ABCF->ab-angle angle

Average Accuracy: 54.0% → 80.7%

Time: 25.0s

Precision: binary64

Cost: 20164

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;A \leq -108:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} \cdot \left(\frac{C}{A} + 1\right)\right)\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)\\ \end{array} \]

FPCore

(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 (<= A -108.0)
   (* 180.0 (/ (atan (* 0.5 (* (/ B A) (+ (/ C A) 1.0)))) PI))
   (* (/ 180.0 PI) (atan (/ (- C (+ A (hypot B (- A C)))) 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 (A <= -108.0) {
		tmp = 180.0 * (atan((0.5 * ((B / A) * ((C / A) + 1.0)))) / ((double) M_PI));
	} else {
		tmp = (180.0 / ((double) M_PI)) * atan(((C - (A + hypot(B, (A - C)))) / B));
	}
	return tmp;
}

Java

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 (A <= -108.0) {
		tmp = 180.0 * (Math.atan((0.5 * ((B / A) * ((C / A) + 1.0)))) / Math.PI);
	} else {
		tmp = (180.0 / Math.PI) * Math.atan(((C - (A + Math.hypot(B, (A - C)))) / B));
	}
	return tmp;
}

Python

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 A <= -108.0:
		tmp = 180.0 * (math.atan((0.5 * ((B / A) * ((C / A) + 1.0)))) / math.pi)
	else:
		tmp = (180.0 / math.pi) * math.atan(((C - (A + math.hypot(B, (A - C)))) / B))
	return tmp

Julia

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 (A <= -108.0)
		tmp = Float64(180.0 * Float64(atan(Float64(0.5 * Float64(Float64(B / A) * Float64(Float64(C / A) + 1.0)))) / pi));
	else
		tmp = Float64(Float64(180.0 / pi) * atan(Float64(Float64(C - Float64(A + hypot(B, Float64(A - C)))) / B)));
	end
	return tmp
end

MATLAB

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 (A <= -108.0)
		tmp = 180.0 * (atan((0.5 * ((B / A) * ((C / A) + 1.0)))) / pi);
	else
		tmp = (180.0 / pi) * atan(((C - (A + hypot(B, (A - C)))) / B));
	end
	tmp_2 = tmp;
end

Wolfram

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[A, -108.0], N[(180.0 * N[(N[ArcTan[N[(0.5 * N[(N[(B / A), $MachinePrecision] * N[(N[(C / A), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(N[(180.0 / Pi), $MachinePrecision] * N[ArcTan[N[(N[(C - N[(A + N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if A < -108

   1. Initial program 25.5%

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

   2. Simplified 56.6%

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

      [Start] 25.5	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      associate-*l/ [=>] 25.5	\[ 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}{B}\right)}}{\pi} \]
      *-lft-identity [=>] 25.5	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}}}{B}\right)}{\pi} \]
      +-commutative [=>] 25.5	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}}{B}\right)}{\pi} \]
      unpow2 [=>] 25.5	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}}{B}\right)}{\pi} \]
      unpow2 [=>] 25.5	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}}{B}\right)}{\pi} \]
      hypot-def [=>] 56.6	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}}{B}\right)}{\pi} \]

   3. Taylor expanded in A around -inf 64.3%

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

   4. Simplified 64.3%

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

      [Start] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{C \cdot B}{{A}^{2}} + 0.5 \cdot \frac{B}{A}\right)}{\pi} \]
      distribute-lft-out [=>] 64.3	\[ 180 \cdot \frac{\tan^{-1} \color{blue}{\left(0.5 \cdot \left(\frac{C \cdot B}{{A}^{2}} + \frac{B}{A}\right)\right)}}{\pi} \]
      *-commutative [=>] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{\color{blue}{B \cdot C}}{{A}^{2}} + \frac{B}{A}\right)\right)}{\pi} \]
      unpow2 [=>] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{B \cdot C}{\color{blue}{A \cdot A}} + \frac{B}{A}\right)\right)}{\pi} \]

   5. Taylor expanded in C around 0 64.3%

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

   6. Simplified 68.4%

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

      [Start] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{C \cdot B}{{A}^{2}} + \frac{B}{A}\right)\right)}{\pi} \]
      *-lft-identity [<=] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{C \cdot B}{{A}^{2}} + \color{blue}{1 \cdot \frac{B}{A}}\right)\right)}{\pi} \]
      metadata-eval [<=] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{C \cdot B}{{A}^{2}} + \color{blue}{\left(--1\right)} \cdot \frac{B}{A}\right)\right)}{\pi} \]
      cancel-sign-sub-inv [<=] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \color{blue}{\left(\frac{C \cdot B}{{A}^{2}} - -1 \cdot \frac{B}{A}\right)}\right)}{\pi} \]
      unpow2 [=>] 64.3	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{C \cdot B}{\color{blue}{A \cdot A}} - -1 \cdot \frac{B}{A}\right)\right)}{\pi} \]
      times-frac [=>] 68.4	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\color{blue}{\frac{C}{A} \cdot \frac{B}{A}} - -1 \cdot \frac{B}{A}\right)\right)}{\pi} \]
      distribute-rgt-out-- [=>] 68.4	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \color{blue}{\left(\frac{B}{A} \cdot \left(\frac{C}{A} - -1\right)\right)}\right)}{\pi} \]

   if -108 < A

   1. Initial program 63.5%

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

   2. Simplified 84.8%

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

      [Start] 63.5	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      associate-*r/ [=>] 63.5	\[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]
      associate-*l/ [<=] 63.5	\[ \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)} \]
      associate-*l/ [=>] 63.5	\[ \frac{180}{\pi} \cdot \tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}{B}\right)} \]

