57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{0} \cdot A}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        3. mul0-lft57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{0}}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        4. div057.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \color{blue}{0} + -0.5 \cdot \frac{B}{C}\right)}} \]
        5. metadata-eval57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{0} + -0.5 \cdot \frac{B}{C}\right)}} \]
        6. metadata-eval57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{\left(0 - 0\right)} + -0.5 \cdot \frac{B}{C}\right)}} \]
        7. associate-+l-57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(0 - \left(0 - -0.5 \cdot \frac{B}{C}\right)\right)}}} \]
        8. neg-sub057.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{\left(--0.5 \cdot \frac{B}{C}\right)}\right)}} \]
        9. distribute-lft-neg-in57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{\left(--0.5\right) \cdot \frac{B}{C}}\right)}} \]
        10. metadata-eval57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{0.5} \cdot \frac{B}{C}\right)}} \]
        11. neg-sub057.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}} \]
        12. distribute-lft-neg-in57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(-0.5\right) \cdot \frac{B}{C}\right)}}} \]
        13. metadata-eval57.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{-0.5} \cdot \frac{B}{C}\right)}} \]
      6. Simplified57.0%

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

      if 3.5e67 < B

      1. Initial program 51.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around inf 74.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{-1}}{\pi} \]
    3. Recombined 5 regimes into one program.
    4. Final simplification57.2%

      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -4.2 \cdot 10^{-91}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -2.1 \cdot 10^{-210}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 3.2 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 2500000000:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 3.5 \cdot 10^{+67}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \]
    5. Add Preprocessing

    Alternative 7: 65.4% accurate, 3.2× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{C - A}{B}\\ \mathbf{if}\;B \leq -4.2 \cdot 10^{-217}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + t\_0\right)}}\\ \mathbf{elif}\;B \leq 5.5 \cdot 10^{-303}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 1.06 \cdot 10^{-252} \lor \neg \left(B \leq 1.85 \cdot 10^{-202}\right):\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(t\_0 + -1\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (/ (- C A) B)))
       (if (<= B -4.2e-217)
         (/ 180.0 (/ PI (atan (+ 1.0 t_0))))
         (if (<= B 5.5e-303)
           (* 180.0 (/ (atan (* 0.5 (/ (* B (+ 1.0 (/ C A))) A))) PI))
           (if (or (<= B 1.06e-252) (not (<= B 1.85e-202)))
             (/ 180.0 (/ PI (atan (+ t_0 -1.0))))
             (* 180.0 (/ (atan (/ 0.0 B)) PI)))))))
    double code(double A, double B, double C) {
    	double t_0 = (C - A) / B;
    	double tmp;
    	if (B <= -4.2e-217) {
    		tmp = 180.0 / (((double) M_PI) / atan((1.0 + t_0)));
    	} else if (B <= 5.5e-303) {
    		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / ((double) M_PI));
    	} else if ((B <= 1.06e-252) || !(B <= 1.85e-202)) {
    		tmp = 180.0 / (((double) M_PI) / atan((t_0 + -1.0)));
    	} else {
    		tmp = 180.0 * (atan((0.0 / B)) / ((double) M_PI));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = (C - A) / B;
    	double tmp;
    	if (B <= -4.2e-217) {
    		tmp = 180.0 / (Math.PI / Math.atan((1.0 + t_0)));
    	} else if (B <= 5.5e-303) {
    		tmp = 180.0 * (Math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / Math.PI);
    	} else if ((B <= 1.06e-252) || !(B <= 1.85e-202)) {
    		tmp = 180.0 / (Math.PI / Math.atan((t_0 + -1.0)));
    	} else {
    		tmp = 180.0 * (Math.atan((0.0 / B)) / Math.PI);
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = (C - A) / B
    	tmp = 0
    	if B <= -4.2e-217:
    		tmp = 180.0 / (math.pi / math.atan((1.0 + t_0)))
    	elif B <= 5.5e-303:
    		tmp = 180.0 * (math.atan((0.5 * ((B * (1.0 + (C / A))) / A))) / math.pi)
    	elif (B <= 1.06e-252) or not (B <= 1.85e-202):
    		tmp = 180.0 / (math.pi / math.atan((t_0 + -1.0)))
    	else:
    		tmp = 180.0 * (math.atan((0.0 / B)) / math.pi)
    	return tmp
    
    function code(A, B, C)
    	t_0 = Float64(Float64(C - A) / B)
    	tmp = 0.0
    	if (B <= -4.2e-217)
    		tmp = Float64(180.0 / Float64(pi / atan(Float64(1.0 + t_0))));
    	elseif (B <= 5.5e-303)
    		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B * Float64(1.0 + Float64(C / A))) / A))) / pi));
    	elseif ((B <= 1.06e-252) || !(B <= 1.85e-202))
    		tmp = Float64(180.0 / Float64(pi / atan(Float64(t_0 + -1.0))));
    	else
    		tmp = Float64(180.0 * Float64(atan(Float64(0.0 / B)) / pi));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = (C - A) / B;
    	tmp = 0.0;
    	if (B <= -4.2e-217)
    		tmp = 180.0 / (pi / atan((1.0 + t_0)));
    	elseif (B <= 5.5e-303)
    		tmp = 180.0 * (atan((0.5 * ((B * (1.0 + (C / A))) / A))) / pi);
    	elseif ((B <= 1.06e-252) || ~((B <= 1.85e-202)))
    		tmp = 180.0 / (pi / atan((t_0 + -1.0)));
    	else
    		tmp = 180.0 * (atan((0.0 / B)) / pi);
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]}, If[LessEqual[B, -4.2e-217], N[(180.0 / N[(Pi / N[ArcTan[N[(1.0 + t$95$0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 5.5e-303], 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[Or[LessEqual[B, 1.06e-252], N[Not[LessEqual[B, 1.85e-202]], $MachinePrecision]], N[(180.0 / N[(Pi / N[ArcTan[N[(t$95$0 + -1.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(0.0 / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := \frac{C - A}{B}\\
    \mathbf{if}\;B \leq -4.2 \cdot 10^{-217}:\\
    \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + t\_0\right)}}\\
    
