n边正多边形的对角线长度

给定一个边长为a的n边规则多边形，任务是找到对角线的长度。
例子：

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

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

方法：

We know that the sum of interior angles of a polygon = (n – 2) * 180 where, n is the no. of sides in the polygon. 
So, each interior angle = (n – 2) * 180/n 
Now, we have to find BC = 2 * x. If we draw a perpendicular AO on BC, we will see that the perpendicular bisects BC in BO and OC, as triangles AOB and AOC are congruent to each other. 
Now, t = (n – 2) * 180/2n 
So, sint = x/a 
Therefore, x = asint 
Hence, diagonal=2x = 2asint = 2asin((n – 2) * 180/2n)

C++

// C++ Program to find the diagonal
// of a regular polygon with given side length
#include 
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

// 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

# 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#

// 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...

Javascript

// 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

输出：

17.119