3. Recombined 2 regimes into one program.

4. Final simplification 80.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -108:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} \cdot \left(\frac{C}{A} + 1\right)\right)\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	80.7%
Cost	20164

\[\begin{array}{l} \mathbf{if}\;A \leq -98:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} \cdot \left(\frac{C}{A} + 1\right)\right)\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \end{array} \]

Alternative 2
Accuracy	46.7%
Cost	14368

\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(\frac{A \cdot -2}{B}\right)}{\pi}\\ t_1 := \frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{if}\;B \leq -5.2 \cdot 10^{+15}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -8.5 \cdot 10^{-84}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \mathbf{elif}\;B \leq -1.5 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq -6.6 \cdot 10^{-245}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq -5.3 \cdot 10^{-297}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 7.5 \cdot 10^{-275}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq 2.45 \cdot 10^{-178}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 1.8 \cdot 10^{-17}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{-2}{\frac{B}{A}}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 3
Accuracy	59.4%
Cost	14036

\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{if}\;A \leq -0.0004:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;A \leq -2 \cdot 10^{-244}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq 5.2 \cdot 10^{-184}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq 2.9 \cdot 10^{-86}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq 1.5 \cdot 10^{+44}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{\left(-B\right) - A}{B}\right)}}\\ \end{array} \]

Alternative 4
Accuracy	46.8%
Cost	13972

\[\begin{array}{l} \mathbf{if}\;B \leq -5.5 \cdot 10^{+15}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -3.6 \cdot 10^{-85}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\ \mathbf{elif}\;B \leq -2.15 \cdot 10^{-203}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 3.1 \cdot 10^{-176}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{elif}\;B \leq 8.4 \cdot 10^{-19}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A \cdot -2}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 5
Accuracy	57.6%
Cost	13972

\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{if}\;A \leq -0.0105:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;A \leq -5 \cdot 10^{-244}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq 6.1 \cdot 10^{-183}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;A \leq 5.8 \cdot 10^{-83}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq 4.5 \cdot 10^{+43}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A \cdot -2}{B}\right)}{\pi}\\ \end{array} \]

Alternative 6
Accuracy	61.1%
Cost	13968

\[\begin{array}{l} \mathbf{if}\;B \leq -2.55 \cdot 10^{-155}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{elif}\;B \leq -5.2 \cdot 10^{-205}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} \cdot \left(\frac{C}{A} + 1\right)\right)\right)}{\pi}\\ \mathbf{elif}\;B \leq -1 \cdot 10^{-244}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A \cdot -2}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq -6 \cdot 10^{-297}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-1 + \frac{C - A}{B}\right)}}\\ \end{array} \]

Alternative 7
Accuracy	46.8%
Cost	13840

\[\begin{array}{l} \mathbf{if}\;B \leq -1.7 \cdot 10^{+16}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -3.6 \cdot 10^{-202}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 1.1 \cdot 10^{-176}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{elif}\;B \leq 2.25 \cdot 10^{-18}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A \cdot -2}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 8
Accuracy	61.0%
Cost	13704

\[\begin{array}{l} \mathbf{if}\;B \leq -1.25 \cdot 10^{-238}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{elif}\;B \leq -7.2 \cdot 10^{-297}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(A + B\right)}{B}\right)\\ \end{array} \]

Alternative 9
Accuracy	61.0%
Cost	13704

\[\begin{array}{l} \mathbf{if}\;B \leq -5.6 \cdot 10^{-239}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{elif}\;B \leq -7.2 \cdot 10^{-297}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(-1 + \frac{C - A}{B}\right)}}\\ \end{array} \]

Alternative 10
Accuracy	45.2%
Cost	13580

\[\begin{array}{l} \mathbf{if}\;B \leq -1.8 \cdot 10^{+16}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -6.8 \cdot 10^{-203}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{elif}\;B \leq 3.4 \cdot 10^{-174}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 11
Accuracy	56.7%
Cost	13576

\[\begin{array}{l} \mathbf{if}\;A \leq -0.00041:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;A \leq 9.5 \cdot 10^{+27}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{A \cdot -2}{B}\right)}{\pi}\\ \end{array} \]

Alternative 12
Accuracy	44.8%
Cost	13448

\[\begin{array}{l} \mathbf{if}\;B \leq -2.1 \cdot 10^{-157}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 1.55 \cdot 10^{-174}:\\ \;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{0}{B}\right)}}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 13
Accuracy	40.3%
Cost	13188

\[\begin{array}{l} \mathbf{if}\;B \leq -4.6 \cdot 10^{-308}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 14
Accuracy	20.7%
Cost	13056

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

Reproduce

herbie shell --seed 2023136 
(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)))