Longitud de la diagonal de un polígono regular de n lados

Dado un polígono regular de n lados de longitud de lado a . La tarea es encontrar la longitud de su diagonal.
Ejemplos: 
 

Input:  a = 9, n = 10
Output:  17.119

Input: a = 4, n = 5
Output: 6.47213

Enfoque :
 

Sabemos que la suma de los ángulos interiores de un polígono = (n – 2) * 180 donde, n es el no. de lados en el polígono. 
Entonces, cada ángulo interior = (n – 2) * 180/n 
Ahora, tenemos que encontrar BC = 2 * x . Si dibujamos una perpendicular AO en BC , veremos que la perpendicular biseca a BC en BO y OC , ya que los triángulos AOB y AOC son congruentes entre sí. 
Ahora, t = (n – 2) * 180/2n 
Entonces, sint = x/a 
Por lo tanto, x = asint 
Por lo tanto, diagonal=2x = 2asint = 2asen((n – 2) * 180/2n)

// C++ Program to find the diagonal
// of a regular polygon with given side length
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the diagonal
// of a regular polygon
float polydiagonal(float n, float a)
{
 
    // Side and side length cannot be negative
    if (a < 0 && n < 0)
        return -1;
 
    // diagonal
    // degree converted to radians
    return 2 * a * sin((((n - 2) * 180) / (2 * n)) * 3.14159 / 180);
}
 
// Driver code
int main()
{
    float a = 9, n = 10;
    cout << polydiagonal(n, a) << endl;
 
    return 0;
}

// Java Program to find the diagonal
// of a regular polygon with given side length
 
class GFG {
 
// Function to find the diagonal
// of a regular polygon
    static float polydiagonal(float n, float a) {
 
        // Side and side length cannot be negative
        if (a < 0 && n < 0) {
            return -1;
        }
 
        // diagonal
        // degree converted to radians
        return (float) (2 * a * Math.sin((((n - 2) * 180) / (2 * n)) * 3.14159 / 180));
    }
 
// Driver code
    public static void main(String[] args) {
        float a = 9, n = 10;
        System.out.printf("%.3f",polydiagonal(n, a));
 
    }
}
 
// This code is contributed by 29AjayKumar

# Python3 Program to find the diagonal
# of a regular polygon with given side length
import math as mt
 
# Function to find the diagonal
# of a regular polygon
def polydiagonal(n, a):
 
    # Side and side length cannot
    # be negative
    if (a < 0 and n < 0):
        return -1
 
    # diagonal degree converted to radians
    return (2 * a * mt.sin((((n - 2) * 180) /
           (2 * n)) * 3.14159 / 180))
 
# Driver code
a, n = 9, 10
print(polydiagonal(n, a))
 
# This code is contributed
# by Mohit kumar 29

// C#  Program to find the diagonal
// of a regular polygon with given side length
using System;
 
public class GFG{
     
// Function to find the diagonal
// of a regular polygon
    static float polydiagonal(float n, float a) {
 
        // Side and side length cannot be negative
        if (a < 0 && n < 0) {
            return -1;
        }
 
        // diagonal
        // degree converted to radians
        return (float) (2 * a * Math.Sin((((n - 2) * 180) / (2 * n)) * 3.14159 / 180));
    }
 
// Driver code
    static public void Main (){
            float a = 9, n = 10;
        Console.WriteLine(polydiagonal(n, a));
 
    }
}
 
// This code is contributed by  @Sachin...

<?php
// PHP Program to find the diagonal of a
// regular polygon with given side length
 
// Function to find the diagonal
// of a regular polygon
function polydiagonal ($n, $a)
{
 
    // Side and side length cannot
    // be negative
    if ($a < 0 && $n < 0)
        return -1;
 
    // diagonal
    // degree converted to radians
    return 2 * $a * sin(((($n - 2) * 180) /
          (2 * $n)) * 3.14159 / 180);
}
 
// Driver code
$a = 9;
$n = 10;
echo polydiagonal($n, $a);
 
// This code is contributed
// by Sach_Code
?>

<script>
// javascript Program to find the diagonal
// of a regular polygon with given side length
 
 
 
// Function to find the diagonal
// of a regular polygon
function polydiagonal(n , a) {
 
    // Side and side length cannot be negative
    if (a < 0 && n < 0) {
        return -1;
    }
 
    // diagonal
    // degree converted to radians
    return  (2 * a * Math.sin((((n - 2) * 180)
    / (2 * n)) * 3.14159 / 180));
}
 
// Driver code
     
var a = 9, n = 10;
document.write(polydiagonal(n, a).toFixed(3));
 
// This code contributed by Princi Singh
</script>

17.119

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)