ABCF->ab-angle angle

Percentage Accurate: 53.2% → 88.1%
Time: 14.2s
Alternatives: 14
Speedup: 0.6×

Specification

?
\[\begin{array}{l} \\ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \end{array} \]
(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)))
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));
}
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);
}
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)
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 tmp = code(A, B, C)
	tmp = 180.0 * (atan(((1.0 / B) * ((C - A) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / pi);
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]
\begin{array}{l}

\\
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}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 14 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 53.2% accurate, 1.0× speedup?

\[\begin{array}{l} \\ 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \end{array} \]
(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)))
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));
}
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);
}
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)
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 tmp = code(A, B, C)
	tmp = 180.0 * (atan(((1.0 / B) * ((C - A) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / pi);
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]
\begin{array}{l}

\\
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}
\end{array}

Alternative 1: 88.1% accurate, 0.5× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\\ \mathbf{if}\;t\_0 \leq -2 \cdot 10^{-51}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \mathsf{hypot}\left(C - A, B\right)\right)\right)}{\pi}\\ \mathbf{elif}\;t\_0 \leq 0:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C - A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}\\ \end{array} \end{array} \]
(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 (<= t_0 -2e-51)
     (* 180.0 (/ (atan (* (/ 1.0 B) (- (- C A) (hypot (- C A) B)))) PI))
     (if (<= t_0 0.0)
       (* 180.0 (/ (atan (* -0.5 (/ B (- C A)))) PI))
       (/ (* 180.0 (atan (/ (- (- C A) (hypot (- A C) B)) B))) 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 <= -2e-51) {
		tmp = 180.0 * (atan(((1.0 / B) * ((C - A) - hypot((C - A), B)))) / ((double) M_PI));
	} else if (t_0 <= 0.0) {
		tmp = 180.0 * (atan((-0.5 * (B / (C - A)))) / ((double) M_PI));
	} else {
		tmp = (180.0 * atan((((C - A) - hypot((A - C), B)) / B))) / ((double) M_PI);
	}
	return tmp;
}
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 <= -2e-51) {
		tmp = 180.0 * (Math.atan(((1.0 / B) * ((C - A) - Math.hypot((C - A), B)))) / Math.PI);
	} else if (t_0 <= 0.0) {
		tmp = 180.0 * (Math.atan((-0.5 * (B / (C - A)))) / Math.PI);
	} else {
		tmp = (180.0 * Math.atan((((C - A) - Math.hypot((A - C), B)) / B))) / Math.PI;
	}
	return tmp;
}
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 <= -2e-51:
		tmp = 180.0 * (math.atan(((1.0 / B) * ((C - A) - math.hypot((C - A), B)))) / math.pi)
	elif t_0 <= 0.0:
		tmp = 180.0 * (math.atan((-0.5 * (B / (C - A)))) / math.pi)
	else:
		tmp = (180.0 * math.atan((((C - A) - math.hypot((A - C), B)) / B))) / math.pi
	return tmp
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 <= -2e-51)
		tmp = Float64(180.0 * Float64(atan(Float64(Float64(1.0 / B) * Float64(Float64(C - A) - hypot(Float64(C - A), B)))) / pi));
	elseif (t_0 <= 0.0)
		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(B / Float64(C - A)))) / pi));
	else
		tmp = Float64(Float64(180.0 * atan(Float64(Float64(Float64(C - A) - hypot(Float64(A - C), B)) / B))) / pi);
	end
	return tmp
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 <= -2e-51)
		tmp = 180.0 * (atan(((1.0 / B) * ((C - A) - hypot((C - A), B)))) / pi);
	elseif (t_0 <= 0.0)
		tmp = 180.0 * (atan((-0.5 * (B / (C - A)))) / pi);
	else
		tmp = (180.0 * atan((((C - A) - hypot((A - C), B)) / B))) / pi;
	end
	tmp_2 = tmp;
end
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[LessEqual[t$95$0, -2e-51], N[(180.0 * N[(N[ArcTan[N[(N[(1.0 / B), $MachinePrecision] * N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(C - A), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, 0.0], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(B / N[(C - A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(N[(180.0 * N[ArcTan[N[(N[(N[(C - A), $MachinePrecision] - N[Sqrt[N[(A - C), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision] / B), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / Pi), $MachinePrecision]]]]
\begin{array}{l}

