ABCF->ab-angle angle

?

Percentage Accurate: 54.0% → 81.7%
Time: 26.6s
Precision: binary64
Cost: 20164

?

\[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
\[\begin{array}{l} \mathbf{if}\;C \leq 4.3 \cdot 10^{+80}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \end{array} \]
(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)))
(FPCore (A B C)
 :precision binary64
 (if (<= C 4.3e+80)
   (* 180.0 (/ (atan (/ (- (- C A) (hypot B (- A C))) B)) PI))
   (* (/ 180.0 PI) (atan (* -0.5 (/ B C))))))
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));
}
double code(double A, double B, double C) {
	double tmp;
	if (C <= 4.3e+80) {
		tmp = 180.0 * (atan((((C - A) - hypot(B, (A - C))) / B)) / ((double) M_PI));
	} else {
		tmp = (180.0 / ((double) M_PI)) * atan((-0.5 * (B / C)));
	}
	return tmp;
}
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);
}
public static double code(double A, double B, double C) {
	double tmp;
	if (C <= 4.3e+80) {
		tmp = 180.0 * (Math.atan((((C - A) - Math.hypot(B, (A - C))) / B)) / Math.PI);
	} else {
		tmp = (180.0 / Math.PI) * Math.atan((-0.5 * (B / C)));
	}
	return tmp;
}
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)
def code(A, B, C):
	tmp = 0
	if C <= 4.3e+80:
		tmp = 180.0 * (math.atan((((C - A) - math.hypot(B, (A - C))) / B)) / math.pi)
	else:
		tmp = (180.0 / math.pi) * math.atan((-0.5 * (B / C)))
	return tmp
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 code(A, B, C)
	tmp = 0.0
	if (C <= 4.3e+80)
		tmp = Float64(180.0 * Float64(atan(Float64(Float64(Float64(C - A) - hypot(B, Float64(A - C))) / B)) / pi));
	else
		tmp = Float64(Float64(180.0 / pi) * atan(Float64(-0.5 * Float64(B / C))));
	end
	return tmp
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
function tmp_2 = code(A, B, C)
	tmp = 0.0;
	if (C <= 4.3e+80)
		tmp = 180.0 * (atan((((C - A) - hypot(B, (A - C))) / B)) / pi);
	else
		tmp = (180.0 / pi) * atan((-0.5 * (B / C)));
	end
	tmp_2 = tmp;
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]
code[A_, B_, C_] := If[LessEqual[C, 4.3e+80], 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[(N[(180.0 / Pi), $MachinePrecision] * N[ArcTan[N[(-0.5 * N[(B / C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]
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}
\begin{array}{l}
\mathbf{if}\;C \leq 4.3 \cdot 10^{+80}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\

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


\end{array}

Local Percentage Accuracy?

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.

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Split input into 2 regimes
  2. if C < 4.30000000000000004e80

    1. Initial program 66.3%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Simplified82.9%

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

      [Start]66.3

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

      associate-*l/ [=>]66.3

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

      *-lft-identity [=>]66.3

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

      +-commutative [=>]66.3

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

      unpow2 [=>]66.3

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

      unpow2 [=>]66.3

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

      hypot-def [=>]82.9

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

    if 4.30000000000000004e80 < C

    1. Initial program 17.5%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Simplified46.4%

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

      [Start]17.5

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

      associate-*r/ [=>]17.5

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

      associate-*l/ [<=]17.5

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

      associate-*l/ [=>]17.5

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

      *-lft-identity [=>]17.5

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

      sub-neg [=>]17.5

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

      associate-+l- [=>]17.5

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

      sub-neg [=>]17.5

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

      remove-double-neg [=>]17.5

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

      +-commutative [=>]17.5

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

      unpow2 [=>]17.5

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

      unpow2 [=>]17.5

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

      hypot-def [=>]46.4

      \[ \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(A + \color{blue}{\mathsf{hypot}\left(B, A - C\right)}\right)}{B}\right) \]
    3. Taylor expanded in A around 0 15.9%

      \[\leadsto \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{\color{blue}{C - \sqrt{{B}^{2} + {C}^{2}}}}{B}\right) \]
    4. Simplified41.8%

