ABCF->ab-angle angle

Average Accuracy: 55.4% → 82.5%

Time: 21.1s

Precision: binary64

Cost: 60488

Math

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

↓

\[\begin{array}{l} t_0 := \frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\\ t_1 := \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)\\ \mathbf{if}\;t_0 \leq -0.5:\\ \;\;\;\;\frac{1}{\frac{\pi \cdot 0.005555555555555556}{t_1}}\\ \mathbf{elif}\;t_0 \leq 0:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \mathbf{else}:\\ \;\;\;\;t_1 \cdot \frac{180}{\pi}\\ \end{array} \]

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
 (let* ((t_0
         (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))
        (t_1 (atan (/ (- (- C A) (hypot B (- A C))) B))))
   (if (<= t_0 -0.5)
     (/ 1.0 (/ (* PI 0.005555555555555556) t_1))
     (if (<= t_0 0.0)
       (* (atan (* 0.5 (/ B A))) (/ 180.0 PI))
       (* t_1 (/ 180.0 PI))))))

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 t_0 = (1.0 / B) * ((C - A) - sqrt((pow((A - C), 2.0) + pow(B, 2.0))));
	double t_1 = atan((((C - A) - hypot(B, (A - C))) / B));
	double tmp;
	if (t_0 <= -0.5) {
		tmp = 1.0 / ((((double) M_PI) * 0.005555555555555556) / t_1);
	} else if (t_0 <= 0.0) {
		tmp = atan((0.5 * (B / A))) * (180.0 / ((double) M_PI));
	} else {
		tmp = t_1 * (180.0 / ((double) M_PI));
	}
	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 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(B, (A - C))) / B));
	double tmp;
	if (t_0 <= -0.5) {
		tmp = 1.0 / ((Math.PI * 0.005555555555555556) / t_1);
	} else if (t_0 <= 0.0) {
		tmp = Math.atan((0.5 * (B / A))) * (180.0 / Math.PI);
	} else {
		tmp = t_1 * (180.0 / Math.PI);
	}
	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):
	t_0 = (1.0 / B) * ((C - A) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0))))
	t_1 = math.atan((((C - A) - math.hypot(B, (A - C))) / B))
	tmp = 0
	if t_0 <= -0.5:
		tmp = 1.0 / ((math.pi * 0.005555555555555556) / t_1)
	elif t_0 <= 0.0:
		tmp = math.atan((0.5 * (B / A))) * (180.0 / math.pi)
	else:
		tmp = t_1 * (180.0 / math.pi)
	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)
	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(B, Float64(A - C))) / B))
	tmp = 0.0
	if (t_0 <= -0.5)
		tmp = Float64(1.0 / Float64(Float64(pi * 0.005555555555555556) / t_1));
	elseif (t_0 <= 0.0)
		tmp = Float64(atan(Float64(0.5 * Float64(B / A))) * Float64(180.0 / pi));
	else
		tmp = Float64(t_1 * Float64(180.0 / pi));
	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)
	t_0 = (1.0 / B) * ((C - A) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0))));
	t_1 = atan((((C - A) - hypot(B, (A - C))) / B));
	tmp = 0.0;
	if (t_0 <= -0.5)
		tmp = 1.0 / ((pi * 0.005555555555555556) / t_1);
	elseif (t_0 <= 0.0)
		tmp = atan((0.5 * (B / A))) * (180.0 / pi);
	else
		tmp = t_1 * (180.0 / pi);
	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_] := 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[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[t$95$0, -0.5], N[(1.0 / N[(N[(Pi * 0.005555555555555556), $MachinePrecision] / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, 0.0], N[(N[ArcTan[N[(0.5 * N[(B / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(180.0 / Pi), $MachinePrecision]), $MachinePrecision], N[(t$95$1 * N[(180.0 / Pi), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (*.f64 (/.f64 1 B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))) < -0.5

   1. Initial program 60.9%

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

   2. Simplified 82.5%

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

      [Start] 60.9	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      associate-*r/ [=>] 60.9	\[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]
      associate-*l/ [<=] 60.9	\[ \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)} \]
      associate-*l/ [=>] 60.9	\[ \frac{180}{\pi} \cdot \tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}{B}\right)} \]

   3. Applied egg-rr 87.6%

      \[\leadsto \color{blue}{\frac{1}{\frac{\pi \cdot 0.005555555555555556}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}}} \]
      Proof