\\
\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 -2 \cdot 10^{-51}:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \mathsf{hypot}\left(C - A, B\right)\right)\right)}{\pi}\\

\mathbf{elif}\;t\_0 \leq 0:\\
\;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C - A}\right)}{\pi}\\

\mathbf{else}:\\
\;\;\;\;\frac{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi}\\


\end{array}
\end{array}
Derivation
  1. Split input into 3 regimes
  2. if (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))) < -2e-51

    1. Initial program 58.4%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Step-by-step derivation
      1. lift-sqrt.f64N/A

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \color{blue}{\sqrt{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)}{\mathsf{PI}\left(\right)} \]
      2. lift-+.f64N/A

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{\color{blue}{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)}{\mathsf{PI}\left(\right)} \]
    4. Applied rewrites87.8%

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

    if -2e-51 < (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))) < -0.0

    1. Initial program 20.6%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Step-by-step derivation
      1. lift-sqrt.f64N/A

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \color{blue}{\sqrt{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)}{\mathsf{PI}\left(\right)} \]
      2. lift-+.f64N/A

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{\color{blue}{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)}{\mathsf{PI}\left(\right)} \]
    4. Applied rewrites20.6%

      \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \color{blue}{\mathsf{hypot}\left(C - A, B\right)}\right)\right)}{\pi} \]
    5. Taylor expanded in B around 0

      \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1}{2} \cdot \frac{B}{C - A}\right)}}{\mathsf{PI}\left(\right)} \]
    6. Step-by-step derivation
      1. lower-*.f64N/A

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1}{2} \cdot \frac{B}{C - A}\right)}}{\mathsf{PI}\left(\right)} \]
      2. lower-/.f64N/A

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1}{2} \cdot \color{blue}{\frac{B}{C - A}}\right)}{\mathsf{PI}\left(\right)} \]
      3. lower--.f6498.7

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{\color{blue}{C - A}}\right)}{\pi} \]
    7. Applied rewrites98.7%

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

    if -0.0 < (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64))))))

    1. Initial program 55.2%

      \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
    2. Add Preprocessing
    3. Taylor expanded in B around 0

      \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{B}\right)}{\mathsf{PI}\left(\right)}} \]
    4. Applied rewrites55.3%

      \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\mathsf{fma}\left(B, B, \left(A - C\right) \cdot \left(A - C\right)\right)}}{B}\right)}{\pi}} \]
    5. Step-by-step derivation
      1. Applied rewrites81.3%

        \[\leadsto \frac{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \mathsf{hypot}\left(A - C, B\right)}{B}\right)}{\pi} \]
    6. Recombined 3 regimes into one program.
    7. Add Preprocessing

    Alternative 2: 78.4% accurate, 0.4× speedup?

    \[\begin{array}{l} \\ \begin{array}{l} t_0 := \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)\\ t_1 := \frac{C - A}{B}\\ \mathbf{if}\;t\_0 \leq -2 \cdot 10^{-51}:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(t\_1 + -1\right)}{\pi}\\ \mathbf{elif}\;t\_0 \leq 0:\\ \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C - A}\right)}{\pi}\\ \mathbf{else}:\\ \;\;\;\;\frac{180 \cdot \tan^{-1} \left(1 + t\_1\right)}{\pi}\\ \end{array} \end{array} \]
    (FPCore (A B C)
     :precision binary64
     (let* ((t_0
             (atan
              (* (/ 1.0 B) (- (- C A) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
            (t_1 (/ (- C A) B)))
       (if (<= t_0 -2e-51)
         (* 180.0 (/ (atan (+ t_1 -1.0)) PI))
         (if (<= t_0 0.0)
           (* 180.0 (/ (atan (* -0.5 (/ B (- C A)))) PI))
           (/ (* 180.0 (atan (+ 1.0 t_1))) PI)))))
    double code(double A, double B, double C) {
    	double t_0 = atan(((1.0 / B) * ((C - A) - sqrt((pow((A - C), 2.0) + pow(B, 2.0))))));
    	double t_1 = (C - A) / B;
    	double tmp;
    	if (t_0 <= -2e-51) {
    		tmp = 180.0 * (atan((t_1 + -1.0)) / ((double) M_PI));
    	} else if (t_0 <= 0.0) {
    		tmp = 180.0 * (atan((-0.5 * (B / (C - A)))) / ((double) M_PI));
    	} else {
    		tmp = (180.0 * atan((1.0 + t_1))) / ((double) M_PI);
    	}
    	return tmp;
    }
    