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

      [Start]15.9

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

      unpow2 [=>]15.9

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

      unpow2 [=>]15.9

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

      hypot-def [=>]41.8

      \[ \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \color{blue}{\mathsf{hypot}\left(B, C\right)}}{B}\right) \]
    5. Taylor expanded in C around inf 77.1%

      \[\leadsto \frac{180}{\pi} \cdot \tan^{-1} \color{blue}{\left(-0.5 \cdot \frac{B}{C}\right)} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification81.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 4.3 \cdot 10^{+80}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \end{array} \]

Alternatives

Alternative 1
Accuracy75.1%
Cost20304
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)\\ \mathbf{if}\;A \leq -4 \cdot 10^{+102}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + \frac{B}{A} \cdot \frac{C}{A}\right)\right)}{\pi}\\ \mathbf{elif}\;A \leq -6.3 \cdot 10^{+63}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq -1.02 \cdot 10^{-16}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + B \cdot \frac{\frac{C}{A}}{A}\right)\right)}{\pi}\\ \mathbf{elif}\;A \leq 2.6 \cdot 10^{+53}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\ \end{array} \]
Alternative 2
Accuracy74.8%
Cost20304
\[\begin{array}{l} t_0 := \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)\\ \mathbf{if}\;A \leq -1.15 \cdot 10^{+97}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + \frac{B}{A} \cdot \frac{C}{A}\right)\right)}{\pi}\\ \mathbf{elif}\;A \leq -4.3 \cdot 10^{+67}:\\ \;\;\;\;\frac{180}{\pi} \cdot t_0\\ \mathbf{elif}\;A \leq -1.65 \cdot 10^{-16}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + B \cdot \frac{\frac{C}{A}}{A}\right)\right)}{\pi}\\ \mathbf{elif}\;A \leq 7 \cdot 10^{+52}:\\ \;\;\;\;\frac{180 \cdot t_0}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\ \end{array} \]
Alternative 3
Accuracy78.3%
Cost20104
\[\begin{array}{l} \mathbf{if}\;C \leq -8 \cdot 10^{-95}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq 1.35 \cdot 10^{+79}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(-A\right) - \mathsf{hypot}\left(A, B\right)}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \end{array} \]
Alternative 4
Accuracy78.3%
Cost20104
\[\begin{array}{l} \mathbf{if}\;C \leq -2.7 \cdot 10^{-95}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq 3.8 \cdot 10^{+79}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{\left(-A\right) - \mathsf{hypot}\left(A, B\right)}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \end{array} \]
Alternative 5
Accuracy57.0%
Cost14632
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\ t_2 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \mathbf{if}\;A \leq -3.5 \cdot 10^{-18}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)\\ \mathbf{elif}\;A \leq -2.2 \cdot 10^{-161}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq -3 \cdot 10^{-191}:\\ \;\;\;\;\frac{180}{\pi} \cdot t_2\\ \mathbf{elif}\;A \leq -1 \cdot 10^{-228}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq -4 \cdot 10^{-305}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{elif}\;A \leq 1.75 \cdot 10^{-235}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq 3.1 \cdot 10^{-115}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq 1.8 \cdot 10^{-25}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq 1.6 \cdot 10^{-13}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq 5.5:\\ \;\;\;\;180 \cdot \frac{t_2}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\ \end{array} \]
Alternative 6
Accuracy59.4%
Cost14236