    \mathbf{elif}\;B \leq 5.5 \cdot 10^{-303}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\
    
    \mathbf{elif}\;B \leq 1.06 \cdot 10^{-252} \lor \neg \left(B \leq 1.85 \cdot 10^{-202}\right):\\
    \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(t\_0 + -1\right)}}\\
    
    \mathbf{else}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 4 regimes
    2. if B < -4.2e-217

      1. Initial program 62.8%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr83.0%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in B around -inf 78.0%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
      5. Step-by-step derivation
        1. associate--l+78.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
        2. div-sub78.8%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
      6. Simplified78.8%

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

      if -4.2e-217 < B < 5.50000000000000018e-303

      1. Initial program 39.8%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in A around -inf 50.3%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. mul-1-neg50.3%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{A}\right)}}{\pi} \]
        2. distribute-neg-frac250.3%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B + -0.5 \cdot \frac{B \cdot C}{A}}{-A}\right)}}{\pi} \]
        3. distribute-lft-out50.3%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{-0.5 \cdot \left(B + \frac{B \cdot C}{A}\right)}}{-A}\right)}{\pi} \]
        4. associate-/l*51.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-0.5 \cdot \left(B + \color{blue}{B \cdot \frac{C}{A}}\right)}{-A}\right)}{\pi} \]
      5. Simplified51.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot \left(B + B \cdot \frac{C}{A}\right)}{-A}\right)}}{\pi} \]
      6. Taylor expanded in B around 0 51.6%

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

      if 5.50000000000000018e-303 < B < 1.06e-252 or 1.84999999999999995e-202 < B

      1. Initial program 58.2%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr77.8%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in B around inf 71.8%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\frac{C}{B} - \left(1 + \frac{A}{B}\right)\right)}}} \]
      5. Step-by-step derivation
        1. +-commutative71.8%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C}{B} - \color{blue}{\left(\frac{A}{B} + 1\right)}\right)}} \]
        2. associate--r+71.8%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(\frac{C}{B} - \frac{A}{B}\right) - 1\right)}}} \]
        3. div-sub73.6%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{\frac{C - A}{B}} - 1\right)}} \]
      6. Simplified73.6%

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

      if 1.06e-252 < B < 1.84999999999999995e-202

      1. Initial program 41.9%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in C around inf 100.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
        2. distribute-rgt1-in100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
        3. metadata-eval100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
        4. mul0-lft100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
        5. metadata-eval100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
      5. Simplified100.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0}{B}\right)}}{\pi} \]
    3. Recombined 4 regimes into one program.
    4. Final simplification74.4%

      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -4.2 \cdot 10^{-217}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \mathbf{elif}\;B \leq 5.5 \cdot 10^{-303}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B \cdot \left(1 + \frac{C}{A}\right)}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 1.06 \cdot 10^{-252} \lor \neg \left(B \leq 1.85 \cdot 10^{-202}\right):\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \end{array} \]
    5. Add Preprocessing

    Alternative 8: 64.2% accurate, 3.2× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{C - A}{B}\\ \mathbf{if}\;B \leq -4.1 \cdot 10^{-213}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + t\_0\right)}}\\ \mathbf{elif}\;B \leq 4.3 \cdot 10^{-303}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 3.2 \cdot 10^{-258} \lor \neg \left(B \leq 1.8 \cdot 10^{-202}\right):\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(t\_0 + -1\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (/ (- C A) B)))
       (if (<= B -4.1e-213)
         (/ 180.0 (/ PI (atan (+ 1.0 t_0))))
         (if (<= B 4.3e-303)
           (* 180.0 (/ (atan (/ (* B 0.5) A)) PI))
           (if (or (<= B 3.2e-258) (not (<= B 1.8e-202)))
             (/ 180.0 (/ PI (atan (+ t_0 -1.0))))
             (* 180.0 (/ (atan (/ 0.0 B)) PI)))))))
    double code(double A, double B, double C) {
    	double t_0 = (C - A) / B;
    	double tmp;
    	if (B <= -4.1e-213) {
    		tmp = 180.0 / (((double) M_PI) / atan((1.0 + t_0)));
    	} else if (B <= 4.3e-303) {
    		tmp = 180.0 * (atan(((B * 0.5) / A)) / ((double) M_PI));
    	} else if ((B <= 3.2e-258) || !(B <= 1.8e-202)) {
    		tmp = 180.0 / (((double) M_PI) / atan((t_0 + -1.0)));
    	} else {
    		tmp = 180.0 * (atan((0.0 / B)) / ((double) M_PI));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = (C - A) / B;
    	double tmp;
    	if (B <= -4.1e-213) {
    		tmp = 180.0 / (Math.PI / Math.atan((1.0 + t_0)));
    	} else if (B <= 4.3e-303) {
    		tmp = 180.0 * (Math.atan(((B * 0.5) / A)) / Math.PI);
    	} else if ((B <= 3.2e-258) || !(B <= 1.8e-202)) {
    		tmp = 180.0 / (Math.PI / Math.atan((t_0 + -1.0)));
    	} else {
    		tmp = 180.0 * (Math.atan((0.0 / B)) / Math.PI);
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = (C - A) / B
    	tmp = 0
    	if B <= -4.1e-213:
    		tmp = 180.0 / (math.pi / math.atan((1.0 + t_0)))
    	elif B <= 4.3e-303:
    		tmp = 180.0 * (math.atan(((B * 0.5) / A)) / math.pi)
    	elif (B <= 3.2e-258) or not (B <= 1.8e-202):
    		tmp = 180.0 / (math.pi / math.atan((t_0 + -1.0)))
    	else:
    		tmp = 180.0 * (math.atan((0.0 / B)) / math.pi)
    	return tmp
    
