?

Average Accuracy: 54.0% → 80.8%
Time: 22.5s
Precision: binary64
Cost: 20164

?

\[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
↓
\[\begin{array}{l} \mathbf{if}\;A \leq -1.1 \cdot 10^{+203}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \end{array} \]
(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 -1.1e+203)
   (* (/ 180.0 PI) (atan (* 0.5 (/ B A))))
   (* 180.0 (/ (atan (/ (- (- C A) (hypot B (- A C))) B)) 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));
}
↓
double code(double A, double B, double C) {
	double tmp;
	if (A <= -1.1e+203) {
		tmp = (180.0 / ((double) M_PI)) * atan((0.5 * (B / A)));
	} 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) {
	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 <= -1.1e+203) {
		tmp = (180.0 / Math.PI) * Math.atan((0.5 * (B / A)));
	} else {
		tmp = 180.0 * (Math.atan((((C - A) - Math.hypot(B, (A - C))) / B)) / Math.PI);
	}
	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 A <= -1.1e+203:
		tmp = (180.0 / math.pi) * math.atan((0.5 * (B / A)))
	else:
		tmp = 180.0 * (math.atan((((C - A) - math.hypot(B, (A - C))) / B)) / math.pi)
	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 (A <= -1.1e+203)
		tmp = Float64(Float64(180.0 / pi) * atan(Float64(0.5 * Float64(B / A))));
	else
		tmp = Float64(180.0 * Float64(atan(Float64(Float64(Float64(C - A) - hypot(B, Float64(A - C))) / B)) / pi));
	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 (A <= -1.1e+203)
		tmp = (180.0 / pi) * atan((0.5 * (B / A)));
	else
		tmp = 180.0 * (atan((((C - A) - hypot(B, (A - C))) / B)) / pi);
	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[A, -1.1e+203], N[(N[(180.0 / Pi), $MachinePrecision] * N[ArcTan[N[(0.5 * N[(B / A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 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]]

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 -1.1 \cdot 10^{+203}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\

\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}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Split input into 2 regimes

