Math FPCore C Java Python Julia MATLAB Wolfram TeX \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi}
\]
↓
\[\begin{array}{l}
\mathbf{if}\;C \leq 4.3 \cdot 10^{+80}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(B, A - C\right)}{B}\right)}{\pi}\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\end{array}
\]
(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}
Alternatives Alternative 1 Accuracy 75.1% Cost 20304
\[\begin{array}{l}
t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)\\
\mathbf{if}\;A \leq -4 \cdot 10^{+102}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + \frac{B}{A} \cdot \frac{C}{A}\right)\right)}{\pi}\\
\mathbf{elif}\;A \leq -6.3 \cdot 10^{+63}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;A \leq -1.02 \cdot 10^{-16}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + B \cdot \frac{\frac{C}{A}}{A}\right)\right)}{\pi}\\
\mathbf{elif}\;A \leq 2.6 \cdot 10^{+53}:\\
\;\;\;\;t_0\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\
\end{array}
\]
Alternative 2 Accuracy 74.8% Cost 20304
\[\begin{array}{l}
t_0 := \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)\\
\mathbf{if}\;A \leq -1.15 \cdot 10^{+97}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + \frac{B}{A} \cdot \frac{C}{A}\right)\right)}{\pi}\\
\mathbf{elif}\;A \leq -4.3 \cdot 10^{+67}:\\
\;\;\;\;\frac{180}{\pi} \cdot t_0\\
\mathbf{elif}\;A \leq -1.65 \cdot 10^{-16}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + B \cdot \frac{\frac{C}{A}}{A}\right)\right)}{\pi}\\
\mathbf{elif}\;A \leq 7 \cdot 10^{+52}:\\
\;\;\;\;\frac{180 \cdot t_0}{\pi}\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\
\end{array}
\]
Alternative 3 Accuracy 78.3% Cost 20104
\[\begin{array}{l}
\mathbf{if}\;C \leq -8 \cdot 10^{-95}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\
\mathbf{elif}\;C \leq 1.35 \cdot 10^{+79}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{\left(-A\right) - \mathsf{hypot}\left(A, B\right)}{B}\right)}{\pi}\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\end{array}
\]
Alternative 4 Accuracy 78.3% Cost 20104
\[\begin{array}{l}
\mathbf{if}\;C \leq -2.7 \cdot 10^{-95}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{C - \mathsf{hypot}\left(B, C\right)}{B}\right)}{\pi}\\
\mathbf{elif}\;C \leq 3.8 \cdot 10^{+79}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{\left(-A\right) - \mathsf{hypot}\left(A, B\right)}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\end{array}
\]
Alternative 5 Accuracy 57.0% Cost 14632
\[\begin{array}{l}
t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\
t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\
t_2 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\mathbf{if}\;A \leq -3.5 \cdot 10^{-18}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)\\
\mathbf{elif}\;A \leq -2.2 \cdot 10^{-161}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;A \leq -3 \cdot 10^{-191}:\\
\;\;\;\;\frac{180}{\pi} \cdot t_2\\
\mathbf{elif}\;A \leq -1 \cdot 10^{-228}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;A \leq -4 \cdot 10^{-305}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C - B}{B}\right)\\
\mathbf{elif}\;A \leq 1.75 \cdot 10^{-235}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;A \leq 3.1 \cdot 10^{-115}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;A \leq 1.8 \cdot 10^{-25}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;A \leq 1.6 \cdot 10^{-13}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;A \leq 5.5:\\
\;\;\;\;180 \cdot \frac{t_2}{\pi}\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\
\end{array}
\]
Alternative 6 Accuracy 59.4% Cost 14236
\[\begin{array}{l}
t_0 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\
t_1 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\mathbf{if}\;A \leq -8.8 \cdot 10^{-17}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{B \cdot 0.5}{A}\right)\\
\mathbf{elif}\;A \leq -3.1 \cdot 10^{-164}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;A \leq -1.22 \cdot 10^{-191}:\\
\;\;\;\;\frac{180}{\pi} \cdot t_1\\
\mathbf{elif}\;A \leq -1.65 \cdot 10^{-233}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;A \leq 1.5 \cdot 10^{-13}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\