    function code(A, B, C)
    	t_0 = Float64(Float64(C - A) / B)
    	tmp = 0.0
    	if (B <= -4.1e-213)
    		tmp = Float64(180.0 / Float64(pi / atan(Float64(1.0 + t_0))));
    	elseif (B <= 4.3e-303)
    		tmp = Float64(180.0 * Float64(atan(Float64(Float64(B * 0.5) / A)) / pi));
    	elseif ((B <= 3.2e-258) || !(B <= 1.8e-202))
    		tmp = Float64(180.0 / Float64(pi / atan(Float64(t_0 + -1.0))));
    	else
    		tmp = Float64(180.0 * Float64(atan(Float64(0.0 / B)) / pi));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = (C - A) / B;
    	tmp = 0.0;
    	if (B <= -4.1e-213)
    		tmp = 180.0 / (pi / atan((1.0 + t_0)));
    	elseif (B <= 4.3e-303)
    		tmp = 180.0 * (atan(((B * 0.5) / A)) / pi);
    	elseif ((B <= 3.2e-258) || ~((B <= 1.8e-202)))
    		tmp = 180.0 / (pi / atan((t_0 + -1.0)));
    	else
    		tmp = 180.0 * (atan((0.0 / B)) / pi);
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]}, If[LessEqual[B, -4.1e-213], N[(180.0 / N[(Pi / N[ArcTan[N[(1.0 + t$95$0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 4.3e-303], N[(180.0 * N[(N[ArcTan[N[(N[(B * 0.5), $MachinePrecision] / A), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[Or[LessEqual[B, 3.2e-258], N[Not[LessEqual[B, 1.8e-202]], $MachinePrecision]], N[(180.0 / N[(Pi / N[ArcTan[N[(t$95$0 + -1.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(180.0 * N[(N[ArcTan[N[(0.0 / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := \frac{C - A}{B}\\
    \mathbf{if}\;B \leq -4.1 \cdot 10^{-213}:\\
    \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + t\_0\right)}}\\
    
    \mathbf{elif}\;B \leq 4.3 \cdot 10^{-303}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\
    
    \mathbf{elif}\;B \leq 3.2 \cdot 10^{-258} \lor \neg \left(B \leq 1.8 \cdot 10^{-202}\right):\\
    \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(t\_0 + -1\right)}}\\
    
    \mathbf{else}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 4 regimes
    2. if B < -4.09999999999999975e-213

      1. Initial program 62.8%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr83.0%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in B around -inf 78.0%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
      5. Step-by-step derivation
        1. associate--l+78.0%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
        2. div-sub78.8%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
      6. Simplified78.8%

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

      if -4.09999999999999975e-213 < B < 4.29999999999999981e-303

      1. Initial program 39.8%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in A around -inf 50.8%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B}{A}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/50.8%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0.5 \cdot B}{A}\right)}}{\pi} \]
      5. Simplified50.8%

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

      if 4.29999999999999981e-303 < B < 3.2000000000000002e-258 or 1.8000000000000001e-202 < B

      1. Initial program 58.2%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr77.8%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in B around inf 71.8%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\frac{C}{B} - \left(1 + \frac{A}{B}\right)\right)}}} \]
      5. Step-by-step derivation
        1. +-commutative71.8%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C}{B} - \color{blue}{\left(\frac{A}{B} + 1\right)}\right)}} \]
        2. associate--r+71.8%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(\frac{C}{B} - \frac{A}{B}\right) - 1\right)}}} \]
        3. div-sub73.6%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{\frac{C - A}{B}} - 1\right)}} \]
      6. Simplified73.6%

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

      if 3.2000000000000002e-258 < B < 1.8000000000000001e-202

      1. Initial program 41.9%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in C around inf 100.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
        2. distribute-rgt1-in100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
        3. metadata-eval100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
        4. mul0-lft100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
        5. metadata-eval100.0%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
      5. Simplified100.0%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0}{B}\right)}}{\pi} \]
    3. Recombined 4 regimes into one program.
    4. Final simplification74.3%

      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -4.1 \cdot 10^{-213}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}}\\ \mathbf{elif}\;B \leq 4.3 \cdot 10^{-303}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 3.2 \cdot 10^{-258} \lor \neg \left(B \leq 1.8 \cdot 10^{-202}\right):\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \end{array} \]
    5. Add Preprocessing

    Alternative 9: 46.2% accurate, 3.2× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\ \mathbf{if}\;B \leq -3.8 \cdot 10^{-91}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.6 \cdot 10^{-210}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;B \leq 2.6 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 2.9 \cdot 10^{-7}:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (* 180.0 (/ (atan (* 2.0 (/ C B))) PI))))
       (if (<= B -3.8e-91)
         (* 180.0 (/ (atan 1.0) PI))
         (if (<= B -1.6e-210)
           t_0
           (if (<= B 2.6e-202)
             (* 180.0 (/ (atan (/ (* B 0.5) A)) PI))
             (if (<= B 2.9e-7) t_0 (* 180.0 (/ (atan -1.0) PI))))))))
    double code(double A, double B, double C) {
    	double t_0 = 180.0 * (atan((2.0 * (C / B))) / ((double) M_PI));
    	double tmp;
    	if (B <= -3.8e-91) {
    		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
    	} else if (B <= -1.6e-210) {
    		tmp = t_0;
    	} else if (B <= 2.6e-202) {
    		tmp = 180.0 * (atan(((B * 0.5) / A)) / ((double) M_PI));
    	} else if (B <= 2.9e-7) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = 180.0 * (Math.atan((2.0 * (C / B))) / Math.PI);
    	double tmp;
    	if (B <= -3.8e-91) {
    		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
    	} else if (B <= -1.6e-210) {
    		tmp = t_0;
    	} else if (B <= 2.6e-202) {
    		tmp = 180.0 * (Math.atan(((B * 0.5) / A)) / Math.PI);
    	} else if (B <= 2.9e-7) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = 180.0 * (math.atan((2.0 * (C / B))) / math.pi)
    	tmp = 0
    	if B <= -3.8e-91:
    		tmp = 180.0 * (math.atan(1.0) / math.pi)
    	elif B <= -1.6e-210:
    		tmp = t_0
    	elif B <= 2.6e-202:
    		tmp = 180.0 * (math.atan(((B * 0.5) / A)) / math.pi)
    	elif B <= 2.9e-7:
    		tmp = t_0
    	else:
    		tmp = 180.0 * (math.atan(-1.0) / math.pi)
    	return tmp
    
