Given an angle where, . The task is to check whether it is possible to make a regular polygon with all of its interior angle equal to . If possible then print “YES”, otherwise print “NO” (without quotes).

**Examples:**

Input:angle = 90Output:YES Polygons with sides 4 is possible with angle 90 degrees.Input:angle = 30Output:NO

**Approach:** The Interior angle is defined as the angle between any two adjacent sides of a regular polygon.

It is given by where, **n** is the number of sides in the polygon.

This can be written as .

On rearranging terms we get, .

Thus, if **n** is an **Integer** the answer is “YES” otherwise, answer is “NO”.

Below is the implementation of the above approach:

## C++

`// C++ implementation of above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to check whether it is possible ` `// to make a regular polygon with a given angle. ` `void` `makePolygon(` `float` `a) ` `{ ` ` ` `// N denotes the number of sides ` ` ` `// of polygons possible ` ` ` `float` `n = 360 / (180 - a); ` ` ` `if` `(n == (` `int` `)n) ` ` ` `cout << ` `"YES"` `; ` ` ` `else` ` ` `cout << ` `"NO"` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `float` `a = 90; ` ` ` ` ` `// function to print the required answer ` ` ` `makePolygon(a); ` ` ` ` ` `return` `0; ` `} ` |

## Java

`class` `GFG ` `{ ` `// Function to check whether ` `// it is possible to make a ` `// regular polygon with a given angle. ` `static` `void` `makePolygon(` `double` `a) ` `{ ` ` ` `// N denotes the number of ` ` ` `// sides of polygons possible ` ` ` `double` `n = ` `360` `/ (` `180` `- a); ` ` ` `if` `(n == (` `int` `)n) ` ` ` `System.out.println(` `"YES"` `); ` ` ` `else` ` ` `System.out.println(` `"NO"` `); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ` `double` `a = ` `90` `; ` ` ` ` ` `// function to print ` ` ` `// the required answer ` ` ` `makePolygon(a); ` `} ` `} ` ` ` `// This code is contributed by Bilal ` |

## Python3

`# Python 3 implementation ` `# of above approach ` ` ` `# Function to check whether ` `# it is possible to make a ` `# regular polygon with a ` `# given angle. ` `def` `makePolygon(a) : ` ` ` ` ` `# N denotes the number of sides ` ` ` `# of polygons possible ` ` ` `n ` `=` `360` `/` `(` `180` `-` `a) ` ` ` ` ` `if` `n ` `=` `=` `int` `(n) : ` ` ` `print` `(` `"YES"` `) ` ` ` ` ` `else` `: ` ` ` `print` `(` `"NO"` `) ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `a ` `=` `90` ` ` ` ` `# function calling ` ` ` `makePolygon(a) ` ` ` `# This code is contributed ` `# by ANKITRAI1 ` |

## C#

`// C# implementation of ` `// above approach ` `using` `System; ` ` ` `class` `GFG ` `{ ` `// Function to check whether ` `// it is possible to make a ` `// regular polygon with a ` `// given angle. ` `static` `void` `makePolygon(` `double` `a) ` `{ ` ` ` `// N denotes the number of ` ` ` `// sides of polygons possible ` ` ` `double` `n = 360 / (180 - a); ` ` ` `if` `(n == (` `int` `)n) ` ` ` `Console.WriteLine(` `"YES"` `); ` ` ` `else` ` ` `Console.WriteLine(` `"NO"` `); ` `} ` ` ` `// Driver code ` `static` `void` `Main() ` `{ ` ` ` `double` `a = 90; ` ` ` ` ` `// function to print ` ` ` `// the required answer ` ` ` `makePolygon(a); ` `} ` `} ` ` ` `// This code is contributed by mits ` |

## PHP

`<?php ` `// PHP implementation of above approach ` ` ` `// Function to check whether it ` `// is possible to make a regular ` `// polygon with a given angle. ` `function` `makePolygon(` `$a` `) ` `{ ` ` ` `// N denotes the number of ` ` ` `// sides of polygons possible ` ` ` `$n` `= 360 / (180 - ` `$a` `); ` ` ` `if` `(` `$n` `== (int)` `$n` `) ` ` ` `echo` `"YES"` `; ` ` ` `else` ` ` `echo` `"NO"` `; ` `} ` ` ` `// Driver code ` `$a` `= 90; ` ` ` `// function to print the ` `// required answer ` `makePolygon(` `$a` `); ` ` ` `// This code is contributed ` `// by ChitraNayal ` `?> ` |

**Output:**

YES

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