ABCF->ab-angle angle

Percentage Accurate: 53.9% → 88.1%

Time: 21.5s

Precision: binary64

Cost: 60489

Math

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

↓

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

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
 (let* ((t_0
         (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
   (if (or (<= t_0 -0.5) (not (<= t_0 0.0)))
     (* (atan (/ (- (- C A) (hypot B (- C A))) B)) (/ 180.0 PI))
     (* (/ 180.0 PI) (atan (/ (* B -0.5) (- C A)))))))

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 tmp;
	if ((t_0 <= -0.5) || !(t_0 <= 0.0)) {
		tmp = atan((((C - A) - hypot(B, (C - A))) / B)) * (180.0 / ((double) M_PI));
	} else {
		tmp = (180.0 / ((double) M_PI)) * atan(((B * -0.5) / (C - A)));
	}
	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 tmp;
	if ((t_0 <= -0.5) || !(t_0 <= 0.0)) {
		tmp = Math.atan((((C - A) - Math.hypot(B, (C - A))) / B)) * (180.0 / Math.PI);
	} else {
		tmp = (180.0 / Math.PI) * Math.atan(((B * -0.5) / (C - A)));
	}
	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))))
	tmp = 0
	if (t_0 <= -0.5) or not (t_0 <= 0.0):
		tmp = math.atan((((C - A) - math.hypot(B, (C - A))) / B)) * (180.0 / math.pi)
	else:
		tmp = (180.0 / math.pi) * math.atan(((B * -0.5) / (C - A)))
	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)))))
	tmp = 0.0
	if ((t_0 <= -0.5) || !(t_0 <= 0.0))
		tmp = Float64(atan(Float64(Float64(Float64(C - A) - hypot(B, Float64(C - A))) / B)) * Float64(180.0 / pi));
	else
		tmp = Float64(Float64(180.0 / pi) * atan(Float64(Float64(B * -0.5) / Float64(C - A))));
	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))));
	tmp = 0.0;
	if ((t_0 <= -0.5) || ~((t_0 <= 0.0)))
		tmp = atan((((C - A) - hypot(B, (C - A))) / B)) * (180.0 / pi);
	else
		tmp = (180.0 / pi) * atan(((B * -0.5) / (C - A)));
	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]}, If[Or[LessEqual[t$95$0, -0.5], N[Not[LessEqual[t$95$0, 0.0]], $MachinePrecision]], N[(N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[B ^ 2 + N[(C - A), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision] * N[(180.0 / Pi), $MachinePrecision]), $MachinePrecision], N[(N[(180.0 / Pi), $MachinePrecision] * N[ArcTan[N[(N[(B * -0.5), $MachinePrecision] / N[(C - A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 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 or 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 60.3%

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

   2. Simplified 87.3%

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

      [Start] 60.3%	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      associate-*r/ [=>] 60.3%	\[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]
      associate-*l/ [<=] 60.4%	\[ \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)} \]
      *-commutative [=>] 60.4%	\[ \color{blue}{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right) \cdot \frac{180}{\pi}} \]

   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 20.4%

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

   2. Simplified 20.4%

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

      [Start] 20.4%	\[ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      associate-*r/ [=>] 20.4%	\[ \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}} \]
      associate-*l/ [<=] 20.4%	\[ \color{blue}{\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)} \]
      *-commutative [=>] 20.4%	\[ \color{blue}{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right) \cdot \frac{180}{\pi}} \]

   3. Taylor expanded in B around 0 99.1%

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

   4. Simplified 99.1%

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

      [Start] 99.1%	\[ \tan^{-1} \left(-0.5 \cdot \frac{B}{C - A}\right) \cdot \frac{180}{\pi} \]
      associate-*r/ [=>] 99.1%	\[ \tan^{-1} \color{blue}{\left(\frac{-0.5 \cdot B}{C - A}\right)} \cdot \frac{180}{\pi} \]

3. Recombined 2 regimes into one program.

4. Final simplification 89.4%

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

Alternatives

Alternative 1
Accuracy	88.1%
Cost	60489

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

Alternative 2
Accuracy	76.7%
Cost	20236

\[\begin{array}{l} \mathbf{if}\;C \leq -3.6 \cdot 10^{-110}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)\\ \mathbf{elif}\;C \leq -3.4 \cdot 10^{-166}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 1050000000:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{\left(-A\right) - \mathsf{hypot}\left(A, B\right)}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C - A}\right)\\ \end{array} \]

Alternative 3
Accuracy	68.3%
Cost	20172

\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)\\ \mathbf{if}\;C \leq -3.6 \cdot 10^{-110}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq -3.4 \cdot 10^{-166}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 105000000:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C - A}\right)\\ \end{array} \]

Alternative 4
Accuracy	48.2%
Cost	14237

\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)}{\pi}\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} 1\\ \mathbf{if}\;C \leq -3800:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq -4.1 \cdot 10^{-275}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 1.3 \cdot 10^{-272}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 9.5 \cdot 10^{-168}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 5.4 \cdot 10^{-115}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \mathbf{elif}\;C \leq 3.8 \cdot 10^{-56} \lor \neg \left(C \leq 62000000\right):\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

Alternative 5
Accuracy	48.4%
Cost	14237

\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} 1\\ \mathbf{if}\;C \leq -2550:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq -3.1 \cdot 10^{-275}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 5.2 \cdot 10^{-272}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 10^{-168}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 3.2 \cdot 10^{-115}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \mathbf{elif}\;C \leq 1.52 \cdot 10^{-55} \lor \neg \left(C \leq 4500\right):\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