    function code(A, B, C)
    	t_0 = Float64(180.0 * Float64(atan(Float64(2.0 * Float64(C / B))) / pi))
    	tmp = 0.0
    	if (B <= -3.8e-91)
    		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
    	elseif (B <= -1.6e-210)
    		tmp = t_0;
    	elseif (B <= 2.6e-202)
    		tmp = Float64(180.0 * Float64(atan(Float64(Float64(B * 0.5) / A)) / pi));
    	elseif (B <= 2.9e-7)
    		tmp = t_0;
    	else
    		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = 180.0 * (atan((2.0 * (C / B))) / pi);
    	tmp = 0.0;
    	if (B <= -3.8e-91)
    		tmp = 180.0 * (atan(1.0) / pi);
    	elseif (B <= -1.6e-210)
    		tmp = t_0;
    	elseif (B <= 2.6e-202)
    		tmp = 180.0 * (atan(((B * 0.5) / A)) / pi);
    	elseif (B <= 2.9e-7)
    		tmp = t_0;
    	else
    		tmp = 180.0 * (atan(-1.0) / pi);
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[(180.0 * N[(N[ArcTan[N[(2.0 * N[(C / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, -3.8e-91], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -1.6e-210], t$95$0, If[LessEqual[B, 2.6e-202], N[(180.0 * N[(N[ArcTan[N[(N[(B * 0.5), $MachinePrecision] / A), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 2.9e-7], t$95$0, N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := 180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\
    \mathbf{if}\;B \leq -3.8 \cdot 10^{-91}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
    
    \mathbf{elif}\;B \leq -1.6 \cdot 10^{-210}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{elif}\;B \leq 2.6 \cdot 10^{-202}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\
    
    \mathbf{elif}\;B \leq 2.9 \cdot 10^{-7}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{else}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 4 regimes
    2. if B < -3.79999999999999978e-91

      1. Initial program 62.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around -inf 66.3%

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

      if -3.79999999999999978e-91 < B < -1.60000000000000014e-210 or 2.60000000000000009e-202 < B < 2.8999999999999998e-7

      1. Initial program 62.1%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in C around -inf 43.5%

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

      if -1.60000000000000014e-210 < B < 2.60000000000000009e-202

      1. Initial program 45.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in A around -inf 44.3%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B}{A}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/44.3%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{0.5 \cdot B}{A}\right)}}{\pi} \]
      5. Simplified44.3%

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

      if 2.8999999999999998e-7 < B

      1. Initial program 55.9%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around inf 65.0%

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

      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq -3.8 \cdot 10^{-91}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.6 \cdot 10^{-210}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 2.6 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 2.9 \cdot 10^{-7}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \]
    5. Add Preprocessing

    Alternative 10: 46.2% accurate, 3.2× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\ \mathbf{if}\;B \leq -5.5 \cdot 10^{-92}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -9.2 \cdot 10^{-211}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;B \leq 1.95 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 5.8 \cdot 10^{-6}:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (* 180.0 (/ (atan (* 2.0 (/ C B))) PI))))
       (if (<= B -5.5e-92)
         (* 180.0 (/ (atan 1.0) PI))
         (if (<= B -9.2e-211)
           t_0
           (if (<= B 1.95e-202)
             (* 180.0 (/ (atan (/ 0.0 B)) PI))
             (if (<= B 5.8e-6) t_0 (* 180.0 (/ (atan -1.0) PI))))))))
    double code(double A, double B, double C) {
    	double t_0 = 180.0 * (atan((2.0 * (C / B))) / ((double) M_PI));
    	double tmp;
    	if (B <= -5.5e-92) {
    		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
    	} else if (B <= -9.2e-211) {
    		tmp = t_0;
    	} else if (B <= 1.95e-202) {
    		tmp = 180.0 * (atan((0.0 / B)) / ((double) M_PI));
    	} else if (B <= 5.8e-6) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = 180.0 * (Math.atan((2.0 * (C / B))) / Math.PI);
    	double tmp;
    	if (B <= -5.5e-92) {
    		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
    	} else if (B <= -9.2e-211) {
    		tmp = t_0;
    	} else if (B <= 1.95e-202) {
    		tmp = 180.0 * (Math.atan((0.0 / B)) / Math.PI);
    	} else if (B <= 5.8e-6) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = 180.0 * (math.atan((2.0 * (C / B))) / math.pi)
    	tmp = 0
    	if B <= -5.5e-92:
    		tmp = 180.0 * (math.atan(1.0) / math.pi)
    	elif B <= -9.2e-211:
    		tmp = t_0
    	elif B <= 1.95e-202:
    		tmp = 180.0 * (math.atan((0.0 / B)) / math.pi)
    	elif B <= 5.8e-6:
    		tmp = t_0
    	else:
    		tmp = 180.0 * (math.atan(-1.0) / math.pi)
    	return tmp
    