\mathbf{elif}\;A \leq 2.9:\\
\;\;\;\;180 \cdot \frac{t_1}{\pi}\\
\mathbf{elif}\;A \leq 2.2 \cdot 10^{+54}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\
\end{array}
\]
Alternative 7 Accuracy 59.8% Cost 14236
\[\begin{array}{l}
t_0 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\
\mathbf{if}\;A \leq -1.9 \cdot 10^{-18}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(0.5 \cdot \left(\frac{B}{A} + \frac{B}{A} \cdot \frac{C}{A}\right)\right)}{\pi}\\
\mathbf{elif}\;A \leq -3.4 \cdot 10^{-165}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;A \leq -2.35 \cdot 10^{-191}:\\
\;\;\;\;\frac{180}{\pi} \cdot t_0\\
\mathbf{elif}\;A \leq -8.2 \cdot 10^{-230}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;A \leq 1.35 \cdot 10^{-13}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + -1\right)}{\pi}\\
\mathbf{elif}\;A \leq 3.5:\\
\;\;\;\;180 \cdot \frac{t_0}{\pi}\\
\mathbf{elif}\;A \leq 1.85 \cdot 10^{+49}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 - \frac{A}{B}\right)\\
\end{array}
\]
Alternative 8 Accuracy 52.1% Cost 14104
\[\begin{array}{l}
t_0 := \frac{180 \cdot \tan^{-1} -1}{\pi}\\
t_1 := \frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\
t_2 := \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\mathbf{if}\;C \leq 1.75 \cdot 10^{-258}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;C \leq 2.45 \cdot 10^{-131}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;C \leq 3.15 \cdot 10^{-118}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;C \leq 1.3 \cdot 10^{-107}:\\
\;\;\;\;180 \cdot \frac{t_2}{\pi}\\
\mathbf{elif}\;C \leq 2.2 \cdot 10^{-53}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;C \leq 18500000:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{-A}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot t_2\\
\end{array}
\]
Alternative 9 Accuracy 47.1% Cost 13580
\[\begin{array}{l}
\mathbf{if}\;B \leq -2.2 \cdot 10^{-129}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\
\mathbf{elif}\;B \leq 1.85 \cdot 10^{-237}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{-A}{B}\right)\\
\mathbf{elif}\;B \leq 1.85 \cdot 10^{-125}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\
\end{array}
\]
Alternative 10 Accuracy 47.1% Cost 13576
\[\begin{array}{l}
\mathbf{if}\;C \leq -3.5 \cdot 10^{-139}:\\
\;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C}{B}\right)}}\\
\mathbf{elif}\;C \leq 3.2 \cdot 10^{-20}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\
\mathbf{else}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)}{\pi}\\
\end{array}
\]
Alternative 11 Accuracy 47.2% Cost 13576
\[\begin{array}{l}
\mathbf{if}\;C \leq -1.1 \cdot 10^{-138}:\\
\;\;\;\;\frac{180}{\frac{\pi}{\tan^{-1} \left(\frac{C}{B}\right)}}\\
\mathbf{elif}\;C \leq 2.4 \cdot 10^{-20}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\end{array}
\]
Alternative 12 Accuracy 59.5% Cost 13576
\[\begin{array}{l}
\mathbf{if}\;C \leq -1.85 \cdot 10^{-139}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(1 + \frac{C}{B}\right)\\
\mathbf{elif}\;C \leq 2.15 \cdot 10^{+48}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\end{array}
\]
Alternative 13 Accuracy 61.4% Cost 13576
\[\begin{array}{l}
\mathbf{if}\;C \leq 5 \cdot 10^{-270}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi}\\
\mathbf{elif}\;C \leq 1.1 \cdot 10^{+46}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-1 - \frac{A}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(-0.5 \cdot \frac{B}{C}\right)\\
\end{array}
\]
Alternative 14 Accuracy 47.4% Cost 13448
\[\begin{array}{l}
\mathbf{if}\;B \leq -8.6 \cdot 10^{-65}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\
\mathbf{elif}\;B \leq 2 \cdot 10^{-125}:\\
\;\;\;\;\frac{180}{\pi} \cdot \tan^{-1} \left(\frac{C}{B}\right)\\
\mathbf{else}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\
\end{array}
\]
Alternative 15 Accuracy 40.5% Cost 13188
\[\begin{array}{l}
\mathbf{if}\;B \leq -5 \cdot 10^{-310}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} 1}{\pi}\\
\mathbf{else}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} -1}{\pi}\\
\end{array}
\]
Alternative 16 Accuracy 21.9% Cost 13056
\[\frac{180 \cdot \tan^{-1} -1}{\pi}
\]