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\ t_1 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \mathbf{if}\;A \leq -8.8 \cdot 10^{-17}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)\\ \mathbf{elif}\;A \leq -3.1 \cdot 10^{-164}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq -1.22 \cdot 10^{-191}:\\ \;\;\;\;\frac{180}{\pi} \cdot t_1\\ \mathbf{elif}\;A \leq -1.65 \cdot 10^{-233}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq 1.5 \cdot 10^{-13}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\ \mathbf{elif}\;A \leq 2.9:\\ \;\;\;\;180 \cdot \frac{t_1}{\pi}\\ \mathbf{elif}\;A \leq 2.2 \cdot 10^{+54}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\ \end{array} \]
Alternative 7
Accuracy59.8%
Cost14236
\[\begin{array}{l} t_0 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\ \mathbf{if}\;A \leq -1.9 \cdot 10^{-18}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + \frac{B}{A} \cdot \frac{C}{A}\right)\right)}{\pi}\\ \mathbf{elif}\;A \leq -3.4 \cdot 10^{-165}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq -2.35 \cdot 10^{-191}:\\ \;\;\;\;\frac{180}{\pi} \cdot t_0\\ \mathbf{elif}\;A \leq -8.2 \cdot 10^{-230}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq 1.35 \cdot 10^{-13}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\ \mathbf{elif}\;A \leq 3.5:\\ \;\;\;\;180 \cdot \frac{t_0}{\pi}\\ \mathbf{elif}\;A \leq 1.85 \cdot 10^{+49}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\ \end{array} \]
Alternative 8
Accuracy52.1%
Cost14104
\[\begin{array}{l} t_0 := \frac{180 \cdot \tan^{-1} -1}{\pi}\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\ t_2 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \mathbf{if}\;C \leq 1.75 \cdot 10^{-258}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 2.45 \cdot 10^{-131}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 3.15 \cdot 10^{-118}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 1.3 \cdot 10^{-107}:\\ \;\;\;\;180 \cdot \frac{t_2}{\pi}\\ \mathbf{elif}\;C \leq 2.2 \cdot 10^{-53}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 18500000:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{-A}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot t_2\\ \end{array} \]
Alternative 9
Accuracy47.1%
Cost13580
\[\begin{array}{l} \mathbf{if}\;B \leq -2.2 \cdot 10^{-129}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 1.85 \cdot 10^{-237}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{-A}{B}\right)\\ \mathbf{elif}\;B \leq 1.85 \cdot 10^{-125}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 10
Accuracy47.1%
Cost13576
\[\begin{array}{l} \mathbf{if}\;C \leq -3.5 \cdot 10^{-139}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C}{B}\right)}}\\ \mathbf{elif}\;C \leq 3.2 \cdot 10^{-20}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \end{array} \]
Alternative 11
Accuracy47.2%
Cost13576
\[\begin{array}{l} \mathbf{if}\;C \leq -1.1 \cdot 10^{-138}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C}{B}\right)}}\\ \mathbf{elif}\;C \leq 2.4 \cdot 10^{-20}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \end{array} \]
Alternative 12
Accuracy59.5%
Cost13576
\[\begin{array}{l} \mathbf{if}\;C \leq -1.85 \cdot 10^{-139}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\ \mathbf{elif}\;C \leq 2.15 \cdot 10^{+48}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \end{array} \]
Alternative 13
Accuracy61.4%
Cost13576
\[\begin{array}{l} \mathbf{if}\;C \leq 5 \cdot 10^{-270}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq 1.1 \cdot 10^{+46}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \end{array} \]
Alternative 14
Accuracy47.4%
Cost13448
\[\begin{array}{l} \mathbf{if}\;B \leq -8.6 \cdot 10^{-65}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 2 \cdot 10^{-125}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 15
Accuracy40.5%
Cost13188
\[\begin{array}{l} \mathbf{if}\;B \leq -5 \cdot 10^{-310}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 16
Accuracy21.9%
Cost13056
\[\frac{180 \cdot \tan^{-1} -1}{\pi} \]

Error

Reproduce?

herbie shell --seed 2023159 
(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)))