    function code(A, B, C)
    	t_0 = Float64(180.0 * Float64(atan(Float64(2.0 * Float64(C / B))) / pi))
    	tmp = 0.0
    	if (B <= -5.5e-92)
    		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
    	elseif (B <= -9.2e-211)
    		tmp = t_0;
    	elseif (B <= 1.95e-202)
    		tmp = Float64(180.0 * Float64(atan(Float64(0.0 / B)) / pi));
    	elseif (B <= 5.8e-6)
    		tmp = t_0;
    	else
    		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = 180.0 * (atan((2.0 * (C / B))) / pi);
    	tmp = 0.0;
    	if (B <= -5.5e-92)
    		tmp = 180.0 * (atan(1.0) / pi);
    	elseif (B <= -9.2e-211)
    		tmp = t_0;
    	elseif (B <= 1.95e-202)
    		tmp = 180.0 * (atan((0.0 / B)) / pi);
    	elseif (B <= 5.8e-6)
    		tmp = t_0;
    	else
    		tmp = 180.0 * (atan(-1.0) / pi);
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[(180.0 * N[(N[ArcTan[N[(2.0 * N[(C / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, -5.5e-92], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -9.2e-211], t$95$0, If[LessEqual[B, 1.95e-202], N[(180.0 * N[(N[ArcTan[N[(0.0 / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 5.8e-6], t$95$0, N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := 180 \cdot \frac{\tan^{-1} \left(2 \cdot \frac{C}{B}\right)}{\pi}\\
    \mathbf{if}\;B \leq -5.5 \cdot 10^{-92}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
    
    \mathbf{elif}\;B \leq -9.2 \cdot 10^{-211}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{elif}\;B \leq 1.95 \cdot 10^{-202}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\
    
    \mathbf{elif}\;B \leq 5.8 \cdot 10^{-6}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{else}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 4 regimes
    2. if B < -5.5000000000000002e-92

      1. Initial program 62.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around -inf 66.3%

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

      if -5.5000000000000002e-92 < B < -9.19999999999999953e-211 or 1.95e-202 < B < 5.8000000000000004e-6

      1. Initial program 62.1%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in C around -inf 43.5%

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

      if -9.19999999999999953e-211 < B < 1.95e-202

      1. Initial program 45.0%

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

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/42.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
        2. distribute-rgt1-in42.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
        3. metadata-eval42.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
        4. mul0-lft42.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
        5. metadata-eval42.4%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
      5. Simplified42.4%

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

      if 5.8000000000000004e-6 < B

      1. Initial program 55.9%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around inf 65.0%

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

    Alternative 11: 45.6% accurate, 3.2× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\ \mathbf{if}\;B \leq -1.18 \cdot 10^{-91}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.1 \cdot 10^{-211}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;B \leq 2.7 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 4.6 \cdot 10^{-111}:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (* 180.0 (/ (atan (* -2.0 (/ A B))) PI))))
       (if (<= B -1.18e-91)
         (* 180.0 (/ (atan 1.0) PI))
         (if (<= B -1.1e-211)
           t_0
           (if (<= B 2.7e-202)
             (* 180.0 (/ (atan (/ 0.0 B)) PI))
             (if (<= B 4.6e-111) t_0 (* 180.0 (/ (atan -1.0) PI))))))))
    double code(double A, double B, double C) {
    	double t_0 = 180.0 * (atan((-2.0 * (A / B))) / ((double) M_PI));
    	double tmp;
    	if (B <= -1.18e-91) {
    		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
    	} else if (B <= -1.1e-211) {
    		tmp = t_0;
    	} else if (B <= 2.7e-202) {
    		tmp = 180.0 * (atan((0.0 / B)) / ((double) M_PI));
    	} else if (B <= 4.6e-111) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 * (atan(-1.0) / ((double) M_PI));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = 180.0 * (Math.atan((-2.0 * (A / B))) / Math.PI);
    	double tmp;
    	if (B <= -1.18e-91) {
    		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
    	} else if (B <= -1.1e-211) {
    		tmp = t_0;
    	} else if (B <= 2.7e-202) {
    		tmp = 180.0 * (Math.atan((0.0 / B)) / Math.PI);
    	} else if (B <= 4.6e-111) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 * (Math.atan(-1.0) / Math.PI);
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = 180.0 * (math.atan((-2.0 * (A / B))) / math.pi)
    	tmp = 0
    	if B <= -1.18e-91:
    		tmp = 180.0 * (math.atan(1.0) / math.pi)
    	elif B <= -1.1e-211:
    		tmp = t_0
    	elif B <= 2.7e-202:
    		tmp = 180.0 * (math.atan((0.0 / B)) / math.pi)
    	elif B <= 4.6e-111:
    		tmp = t_0
    	else:
    		tmp = 180.0 * (math.atan(-1.0) / math.pi)
    	return tmp
    
    function code(A, B, C)
    	t_0 = Float64(180.0 * Float64(atan(Float64(-2.0 * Float64(A / B))) / pi))
    	tmp = 0.0
    	if (B <= -1.18e-91)
    		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
    	elseif (B <= -1.1e-211)
    		tmp = t_0;
    	elseif (B <= 2.7e-202)
    		tmp = Float64(180.0 * Float64(atan(Float64(0.0 / B)) / pi));
    	elseif (B <= 4.6e-111)
    		tmp = t_0;
    	else
    		tmp = Float64(180.0 * Float64(atan(-1.0) / pi));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = 180.0 * (atan((-2.0 * (A / B))) / pi);
    	tmp = 0.0;
    	if (B <= -1.18e-91)
    		tmp = 180.0 * (atan(1.0) / pi);
    	elseif (B <= -1.1e-211)
    		tmp = t_0;
    	elseif (B <= 2.7e-202)
    		tmp = 180.0 * (atan((0.0 / B)) / pi);
    	elseif (B <= 4.6e-111)
    		tmp = t_0;
    	else
    		tmp = 180.0 * (atan(-1.0) / pi);
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[(180.0 * N[(N[ArcTan[N[(-2.0 * N[(A / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, -1.18e-91], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, -1.1e-211], t$95$0, If[LessEqual[B, 2.7e-202], N[(180.0 * N[(N[ArcTan[N[(0.0 / B), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 4.6e-111], t$95$0, N[(180.0 * N[(N[ArcTan[-1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := 180 \cdot \frac{\tan^{-1} \left(-2 \cdot \frac{A}{B}\right)}{\pi}\\
    \mathbf{if}\;B \leq -1.18 \cdot 10^{-91}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
    
    \mathbf{elif}\;B \leq -1.1 \cdot 10^{-211}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{elif}\;B \leq 2.7 \cdot 10^{-202}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\
    