    public static double code(double A, double B, double C) {
    	double t_0 = Math.atan(((1.0 / B) * ((C - A) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0))))));
    	double t_1 = (C - A) / B;
    	double tmp;
    	if (t_0 <= -2e-51) {
    		tmp = 180.0 * (Math.atan((t_1 + -1.0)) / Math.PI);
    	} else if (t_0 <= 0.0) {
    		tmp = 180.0 * (Math.atan((-0.5 * (B / (C - A)))) / Math.PI);
    	} else {
    		tmp = (180.0 * Math.atan((1.0 + t_1))) / Math.PI;
    	}
    	return tmp;
    }
    
    def code(A, B, C):
    	t_0 = math.atan(((1.0 / B) * ((C - A) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0))))))
    	t_1 = (C - A) / B
    	tmp = 0
    	if t_0 <= -2e-51:
    		tmp = 180.0 * (math.atan((t_1 + -1.0)) / math.pi)
    	elif t_0 <= 0.0:
    		tmp = 180.0 * (math.atan((-0.5 * (B / (C - A)))) / math.pi)
    	else:
    		tmp = (180.0 * math.atan((1.0 + t_1))) / math.pi
    	return tmp
    
    function code(A, B, C)
    	t_0 = atan(Float64(Float64(1.0 / B) * Float64(Float64(C - A) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))
    	t_1 = Float64(Float64(C - A) / B)
    	tmp = 0.0
    	if (t_0 <= -2e-51)
    		tmp = Float64(180.0 * Float64(atan(Float64(t_1 + -1.0)) / pi));
    	elseif (t_0 <= 0.0)
    		tmp = Float64(180.0 * Float64(atan(Float64(-0.5 * Float64(B / Float64(C - A)))) / pi));
    	else
    		tmp = Float64(Float64(180.0 * atan(Float64(1.0 + t_1))) / pi);
    	end
    	return tmp
    end
    
    function tmp_2 = code(A, B, C)
    	t_0 = atan(((1.0 / B) * ((C - A) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0))))));
    	t_1 = (C - A) / B;
    	tmp = 0.0;
    	if (t_0 <= -2e-51)
    		tmp = 180.0 * (atan((t_1 + -1.0)) / pi);
    	elseif (t_0 <= 0.0)
    		tmp = 180.0 * (atan((-0.5 * (B / (C - A)))) / pi);
    	else
    		tmp = (180.0 * atan((1.0 + t_1))) / pi;
    	end
    	tmp_2 = tmp;
    end
    
    code[A_, B_, C_] := Block[{t$95$0 = 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]}, Block[{t$95$1 = N[(N[(C - A), $MachinePrecision] / B), $MachinePrecision]}, If[LessEqual[t$95$0, -2e-51], N[(180.0 * N[(N[ArcTan[N[(t$95$1 + -1.0), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$0, 0.0], N[(180.0 * N[(N[ArcTan[N[(-0.5 * N[(B / N[(C - A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision], N[(N[(180.0 * N[ArcTan[N[(1.0 + t$95$1), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / Pi), $MachinePrecision]]]]]
    
    \begin{array}{l}
    
    \\
    \begin{array}{l}
    t_0 := \tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)\\
    t_1 := \frac{C - A}{B}\\
    \mathbf{if}\;t\_0 \leq -2 \cdot 10^{-51}:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(t\_1 + -1\right)}{\pi}\\
    
    \mathbf{elif}\;t\_0 \leq 0:\\
    \;\;\;\;180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{C - A}\right)}{\pi}\\
    
    \mathbf{else}:\\
    \;\;\;\;\frac{180 \cdot \tan^{-1} \left(1 + t\_1\right)}{\pi}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 3 regimes
    2. if (atan.f64 (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64))))))) < -2e-51

