Nth angle of a Polygon whose initial angle and per angle increment is given - Coding

Given four integers N, A, K, n where N represents the number of sides the polygon, A represents the initial angle of the polygon, K represents the per angle increase, the task is to find the n^th angle of the polygon having N sides. If it is not possible then print -1.

Examples:

Input: N = 3, A = 30, K = 30, n = 3
Output: 90
Explanation:
The 3rd angle of the polygon having three sides is 90 degree as the all angles are 30, 60, 90.

Input: N = 4, A = 40, K = 30, n = 3
Output: -1
Explanation:
It is not possible to create that polygon whose initial angle is 40 and no. of sides are 4.

Approach: The idea is to observe that the angles of the polygon are increasing in the terms of Arithmetic Progression by a difference of K degree.

Find out the angular sum of the N sided polygon and the sum of the angles of the given polygon.
Check if both the values are equal or not. If yes, then the nth angle is possible hence find the nth angle.
Otherwise, print -1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to check if the angle
// is possible or not
bool possible(int N, int a, int b, int n)
{
    // Angular sum of a N-sided polygon
    int sum_of_angle = 180 * (N - 2);
 
    // Angular sum of N-sided given polygon
    int Total_angle = (N * ((2 * a) + (N - 1) * b)) / 2;
 
    // Check if both sum are equal
    if (sum_of_angle != Total_angle)
        return false;
    else
        return true;
}
 
// Function to find the nth angle
int nth_angle(int N, int a,
              int b, int n)
{
    int nth = 0;
 
    // Calculate nth angle
    nth = a + (n - 1) * b;
 
    // Return the nth angle
    return nth;
}
 
// Driver Code
int main()
{
 
    int N = 3, a = 30, b = 30, n = 3;
 
    // Checks the possibility of the
    // polygon exist or not
    if (possible(N, a, b, n))
 
        // Print nth angle
        // of the polygon
        cout << nth_angle(N, a, b, n);
    else
        cout << "Not Possible";
 
    return 0;
}

Java

// Java program for the above approach 
class GFG{ 
 
// Function to check if the angle 
// is possible or not 
static boolean possible(int N, int a,
                        int b, int n) 
{ 
     
    // Angular sum of a N-sided polygon 
    int sum_of_angle = 180 * (N - 2); 
 
    // Angular sum of N-sided given polygon 
    int Total_angle = (N * ((2 * a) + 
                      (N - 1) * b)) / 2; 
 
    // Check if both sum are equal 
    if (sum_of_angle != Total_angle) 
        return false; 
    else
        return true; 
} 
 
// Function to find the nth angle 
static int nth_angle(int N, int a, 
                     int b, int n) 
{ 
    int nth = 0; 
 
    // Calculate nth angle 
    nth = a + (n - 1) * b; 
 
    // Return the nth angle 
    return nth; 
} 
 
// Driver Code 
public static void main(String[] args) 
{ 
    int N = 3, a = 30, b = 30, n = 3; 
 
    // Checks the possibility of the 
    // polygon exist or not 
    if (possible(N, a, b, n)) 
         
        // Print nth angle 
        // of the polygon 
        System.out.print(nth_angle(N, a, b, n)); 
    else
        System.out.print("Not Possible"); 
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 program for the above approach
 
# Function to check if the angle
# is possible or not
def possible(N, a, b, n):
     
    # Angular sum of a N-sided polygon
    sum_of_angle = 180 * (N - 2)
 
    # Angular sum of N-sided given polygon
    Total_angle = (N * ((2 * a) +
                  (N - 1) * b)) / 2
 
    # Check if both sum are equal
    if (sum_of_angle != Total_angle):
        return False
    else:
        return True
 
# Function to find the nth angle
def nth_angle(N, a, b, n):
    nth = 0
 
    # Calculate nth angle
    nth = a + (n - 1) * b
 
    # Return the nth angle
    return nth
 
# Driver Code
if __name__ == '__main__':
 
    N = 3
    a = 30
    b = 30
    n = 3
 
    # Checks the possibility of the
    # polygon exist or not
    if (possible(N, a, b, n)):
 
        # Print nth angle
        # of the polygon
        print(nth_angle(N, a, b, n))
    else:
        print("Not Possible")
 
# This code is contributed by Mohit Kumar

C#

// C# program for the above approach 
using System;
class GFG{ 
  
// Function to check if the angle 
// is possible or not 
static bool possible(int N, int a,
                     int b, int n) 
{ 
      
    // Angular sum of a N-sided polygon 
    int sum_of_angle = 180 * (N - 2); 
  
    // Angular sum of N-sided given polygon 
    int Total_angle = (N * ((2 * a) + 
                      (N - 1) * b)) / 2; 
  
    // Check if both sum are equal 
    if (sum_of_angle != Total_angle) 
        return false; 
    else
        return true; 
} 
  
// Function to find the nth angle 
static int nth_angle(int N, int a, 
                     int b, int n) 
{ 
    int nth = 0; 
  
    // Calculate nth angle 
    nth = a + (n - 1) * b; 
  
    // Return the nth angle 
    return nth; 
} 
  
// Driver Code 
public static void Main(string[] args) 
{ 
    int N = 3, a = 30, b = 30, n = 3; 
  
    // Checks the possibility of the 
    // polygon exist or not 
    if (possible(N, a, b, n)) 
          
        // Print nth angle 
        // of the polygon 
        Console.Write(nth_angle(N, a, b, n)); 
    else
        Console.Write("Not Possible"); 
}
}
  
// This code is contributed by Ritik Bansal

Javascript

<script>
// JavaScript program for the above approach
 
// Function to check if the angle
// is possible or not
function possible(N, a, b, n)
{
    // Angular sum of a N-sided polygon
    let sum_of_angle = 180 * (N - 2);
 
    // Angular sum of N-sided given polygon
    let Total_angle = Math.floor((N * ((2 * a) + (N - 1) * b)) / 2);
 
    // Check if both sum are equal
    if (sum_of_angle != Total_angle)
        return false;
    else
        return true;
}
 
// Function to find the nth angle
function nth_angle(N, a, b, n)
{
    let nth = 0;
 
    // Calculate nth angle
    nth = a + (n - 1) * b;
 
    // Return the nth angle
    return nth;
}
 
// Driver Code
 
    let N = 3, a = 30, b = 30, n = 3;
 
    // Checks the possibility of the
    // polygon exist or not
    if (possible(N, a, b, n))
 
        // Print nth angle
        // of the polygon
        document.write(nth_angle(N, a, b, n));
    else
        document.write("Not Possible");
 
 
 
 
// This code is contributed by Surbhi Tyagi.
</script>

Output:

Time Complexity: O(1)
Auxiliary Space: O(1)

Reffered: https://www.geeksforgeeks.org

Geometric

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Category:	Coding
Sub Category:	Tutorial
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