    \mathbf{elif}\;B \leq 4.6 \cdot 10^{-111}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{else}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 4 regimes
    2. if B < -1.18e-91

      1. Initial program 62.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around -inf 66.3%

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

      if -1.18e-91 < B < -1.09999999999999999e-211 or 2.6999999999999999e-202 < B < 4.6e-111

      1. Initial program 63.4%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in A around inf 38.6%

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

      if -1.09999999999999999e-211 < B < 2.6999999999999999e-202

      1. Initial program 43.4%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in C around inf 43.6%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/43.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
        2. distribute-rgt1-in43.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
        3. metadata-eval43.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
        4. mul0-lft43.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
        5. metadata-eval43.6%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
      5. Simplified43.6%

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

      if 4.6e-111 < B

      1. Initial program 57.1%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around inf 54.3%

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

    Alternative 12: 60.9% accurate, 3.3× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := \tan^{-1} \left(1 + \frac{C - A}{B}\right)\\ \mathbf{if}\;C \leq -1.2 \cdot 10^{-262}:\\ \;\;\;\;180 \cdot \frac{t\_0}{\pi}\\ \mathbf{elif}\;C \leq -2.9 \cdot 10^{-303}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 3.3 \cdot 10^{+19}:\\ \;\;\;\;\frac{180}{\frac{\pi}{t\_0}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (atan (+ 1.0 (/ (- C A) B)))))
       (if (<= C -1.2e-262)
         (* 180.0 (/ t_0 PI))
         (if (<= C -2.9e-303)
           (* (/ 180.0 PI) (atan (* 0.5 (/ B A))))
           (if (<= C 3.3e+19)
             (/ 180.0 (/ PI t_0))
             (/ 180.0 (/ PI (atan (* -0.5 (/ B C))))))))))
    double code(double A, double B, double C) {
    	double t_0 = atan((1.0 + ((C - A) / B)));
    	double tmp;
    	if (C <= -1.2e-262) {
    		tmp = 180.0 * (t_0 / ((double) M_PI));
    	} else if (C <= -2.9e-303) {
    		tmp = (180.0 / ((double) M_PI)) * atan((0.5 * (B / A)));
    	} else if (C <= 3.3e+19) {
    		tmp = 180.0 / (((double) M_PI) / t_0);
    	} else {
    		tmp = 180.0 / (((double) M_PI) / atan((-0.5 * (B / C))));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = Math.atan((1.0 + ((C - A) / B)));
    	double tmp;
    	if (C <= -1.2e-262) {
    		tmp = 180.0 * (t_0 / Math.PI);
    	} else if (C <= -2.9e-303) {
    		tmp = (180.0 / Math.PI) * Math.atan((0.5 * (B / A)));
    	} else if (C <= 3.3e+19) {
    		tmp = 180.0 / (Math.PI / t_0);
    	} else {
    		tmp = 180.0 / (Math.PI / Math.atan((-0.5 * (B / C))));
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = math.atan((1.0 + ((C - A) / B)))
    	tmp = 0
    	if C <= -1.2e-262:
    		tmp = 180.0 * (t_0 / math.pi)
    	elif C <= -2.9e-303:
    		tmp = (180.0 / math.pi) * math.atan((0.5 * (B / A)))
    	elif C <= 3.3e+19:
    		tmp = 180.0 / (math.pi / t_0)
    	else:
    		tmp = 180.0 / (math.pi / math.atan((-0.5 * (B / C))))
    	return tmp
    
    function code(A, B, C)
    	t_0 = atan(Float64(1.0 + Float64(Float64(C - A) / B)))
    	tmp = 0.0
    	if (C <= -1.2e-262)
    		tmp = Float64(180.0 * Float64(t_0 / pi));
    	elseif (C <= -2.9e-303)
    		tmp = Float64(Float64(180.0 / pi) * atan(Float64(0.5 * Float64(B / A))));
    	elseif (C <= 3.3e+19)
    		tmp = Float64(180.0 / Float64(pi / t_0));
    	else
    		tmp = Float64(180.0 / Float64(pi / atan(Float64(-0.5 * Float64(B / C)))));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = atan((1.0 + ((C - A) / B)));
    	tmp = 0.0;
    	if (C <= -1.2e-262)
    		tmp = 180.0 * (t_0 / pi);
    	elseif (C <= -2.9e-303)
    		tmp = (180.0 / pi) * atan((0.5 * (B / A)));
    	elseif (C <= 3.3e+19)
    		tmp = 180.0 / (pi / t_0);
    	else
    		tmp = 180.0 / (pi / atan((-0.5 * (B / C))));
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[ArcTan[N[(1.0 + N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[C, -1.2e-262], N[(180.0 * N[(t$95$0 / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[C, -2.9e-303], N[(N[(180.0 / Pi), $MachinePrecision] * N[ArcTan[N[(0.5 * N[(B / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[C, 3.3e+19], N[(180.0 / N[(Pi / t$95$0), $MachinePrecision]), $MachinePrecision], N[(180.0 / N[(Pi / N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := \tan^{-1} \left(1 + \frac{C - A}{B}\right)\\
    \mathbf{if}\;C \leq -1.2 \cdot 10^{-262}:\\
    \;\;\;\;180 \cdot \frac{t\_0}{\pi}\\
    
    \mathbf{elif}\;C \leq -2.9 \cdot 10^{-303}:\\
    \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\
    
    \mathbf{elif}\;C \leq 3.3 \cdot 10^{+19}:\\
    \;\;\;\;\frac{180}{\frac{\pi}{t\_0}}\\
    
    \mathbf{else}:\\
    \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 4 regimes
    2. if C < -1.2e-262

      1. Initial program 77.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around -inf 71.5%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate--l+71.5%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}{\pi} \]
        2. div-sub73.2%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}{\pi} \]
      5. Simplified73.2%

