# Angle between 3 given vertices in a n-sided regular polygon

Given a n-sided regular polygon and three vertices a1, a2 and a3, the task is to find the angle suspended at vertex a1 by vertex a2 and vertex a3.

Examples:

```Input: n = 6, a1 = 1, a2 = 2, a3 = 4
Output: 90

Input: n = 5, a1 = 1, a2 = 2, a3 = 5
Output: 36
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. The angle subtended by an edge on the center of n sided regular polygon is 360/n.
2. The angle subtended by vertices seperated by k edges becomes (360*k)/n.
3. The chord between the vertices subtends an angle with half the value of the angle subtended at the center at the third vertex which is a point on the circumference on the circumcircle.
4. Let the angle obtained in this manner be a = (180*x)/n where k is number of edges between i and k.
5. Similarily for the opposite vertex we get the angle to be b = (180*y)/n where l is number of edges between j and k.
6. The angle between the three vertices thus equals 180-a-b.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function that checks whether given angle ` `// can be created using any 3 sides ` `double` `calculate_angle(``int` `n, ``int` `i, ``int` `j, ``int` `k) ` `{ ` `    ``// Initialize x and y ` `    ``int` `x, y; ` ` `  `    ``// Calculate the number of vertices ` `    ``// between i and j, j and k ` `    ``if` `(i < j) ` `        ``x = j - i; ` `    ``else` `        ``x = j + n - i; ` `    ``if` `(j < k) ` `        ``y = k - j; ` `    ``else` `        ``y = k + n - j; ` ` `  `    ``// Calculate the angle subtended ` `    ``// at the circumference ` `    ``double` `ang1 = (180 * x) / n; ` `    ``double` `ang2 = (180 * y) / n; ` ` `  `    ``// Angle subtended at j can be found ` `    ``// using the fact that the sum of angles ` `    ``// of a triangle is equal to 180 degrees ` `    ``double` `ans = 180 - ang1 - ang2; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5; ` `    ``int` `a1 = 1; ` `    ``int` `a2 = 2; ` `    ``int` `a3 = 5; ` ` `  `    ``cout << calculate_angle(n, a1, a2, a3); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `// Function that checks whether given angle ` `// can be created using any 3 sides ` `static` `double` `calculate_angle(``int` `n, ``int` `i,  ` `                              ``int` `j, ``int` `k) ` `{ ` `    ``// Initialize x and y ` `    ``int` `x, y; ` ` `  `    ``// Calculate the number of vertices ` `    ``// between i and j, j and k ` `    ``if` `(i < j) ` `        ``x = j - i; ` `    ``else` `        ``x = j + n - i; ` `    ``if` `(j < k) ` `        ``y = k - j; ` `    ``else` `        ``y = k + n - j; ` ` `  `    ``// Calculate the angle subtended ` `    ``// at the circumference ` `    ``double` `ang1 = (``180` `* x) / n; ` `    ``double` `ang2 = (``180` `* y) / n; ` ` `  `    ``// Angle subtended at j can be found ` `    ``// using the fact that the sum of angles ` `    ``// of a triangle is equal to 180 degrees ` `    ``double` `ans = ``180` `- ang1 - ang2; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `n = ``5``; ` `    ``int` `a1 = ``1``; ` `    ``int` `a2 = ``2``; ` `    ``int` `a3 = ``5``; ` ` `  `    ``System.out.println((``int``)calculate_angle(n, a1, a2, a3)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function that checks whether given angle ` `# can be created using any 3 sides ` `def` `calculate_angle(n, i, j, k): ` `     `  `    ``# Initialize x and y ` `    ``x, y ``=` `0``, ``0` `  `  `    ``# Calculate the number of vertices ` `    ``# between i and j, j and k ` `    ``if` `(i < j): ` `        ``x ``=` `j ``-` `i ` `    ``else``: ` `        ``x ``=` `j ``+` `n ``-` `i ` `    ``if` `(j < k): ` `        ``y ``=` `k ``-` `j ` `    ``else``: ` `        ``y ``=` `k ``+` `n ``-` `j ` ` `  `    ``# Calculate the angle subtended ` `    ``# at the circumference ` `    ``ang1 ``=` `(``180` `*` `x) ``/``/` `n ` `    ``ang2 ``=` `(``180` `*` `y) ``/``/` `n ` ` `  `    ``# Angle subtended at j can be found ` `    ``# using the fact that the sum of angles ` `    ``# of a triangle is equal to 180 degrees ` `    ``ans ``=` `180` `-` `ang1 ``-` `ang2 ` `    ``return` `ans ` ` `  `# Driver code ` `n ``=` `5` `a1 ``=` `1` `a2 ``=` `2` `a3 ``=` `5` ` `  `print``(calculate_angle(n, a1, a2, a3)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function that checks whether given angle ` `// can be created using any 3 sides ` `static` `double` `calculate_angle(``int` `n, ``int` `i,  ` `                              ``int` `j, ``int` `k) ` `{ ` `    ``// Initialize x and y ` `    ``int` `x, y; ` ` `  `    ``// Calculate the number of vertices ` `    ``// between i and j, j and k ` `    ``if` `(i < j) ` `        ``x = j - i; ` `    ``else` `        ``x = j + n - i; ` `    ``if` `(j < k) ` `        ``y = k - j; ` `    ``else` `        ``y = k + n - j; ` ` `  `    ``// Calculate the angle subtended ` `    ``// at the circumference ` `    ``double` `ang1 = (180 * x) / n; ` `    ``double` `ang2 = (180 * y) / n; ` ` `  `    ``// Angle subtended at j can be found ` `    ``// using the fact that the sum of angles ` `    ``// of a triangle is equal to 180 degrees ` `    ``double` `ans = 180 - ang1 - ang2; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main () ` `{ ` `    ``int` `n = 5; ` `    ``int` `a1 = 1; ` `    ``int` `a2 = 2; ` `    ``int` `a3 = 5; ` ` `  `    ``Console.WriteLine((``int``)calculate_angle(n, a1, a2, a3)); ` `} ` `} ` ` `  `// This code is contributed by ihritik `

Output:

```36
```

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.