Alternative 6
Accuracy	48.3%
Cost	14237

\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} 1\\ \mathbf{if}\;C \leq -3500:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq -3.7 \cdot 10^{-275}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 1.12 \cdot 10^{-272}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 8.5 \cdot 10^{-172}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 1.9 \cdot 10^{-115}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \mathbf{elif}\;C \leq 5.2 \cdot 10^{-56} \lor \neg \left(C \leq 1400000\right):\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C}\right)\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

Alternative 7
Accuracy	48.5%
Cost	14237

\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} 1\\ \mathbf{if}\;C \leq -2400:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C \cdot 2}{B}\right)\\ \mathbf{elif}\;C \leq -4.5 \cdot 10^{-275}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 2.8 \cdot 10^{-272}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 6.2 \cdot 10^{-172}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 4.8 \cdot 10^{-116}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \mathbf{elif}\;C \leq 2.7 \cdot 10^{-55} \lor \neg \left(C \leq 700\right):\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C}\right)\\ \mathbf{else}:\\ \;\;\;\;t_1\\ \end{array} \]

Alternative 8
Accuracy	51.9%
Cost	14237

\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ t_1 := \frac{180 \cdot \tan^{-1} \left(\frac{B + C}{B}\right)}{\pi}\\ \mathbf{if}\;C \leq -1.65 \cdot 10^{-64}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq -6.2 \cdot 10^{-251}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 1.6 \cdot 10^{-271}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 1.1 \cdot 10^{-171}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 3.8 \cdot 10^{-116}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \mathbf{elif}\;C \leq 4.8 \cdot 10^{-56} \lor \neg \left(C \leq 700\right):\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} 1\\ \end{array} \]

Alternative 9
Accuracy	62.2%
Cost	13969

\[\begin{array}{l} \mathbf{if}\;B \leq -1.28 \cdot 10^{-54}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B + C}{B}\right)}{\pi}\\ \mathbf{elif}\;B \leq 9.6 \cdot 10^{-301} \lor \neg \left(B \leq 4.4 \cdot 10^{-163}\right) \land B \leq 4.5 \cdot 10^{-101}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C - A}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(B + A\right)}{B}\right)\\ \end{array} \]

Alternative 10
Accuracy	64.3%
Cost	13969

\[\begin{array}{l} \mathbf{if}\;B \leq -5 \cdot 10^{-53}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C + \left(B - A\right)}{B}\right)\\ \mathbf{elif}\;B \leq 9.2 \cdot 10^{-297} \lor \neg \left(B \leq 1.05 \cdot 10^{-162}\right) \land B \leq 2.7 \cdot 10^{-103}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C - A}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \left(B + A\right)}{B}\right)\\ \end{array} \]

Alternative 11
Accuracy	56.4%
Cost	13836

\[\begin{array}{l} \mathbf{if}\;C \leq -1.65 \cdot 10^{-64}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B + C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq -5.4 \cdot 10^{-275}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 2.15 \cdot 10^{-271}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B - A}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(B \cdot \frac{0.5}{A - C}\right)}{\pi}\\ \end{array} \]

Alternative 12
Accuracy	56.4%
Cost	13836

\[\begin{array}{l} \mathbf{if}\;C \leq -2.1 \cdot 10^{-64}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B + C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq -7.6 \cdot 10^{-275}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 2.15 \cdot 10^{-271}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B - A}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{B \cdot 0.5}{A - C}\right)}{\pi}\\ \end{array} \]

Alternative 13
Accuracy	56.5%
Cost	13836

\[\begin{array}{l} \mathbf{if}\;C \leq -2.7 \cdot 10^{-64}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B + C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq -5.8 \cdot 10^{-275}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 2.15 \cdot 10^{-271}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B - A}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C - A}\right)\\ \end{array} \]

Alternative 14
Accuracy	47.4%
Cost	13712

\[\begin{array}{l} t_0 := \frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{if}\;B \leq -7.5 \cdot 10^{-35}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} 1\\ \mathbf{elif}\;B \leq -4 \cdot 10^{-292}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq 5.3 \cdot 10^{-289}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{0}{B}\right)\\ \mathbf{elif}\;B \leq 4.2 \cdot 10^{-69}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \end{array} \]

Alternative 15
Accuracy	57.1%
Cost	13708

\[\begin{array}{l} \mathbf{if}\;C \leq -5.5 \cdot 10^{-64}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B + C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq -7.6 \cdot 10^{-275}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq 40000:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{B - A}{B}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot -0.5}{C}\right)\\ \end{array} \]

Alternative 16
Accuracy	48.0%
Cost	13576

\[\begin{array}{l} \mathbf{if}\;C \leq -1.75 \cdot 10^{-18}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C}{B}\right)}{\pi}\\ \mathbf{elif}\;C \leq 8.4 \cdot 10^{-116}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\ \end{array} \]

Alternative 17
Accuracy	44.5%
Cost	13448

\[\begin{array}{l} \mathbf{if}\;B \leq -1.8 \cdot 10^{-60}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} 1\\ \mathbf{elif}\;B \leq 1.2 \cdot 10^{-135}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{0}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \end{array} \]

Alternative 18
Accuracy	39.9%
Cost	13188

\[\begin{array}{l} \mathbf{if}\;B \leq -1.45 \cdot 10^{-301}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} 1\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} -1\\ \end{array} \]

Alternative 19
Accuracy	21.0%
Cost	13056

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

Reproduce

herbie shell --seed 2023178 
(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)))