      [Start] 82.5	\[ \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right) \]
      clear-num [=>] 82.5	\[ \color{blue}{\frac{1}{\frac{\pi}{180}}} \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right) \]
      associate-*l/ [=>] 82.5	\[ \color{blue}{\frac{1 \cdot \tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}{\frac{\pi}{180}}} \]
      associate-/l* [=>] 82.5	\[ \color{blue}{\frac{1}{\frac{\frac{\pi}{180}}{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}}} \]
      div-inv [=>] 82.5	\[ \frac{1}{\frac{\color{blue}{\pi \cdot \frac{1}{180}}}{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}} \]
      metadata-eval [=>] 82.5	\[ \frac{1}{\frac{\pi \cdot \color{blue}{0.005555555555555556}}{\tan^{-1} \left(\frac{C - \left(A + \mathsf{hypot}\left(B, A - C\right)\right)}{B}\right)}} \]
      associate--r+ [=>] 87.6	\[ \frac{1}{\frac{\pi \cdot 0.005555555555555556}{\tan^{-1} \left(\frac{\color{blue}{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}}{B}\right)}} \]

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

   1. Initial program 21.7%

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

   2. Simplified 21.7%

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

      [Start] 21.7	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      associate-*r/ [=>] 21.7	\[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]
      sub-neg [=>] 21.7	\[ \frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(\left(C - A\right) + \left(-\sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}\right)}{\pi} \]
      sub-neg [<=] 21.7	\[ \frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \color{blue}{\left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}\right)}{\pi} \]
      unpow2 [=>] 21.7	\[ \frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + \color{blue}{B \cdot B}}\right)\right)}{\pi} \]

   3. Taylor expanded in A around -inf 52.2%

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

   4. Taylor expanded in B around 0 52.2%

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

   5. Simplified 52.3%

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

      [Start] 52.2	\[ 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi} \]
      associate-*r/ [=>] 52.2	\[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}} \]
      associate-*r/ [=>] 52.2	\[ \frac{180 \cdot \tan^{-1} \color{blue}{\left(\frac{0.5 \cdot B}{A}\right)}}{\pi} \]
      *-commutative [=>] 52.2	\[ \frac{180 \cdot \tan^{-1} \left(\frac{\color{blue}{B \cdot 0.5}}{A}\right)}{\pi} \]
      associate-*r/ [<=] 52.2	\[ \frac{180 \cdot \tan^{-1} \color{blue}{\left(B \cdot \frac{0.5}{A}\right)}}{\pi} \]
      associate-*l/ [<=] 52.2	\[ \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(B \cdot \frac{0.5}{A}\right)} \]
      *-commutative [=>] 52.2	\[ \color{blue}{\tan^{-1} \left(B \cdot \frac{0.5}{A}\right) \cdot \frac{180}{\pi}} \]
      associate-*r/ [=>] 52.3	\[ \tan^{-1} \color{blue}{\left(\frac{B \cdot 0.5}{A}\right)} \cdot \frac{180}{\pi} \]
      *-commutative [<=] 52.3	\[ \tan^{-1} \left(\frac{\color{blue}{0.5 \cdot B}}{A}\right) \cdot \frac{180}{\pi} \]
      associate-*r/ [<=] 52.3	\[ \tan^{-1} \color{blue}{\left(0.5 \cdot \frac{B}{A}\right)} \cdot \frac{180}{\pi} \]

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

   1. Initial program 61.1%

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

   2. Simplified 87.5%

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

      [Start] 61.1	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      associate-*r/ [=>] 61.0	\[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]
      associate-*l/ [<=] 61.1	\[ \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)} \]
      *-commutative [=>] 61.1	\[ \color{blue}{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right) \cdot \frac{180}{\pi}} \]
      associate-*l/ [=>] 61.1	\[ \tan^{-1} \color{blue}{\left(\frac{1 \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}{B}\right)} \cdot \frac{180}{\pi} \]
      *-lft-identity [=>] 61.1	\[ \tan^{-1} \left(\frac{\color{blue}{\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}}}{B}\right) \cdot \frac{180}{\pi} \]
      +-commutative [=>] 61.1	\[ \tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{{B}^{2} + {\left(A - C\right)}^{2}}}}{B}\right) \cdot \frac{180}{\pi} \]
      unpow2 [=>] 61.1	\[ \tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\color{blue}{B \cdot B} + {\left(A - C\right)}^{2}}}{B}\right) \cdot \frac{180}{\pi} \]
      unpow2 [=>] 61.1	\[ \tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{B \cdot B + \color{blue}{\left(A - C\right) \cdot \left(A - C\right)}}}{B}\right) \cdot \frac{180}{\pi} \]
      hypot-def [=>] 87.5	\[ \tan^{-1} \left(\frac{\left(C - A\right) - \color{blue}{\mathsf{hypot}\left(B, A - C\right)}}{B}\right) \cdot \frac{180}{\pi} \]