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

      if -1.2e-262 < C < -2.90000000000000014e-303

      1. Initial program 57.2%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr68.2%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in A around -inf 57.9%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B}{A}\right)}}} \]
      5. Step-by-step derivation
        1. associate-/r/58.2%

          \[\leadsto \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)} \]
      6. Applied egg-rr58.2%

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

      if -2.90000000000000014e-303 < C < 3.3e19

      1. Initial program 56.8%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr81.4%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in B around -inf 57.6%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}} \]
      5. Step-by-step derivation
        1. associate--l+57.6%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(1 + \left(\frac{C}{B} - \frac{A}{B}\right)\right)}}} \]
        2. div-sub57.6%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}} \]
      6. Simplified57.6%

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

      if 3.3e19 < C

      1. Initial program 22.2%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr52.7%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in C around inf 71.7%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B} + -0.5 \cdot \frac{B}{C}\right)}}} \]
      5. Step-by-step derivation
        1. distribute-rgt1-in71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{\left(-1 + 1\right) \cdot A}}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        2. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{0} \cdot A}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        3. mul0-lft71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{0}}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        4. div071.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \color{blue}{0} + -0.5 \cdot \frac{B}{C}\right)}} \]
        5. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{0} + -0.5 \cdot \frac{B}{C}\right)}} \]
        6. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{\left(0 - 0\right)} + -0.5 \cdot \frac{B}{C}\right)}} \]
        7. associate-+l-71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(0 - \left(0 - -0.5 \cdot \frac{B}{C}\right)\right)}}} \]
        8. neg-sub071.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{\left(--0.5 \cdot \frac{B}{C}\right)}\right)}} \]
        9. distribute-lft-neg-in71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{\left(--0.5\right) \cdot \frac{B}{C}}\right)}} \]
        10. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{0.5} \cdot \frac{B}{C}\right)}} \]
        11. neg-sub071.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}} \]
        12. distribute-lft-neg-in71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(-0.5\right) \cdot \frac{B}{C}\right)}}} \]
        13. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{-0.5} \cdot \frac{B}{C}\right)}} \]
      6. Simplified71.7%

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

    Alternative 13: 60.9% accurate, 3.3× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi}\\ \mathbf{if}\;C \leq -1.2 \cdot 10^{-262}:\\ \;\;\;\;t\_0\\ \mathbf{elif}\;C \leq -1.12 \cdot 10^{-303}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 2.9 \cdot 10^{+19}:\\ \;\;\;\;t\_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0 (* 180.0 (/ (atan (+ 1.0 (/ (- C A) B))) PI))))
       (if (<= C -1.2e-262)
         t_0
         (if (<= C -1.12e-303)
           (* (/ 180.0 PI) (atan (* 0.5 (/ B A))))
           (if (<= C 2.9e+19) t_0 (/ 180.0 (/ PI (atan (* -0.5 (/ B C))))))))))
    double code(double A, double B, double C) {
    	double t_0 = 180.0 * (atan((1.0 + ((C - A) / B))) / ((double) M_PI));
    	double tmp;
    	if (C <= -1.2e-262) {
    		tmp = t_0;
    	} else if (C <= -1.12e-303) {
    		tmp = (180.0 / ((double) M_PI)) * atan((0.5 * (B / A)));
    	} else if (C <= 2.9e+19) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 / (((double) M_PI) / atan((-0.5 * (B / C))));
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = 180.0 * (Math.atan((1.0 + ((C - A) / B))) / Math.PI);
    	double tmp;
    	if (C <= -1.2e-262) {
    		tmp = t_0;
    	} else if (C <= -1.12e-303) {
    		tmp = (180.0 / Math.PI) * Math.atan((0.5 * (B / A)));
    	} else if (C <= 2.9e+19) {
    		tmp = t_0;
    	} else {
    		tmp = 180.0 / (Math.PI / Math.atan((-0.5 * (B / C))));
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = 180.0 * (math.atan((1.0 + ((C - A) / B))) / math.pi)
    	tmp = 0
    	if C <= -1.2e-262:
    		tmp = t_0
    	elif C <= -1.12e-303:
    		tmp = (180.0 / math.pi) * math.atan((0.5 * (B / A)))
    	elif C <= 2.9e+19:
    		tmp = t_0
    	else:
    		tmp = 180.0 / (math.pi / math.atan((-0.5 * (B / C))))
    	return tmp
    
    function code(A, B, C)
    	t_0 = Float64(180.0 * Float64(atan(Float64(1.0 + Float64(Float64(C - A) / B))) / pi))
    	tmp = 0.0
    	if (C <= -1.2e-262)
    		tmp = t_0;
    	elseif (C <= -1.12e-303)
    		tmp = Float64(Float64(180.0 / pi) * atan(Float64(0.5 * Float64(B / A))));
    	elseif (C <= 2.9e+19)
    		tmp = t_0;
    	else
    		tmp = Float64(180.0 / Float64(pi / atan(Float64(-0.5 * Float64(B / C)))));
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = 180.0 * (atan((1.0 + ((C - A) / B))) / pi);
    	tmp = 0.0;
    	if (C <= -1.2e-262)
    		tmp = t_0;
    	elseif (C <= -1.12e-303)
    		tmp = (180.0 / pi) * atan((0.5 * (B / A)));
    	elseif (C <= 2.9e+19)
    		tmp = t_0;
    	else
    		tmp = 180.0 / (pi / atan((-0.5 * (B / C))));
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = N[(180.0 * N[(N[ArcTan[N[(1.0 + N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[C, -1.2e-262], t$95$0, If[LessEqual[C, -1.12e-303], N[(N[(180.0 / Pi), $MachinePrecision] * N[ArcTan[N[(0.5 * N[(B / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[C, 2.9e+19], t$95$0, N[(180.0 / N[(Pi / N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := 180 \cdot \frac{\tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi}\\
    \mathbf{if}\;C \leq -1.2 \cdot 10^{-262}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{elif}\;C \leq -1.12 \cdot 10^{-303}:\\
    \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\
    