      1. Initial program 58.2%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around inf

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{C}{B} - \left(1 + \frac{A}{B}\right)\right)}}{\mathsf{PI}\left(\right)} \]
      4. Step-by-step derivation
        1. +-commutativeN/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C}{B} - \color{blue}{\left(\frac{A}{B} + 1\right)}\right)}{\mathsf{PI}\left(\right)} \]
        2. associate--r+N/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\left(\frac{C}{B} - \frac{A}{B}\right) - 1\right)}}{\mathsf{PI}\left(\right)} \]
        3. div-subN/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\color{blue}{\frac{C - A}{B}} - 1\right)}{\mathsf{PI}\left(\right)} \]
        4. sub-negN/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{C - A}{B} + \left(\mathsf{neg}\left(1\right)\right)\right)}}{\mathsf{PI}\left(\right)} \]
        5. metadata-evalN/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{C - A}{B} + \color{blue}{-1}\right)}{\mathsf{PI}\left(\right)} \]
        6. lower-+.f64N/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{C - A}{B} + -1\right)}}{\mathsf{PI}\left(\right)} \]
        7. lower-/.f64N/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\color{blue}{\frac{C - A}{B}} + -1\right)}{\mathsf{PI}\left(\right)} \]
        8. lower--.f6474.4

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{\color{blue}{C - A}}{B} + -1\right)}{\pi} \]
      5. Applied rewrites74.4%

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

      if -2e-51 < (atan.f64 (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64))))))) < -0.0

      1. Initial program 18.4%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Step-by-step derivation
        1. lift-sqrt.f64N/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \color{blue}{\sqrt{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)}{\mathsf{PI}\left(\right)} \]
        2. lift-+.f64N/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{\color{blue}{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)}{\mathsf{PI}\left(\right)} \]
      4. Applied rewrites18.4%

        \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \color{blue}{\mathsf{hypot}\left(C - A, B\right)}\right)\right)}{\pi} \]
      5. Taylor expanded in B around 0

        \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1}{2} \cdot \frac{B}{C - A}\right)}}{\mathsf{PI}\left(\right)} \]
      6. Step-by-step derivation
        1. lower-*.f64N/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \color{blue}{\left(\frac{-1}{2} \cdot \frac{B}{C - A}\right)}}{\mathsf{PI}\left(\right)} \]
        2. lower-/.f64N/A

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(\frac{-1}{2} \cdot \color{blue}{\frac{B}{C - A}}\right)}{\mathsf{PI}\left(\right)} \]
        3. lower--.f6499.2

          \[\leadsto 180 \cdot \frac{\tan^{-1} \left(-0.5 \cdot \frac{B}{\color{blue}{C - A}}\right)}{\pi} \]
      7. Applied rewrites99.2%

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

      if -0.0 < (atan.f64 (*.f64 (/.f64 #s(literal 1 binary64) B) (-.f64 (-.f64 C A) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))

      1. Initial program 58.9%

        \[180 \cdot \frac{\tan^{-1} \left(\frac{1}{B} \cdot \left(\left(C - A\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)\right)}{\pi} \]
      2. Add Preprocessing
      3. Taylor expanded in B around 0

        \[\leadsto \color{blue}{180 \cdot \frac{\tan^{-1} \left(\frac{C - \left(A + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{B}\right)}{\mathsf{PI}\left(\right)}} \]
      4. Applied rewrites58.9%

        \[\leadsto \color{blue}{\frac{180 \cdot \tan^{-1} \left(\frac{\left(C - A\right) - \sqrt{\mathsf{fma}\left(B, B, \left(A - C\right) \cdot \left(A - C\right)\right)}}{B}\right)}{\pi}} \]
      5. Taylor expanded in B around -inf

        \[\leadsto \frac{180 \cdot \tan^{-1} \left(\left(1 + \frac{C}{B}\right) - \frac{A}{B}\right)}{\mathsf{PI}\left(\right)} \]
      6. Step-by-step derivation
        1. Applied rewrites76.0%

          \[\leadsto \frac{180 \cdot \tan^{-1} \left(1 + \frac{C - A}{B}\right)}{\pi} \]
      7. Recombined 3 regimes into one program.
      8. Add Preprocessing

      Reproduce

      ?
      herbie shell --seed 2024223 
      (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)))