3. Recombined 3 regimes into one program.

4. Final simplification 82.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right) \leq -0.5:\\ \;\;\;\;\frac{1}{\frac{\pi \cdot 0.005555555555555556}{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}}\\ \mathbf{elif}\;\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right) \leq 0:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right) \cdot \frac{180}{\pi}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	80.9%
Cost	20164

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

Alternative 2
Accuracy	57.3%
Cost	13965

\[\begin{array}{l} \mathbf{if}\;B \leq -2.3 \cdot 10^{-13}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 1.55 \cdot 10^{+100} \lor \neg \left(B \leq 1.58 \cdot 10^{+171}\right):\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{1}{\frac{B}{C - \left(B + A\right)}}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \end{array} \]

Alternative 3
Accuracy	44.2%
Cost	13841

\[\begin{array}{l} \mathbf{if}\;B \leq -2.3 \cdot 10^{-22}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 2.1 \cdot 10^{-144}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 6.5 \cdot 10^{+92} \lor \neg \left(B \leq 1.58 \cdot 10^{+171}\right):\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(B \cdot \frac{0.5}{A}\right)}{\pi}\\ \end{array} \]

Alternative 4
Accuracy	44.2%
Cost	13841

\[\begin{array}{l} \mathbf{if}\;B \leq -2.4 \cdot 10^{-22}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 1.85 \cdot 10^{-144}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 6.6 \cdot 10^{+98} \lor \neg \left(B \leq 1.58 \cdot 10^{+171}\right):\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \end{array} \]

Alternative 5
Accuracy	44.3%
Cost	13841

\[\begin{array}{l} \mathbf{if}\;B \leq -2.8 \cdot 10^{-22}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 2.1 \cdot 10^{-144}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C \cdot 2}{B}\right)\\ \mathbf{elif}\;B \leq 1.36 \cdot 10^{+100} \lor \neg \left(B \leq 1.58 \cdot 10^{+171}\right):\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \end{array} \]

Alternative 6
Accuracy	59.7%
Cost	13704

\[\begin{array}{l} \mathbf{if}\;A \leq -5 \cdot 10^{+30}:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \mathbf{elif}\;A \leq 8.6 \cdot 10^{-157}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{1}{\frac{B}{C - B}}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{\left(-A\right) - B}{B}\right)}{\pi}\\ \end{array} \]

Alternative 7
Accuracy	59.7%
Cost	13640

\[\begin{array}{l} \mathbf{if}\;A \leq -6.2 \cdot 10^{+32}:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \mathbf{elif}\;A \leq 9.5 \cdot 10^{-157}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{\left(-A\right) - B}{B}\right)}{\pi}\\ \end{array} \]

Alternative 8
Accuracy	57.3%
Cost	13576

\[\begin{array}{l} \mathbf{if}\;A \leq -2.7 \cdot 10^{+31}:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \mathbf{elif}\;A \leq 5.8 \cdot 10^{+69}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{-A}{B}\right)}{\pi}\\ \end{array} \]

Alternative 9
Accuracy	57.3%
Cost	13576

\[\begin{array}{l} \mathbf{if}\;A \leq -3.8 \cdot 10^{+30}:\\ \;\;\;\;\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right) \cdot \frac{180}{\pi}\\ \mathbf{elif}\;A \leq 1.15 \cdot 10^{+70}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{A \cdot -2}{B}\right)}{\pi}\\ \end{array} \]

Alternative 10
Accuracy	44.5%
Cost	13448

\[\begin{array}{l} \mathbf{if}\;B \leq -2.7 \cdot 10^{-142}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 8.2 \cdot 10^{-165}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{0}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 11
Accuracy	46.0%
Cost	13448

\[\begin{array}{l} \mathbf{if}\;B \leq -3.1 \cdot 10^{-22}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 2.1 \cdot 10^{-144}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 12
Accuracy	39.5%
Cost	13188

\[\begin{array}{l} \mathbf{if}\;B \leq -4 \cdot 10^{-310}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]

Alternative 13
Accuracy	20.5%
Cost	13056

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

Reproduce

herbie shell --seed 2023138 
(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)))