    \mathbf{elif}\;C \leq 2.9 \cdot 10^{+19}:\\
    \;\;\;\;t\_0\\
    
    \mathbf{else}:\\
    \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 3 regimes
    2. if C < -1.2e-262 or -1.1199999999999999e-303 < C < 2.9e19

      1. Initial program 69.3%

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

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate--l+66.2%

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

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(1 + \color{blue}{\frac{C - A}{B}}\right)}{\pi} \]
      5. Simplified67.2%

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

      if -1.2e-262 < C < -1.1199999999999999e-303

      1. Initial program 57.2%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr68.2%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in A around -inf 57.9%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B}{A}\right)}}} \]
      5. Step-by-step derivation
        1. associate-/r/58.2%

          \[\leadsto \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)} \]
      6. Applied egg-rr58.2%

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

      if 2.9e19 < C

      1. Initial program 22.2%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Applied egg-rr52.7%

        \[\leadsto \color{blue}{\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}}} \]
      4. Taylor expanded in C around inf 71.7%

        \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B} + -0.5 \cdot \frac{B}{C}\right)}}} \]
      5. Step-by-step derivation
        1. distribute-rgt1-in71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{\left(-1 + 1\right) \cdot A}}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        2. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{0} \cdot A}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        3. mul0-lft71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \frac{\color{blue}{0}}{B} + -0.5 \cdot \frac{B}{C}\right)}} \]
        4. div071.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(-1 \cdot \color{blue}{0} + -0.5 \cdot \frac{B}{C}\right)}} \]
        5. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{0} + -0.5 \cdot \frac{B}{C}\right)}} \]
        6. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{\left(0 - 0\right)} + -0.5 \cdot \frac{B}{C}\right)}} \]
        7. associate-+l-71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(0 - \left(0 - -0.5 \cdot \frac{B}{C}\right)\right)}}} \]
        8. neg-sub071.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{\left(--0.5 \cdot \frac{B}{C}\right)}\right)}} \]
        9. distribute-lft-neg-in71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{\left(--0.5\right) \cdot \frac{B}{C}}\right)}} \]
        10. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(0 - \color{blue}{0.5} \cdot \frac{B}{C}\right)}} \]
        11. neg-sub071.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)}}} \]
        12. distribute-lft-neg-in71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \color{blue}{\left(\left(-0.5\right) \cdot \frac{B}{C}\right)}}} \]
        13. metadata-eval71.7%

          \[\leadsto \frac{180}{\frac{\pi}{\tan^{-1} \left(\color{blue}{-0.5} \cdot \frac{B}{C}\right)}} \]
      6. Simplified71.7%

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

    Alternative 14: 44.3% accurate, 3.6× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;B \leq -5.3 \cdot 10^{-123}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 3 \cdot 10^{-93}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{0}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (if (<= B -5.3e-123)
       (* 180.0 (/ (atan 1.0) PI))
       (if (<= B 3e-93)
         (* 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 <= -5.3e-123) {
    		tmp = 180.0 * (atan(1.0) / ((double) M_PI));
    	} else if (B <= 3e-93) {
    		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 <= -5.3e-123) {
    		tmp = 180.0 * (Math.atan(1.0) / Math.PI);
    	} else if (B <= 3e-93) {
    		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 <= -5.3e-123:
    		tmp = 180.0 * (math.atan(1.0) / math.pi)
    	elif B <= 3e-93:
    		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 <= -5.3e-123)
    		tmp = Float64(180.0 * Float64(atan(1.0) / pi));
    	elseif (B <= 3e-93)
    		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 <= -5.3e-123)
    		tmp = 180.0 * (atan(1.0) / pi);
    	elseif (B <= 3e-93)
    		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, -5.3e-123], N[(180.0 * N[(N[ArcTan[1.0], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 3e-93], 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 -5.3 \cdot 10^{-123}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\
    
    \mathbf{elif}\;B \leq 3 \cdot 10^{-93}:\\
    \;\;\;\;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 < -5.29999999999999971e-123

      1. Initial program 63.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around -inf 60.8%

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

      if -5.29999999999999971e-123 < B < 3.0000000000000001e-93

      1. Initial program 53.0%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in C around inf 29.2%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(-1 \cdot \frac{A + -1 \cdot A}{B}\right)}}{\pi} \]
      4. Step-by-step derivation
        1. associate-*r/29.2%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1 \cdot \left(A + -1 \cdot A\right)}{B}\right)}}{\pi} \]
        2. distribute-rgt1-in29.2%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{\left(\left(-1 + 1\right) \cdot A\right)}}{B}\right)}{\pi} \]
        3. metadata-eval29.2%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \left(\color{blue}{0} \cdot A\right)}{B}\right)}{\pi} \]
        4. mul0-lft29.2%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1 \cdot \color{blue}{0}}{B}\right)}{\pi} \]
        5. metadata-eval29.2%

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{0}}{B}\right)}{\pi} \]
      5. Simplified29.2%

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

      if 3.0000000000000001e-93 < B

      1. Initial program 57.9%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around inf 55.9%

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

    Alternative 15: 39.5% accurate, 3.8× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;B \leq -5 \cdot 10^{-310}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} -1}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (if (<= B -5e-310) (* 180.0 (/ (atan 1.0) PI)) (* 180.0 (/ (atan -1.0) PI))))
    double code(double A, double B, double C) {
    	double tmp;
    	if (B <= -5e-310) {
    		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 <= -5e-310) {
    		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 <= -5e-310:
    		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 <= -5e-310)
    		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 <= -5e-310)
    		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, -5e-310], 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 -5 \cdot 10^{-310}:\\
    \;\;\;\;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 < -4.999999999999985e-310

      1. Initial program 59.7%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around -inf 44.8%

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

      if -4.999999999999985e-310 < B

      1. Initial program 56.5%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around inf 41.8%

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

    Alternative 16: 20.5% 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 58.2%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf 20.2%

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

    Reproduce

    ?
    herbie shell --seed 2024087 
    (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)))