  2. if A < -1.10000000000000002e203

    1. Initial program 9.5%

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

    2. Simplified18.1%

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

      [Start]9.5

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

      associate-*r/ [=>]9.5

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

      associate-*l/ [<=]9.5

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

      associate-*l/ [=>]9.5

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

    3. Taylor expanded in A around -inf 86.6%

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

    if -1.10000000000000002e203 < A

    1. Initial program 58.3%

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

    2. Simplified80.3%

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

      [Start]58.3

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

      associate-*l/ [=>]58.3

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

      *-lft-identity [=>]58.3

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

      +-commutative [=>]58.3

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

      unpow2 [=>]58.3

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

      unpow2 [=>]58.3

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

      hypot-def [=>]80.3

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

  4. Final simplification80.8%

    \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -1.1 \cdot 10^{+203}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy58.5%
Cost14629
\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{\left(B + C\right) - A}{B}\right)\\ \mathbf{if}\;C \leq -9:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq -1.6 \cdot 10^{-75}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq -4.8 \cdot 10^{-152}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{elif}\;C \leq -2.25 \cdot 10^{-215}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 2.1 \cdot 10^{-245}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 10^{-133}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 2.05 \cdot 10^{-71}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;C \leq 2.8 \cdot 10^{-5} \lor \neg \left(C \leq 2.4 \cdot 10^{+52}\right):\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 2
Accuracy58.5%
Cost14629
\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\ t_1 := \tan^{-1} \left(\frac{\left(B + C\right) - A}{B}\right)\\ t_2 := \frac{180}{\pi} \cdot t_1\\ \mathbf{if}\;C \leq -11.2:\\ \;\;\;\;\frac{180}{\frac{\pi}{t_1}}\\ \mathbf{elif}\;C \leq -1.7 \cdot 10^{-75}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq -4.6 \cdot 10^{-152}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{elif}\;C \leq -2.5 \cdot 10^{-215}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 1.05 \cdot 10^{-245}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;C \leq 3.3 \cdot 10^{-131}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 2.65 \cdot 10^{-71}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;C \leq 8.6 \cdot 10^{-8} \lor \neg \left(C \leq 8.8 \cdot 10^{+54}\right):\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]
Alternative 3
Accuracy58.2%
Cost14628
\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\ t_1 := \tan^{-1} \left(\frac{\left(B + C\right) - A}{B}\right)\\ t_2 := \frac{180}{\pi} \cdot t_1\\ \mathbf{if}\;C \leq -13.5:\\ \;\;\;\;\frac{180}{\frac{\pi}{t_1}}\\ \mathbf{elif}\;C \leq -1.2 \cdot 10^{-76}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;C \leq -3.2 \cdot 10^{-152}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{elif}\;C \leq -2.2 \cdot 10^{-215}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 10^{-245}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;C \leq 2.5 \cdot 10^{-132}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 4.05 \cdot 10^{-70}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;C \leq 0.245:\\ \;\;\;\;\frac{\frac{\tan^{-1} \left(\frac{B \cdot \frac{B}{C}}{\frac{B}{-0.5}}\right)}{0.005555555555555556}}{\pi}\\ \mathbf{elif}\;C \leq 2.2 \cdot 10^{+52}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ \end{array} \]
Alternative 4
Accuracy58.9%
Cost14628
\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\ t_1 := \tan^{-1} \left(\frac{\left(B + C\right) - A}{B}\right)\\ t_2 := \frac{180}{\pi} \cdot t_1\\ \mathbf{if}\;C \leq -18:\\ \;\;\;\;\frac{180}{\frac{\pi}{t_1}}\\ \mathbf{elif}\;C \leq -1.2 \cdot 10^{-77}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + \frac{C}{\frac{A \cdot A}{B}}\right)\right)\\ \mathbf{elif}\;C \leq -5.5 \cdot 10^{-152}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{elif}\;C \leq -2.5 \cdot 10^{-215}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 1.75 \cdot 10^{-245}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;C \leq 4.5 \cdot 10^{-132}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;C \leq 4.6 \cdot 10^{-70}:\\ \;\;\;\;t_2\\ \mathbf{elif}\;C \leq 0.13:\\ \;\;\;\;\frac{\frac{\tan^{-1} \left(\frac{B \cdot \frac{B}{C}}{\frac{B}{-0.5}}\right)}{0.005555555555555556}}{\pi}\\ \mathbf{elif}\;C \leq 1.9 \cdot 10^{+53}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ \end{array} \]
Alternative 5
Accuracy45.8%
Cost14104
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ t_1 := 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{if}\;B \leq -1.1 \cdot 10^{+21}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -1.35 \cdot 10^{-188}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq -2.1 \cdot 10^{-228}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 5.4 \cdot 10^{-299}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq 3.65 \cdot 10^{-109}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\ \mathbf{elif}\;B \leq 2.4 \cdot 10^{+33}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 6
Accuracy45.8%
Cost14104
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{if}\;B \leq -5.5 \cdot 10^{+26}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -7.2 \cdot 10^{-188}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq -8.8 \cdot 10^{-229}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 6.5 \cdot 10^{-299}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq 3.5 \cdot 10^{-109}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\ \mathbf{elif}\;B \leq 7 \cdot 10^{+31}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 7
Accuracy45.8%
Cost14104
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{if}\;B \leq -4.4 \cdot 10^{+26}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -9.5 \cdot 10^{-188}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq -1.4 \cdot 10^{-228}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 7 \cdot 10^{-299}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq 5.8 \cdot 10^{-113}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C \cdot 2}{B}\right)\\ \mathbf{elif}\;B \leq 1.35 \cdot 10^{+32}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 8
Accuracy61.0%
Cost13968
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{if}\;A \leq -3.1 \cdot 10^{+16}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;A \leq -2 \cdot 10^{-47}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq -5 \cdot 10^{-248}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{elif}\;A \leq 9.5 \cdot 10^{-69}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\ \end{array} \]
Alternative 9
Accuracy45.8%
Cost13840
\[\begin{array}{l} t_0 := 180 \cdot \frac{\tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)}{\pi}\\ \mathbf{if}\;B \leq -3.05 \cdot 10^{+29}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq -2.2 \cdot 10^{-222}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;B \leq 2.35 \cdot 10^{-113}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\ \mathbf{elif}\;B \leq 4 \cdot 10^{+33}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 10
Accuracy56.6%
Cost13840
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{if}\;A \leq -5.5 \cdot 10^{+15}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;A \leq -1.1 \cdot 10^{-188}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq -2.3 \cdot 10^{-218}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B}{C} \cdot -0.5\right)\\ \mathbf{elif}\;A \leq 9.6 \cdot 10^{+111}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{A \cdot -2}{B}\right)\\ \end{array} \]
Alternative 11
Accuracy57.2%
Cost13840
\[\begin{array}{l} t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B + C}{B}\right)\\ \mathbf{if}\;A \leq -3.6 \cdot 10^{+16}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(0.5 \cdot \frac{B}{A}\right)\\ \mathbf{elif}\;A \leq -6.5 \cdot 10^{-43}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;A \leq -9.8 \cdot 10^{-248}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\ \mathbf{elif}\;A \leq 9.6 \cdot 10^{+111}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{A \cdot -2}{B}\right)\\ \end{array} \]
Alternative 12
Accuracy45.8%
Cost13448
\[\begin{array}{l} \mathbf{if}\;B \leq -8.5 \cdot 10^{+28}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{elif}\;B \leq 2.9 \cdot 10^{+31}:\\ \;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 13
Accuracy39.0%
Cost13188
\[\begin{array}{l} \mathbf{if}\;B \leq -9.8 \cdot 10^{-296}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\ \end{array} \]
Alternative 14
Accuracy20.2%
Cost13056
\[\frac{180 \cdot \tan^{-1} -1}{\pi} \]

Error

Reproduce?

herbie shell --seed 2023137 
(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)))