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# Find interior angles for each side of a given Cyclic Quadrilateral

• Last Updated : 21 Apr, 2021

Given four positive integers A, B, C, and D representing the sides of a Cyclic Quadrilateral, the task is to find all the interior angles of the cyclic quadrilateral. Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

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A cyclic quadrilateral is a quadrilateral whose vertices lie on a single circle.
This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic(A, B, C, and D).
( In the figure, r is the circumradius and a, b, c, and d are length of AB, BC, CD, and DA respectively).

Examples:

Input: A = 10, B = 15, C = 20, D = 25
Output:
∠A: 85.59 degrees
∠B: 122.58 degrees
∠C: 94.41 degrees
∠D: 57.42 degrees

Input: A = 10, B = 10, C = 10, D = 10
Output:
∠A: 90.00 degrees
∠B: 90.00 degrees
∠C: 90.00 degrees
∠D: 90.00 degrees

Approach: The given problem can be solved by using the formula to calculate the cosine of the interior angle of a cyclic quadrilateral. The formula is given by:     Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Function to find the interior angles// of the cyclic quadrilateralvoid findAngles(double a, double b,                double c, double d){    // Stores the numerator and the    // denominator to find angle A    double numerator = a * a + d * d                       - b * b - c * c;     double denominator = 2 * (a * b + c * d);     double x = numerator / denominator;     cout << fixed << setprecision(2)         << "A: " << (acos(x) * 180) / 3.141592         << " degrees";     // Stores the numerator and the    // denominator to find angle B    numerator = a * a + b * b                - c * c - d * d;     x = numerator / denominator;     cout << fixed << setprecision(2)         << "\nB: " << (acos(x) * 180) / 3.141592         << " degrees";     // Stores the numerator and the    // denominator to find angle C:    numerator = c * c + b * b                - a * a - d * d;     x = numerator / denominator;     cout << fixed << setprecision(2)         << "\nC: " << (acos(x) * 180) / 3.141592         << " degrees";     // Stores the numerator and the    // denominator to find angle D:    numerator = d * d + c * c                - a * a - b * b;     x = numerator / denominator;     cout << fixed << setprecision(2)         << "\nD: " << (acos(x) * 180) / 3.141592         << " degrees";} // Driver Codeint main(){    double A = 10, B = 15, C = 20, D = 25;    findAngles(A, B, C, D);     return 0;}

## Java

 // Java program for the above approachclass GFG{     // Function to find the interior angles// of the cyclic quadrilateralstatic void findAngles(double a, double b,                       double c, double d){         // Stores the numerator and the    // denominator to find angle A    double numerator = a * a + d * d -                       b * b - c * c;     double denominator = 2 * (a * b + c * d);     double x = numerator / denominator;     System.out.println("A: " +       Math.round(((Math.acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");     // Stores the numerator and the    // denominator to find angle B    numerator = a * a + b * b - c * c - d * d;     x = numerator / denominator;     System.out.println("B: " +       Math.round(((Math.acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");     // Stores the numerator and the    // denominator to find angle C:    numerator = c * c + b * b -                a * a - d * d;     x = numerator / denominator;     System.out.println("C: " +       Math.round(((Math.acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");     // Stores the numerator and the    // denominator to find angle D:    numerator = d * d + c * c -                a * a - b * b;     x = numerator / denominator;     System.out.println("D: " +       Math.round(((Math.acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");} // Driver Codepublic static void main (String[] args){    double A = 10, B = 15, C = 20, D = 25;         findAngles(A, B, C, D);}} // This code is contributed by AnkThon

## Python3

 # Python3 program for the above approachimport math # Function to find the interior angles# of the cyclic quadrilateraldef findAngles(a, b, c, d):         # Stores the numerator and the    # denominator to find angle A    numerator = a * a + d * d - b * b - c * c    denominator = 2 * (a * b + c * d)    x = numerator / denominator    print("A: ", '%.2f' % ((math.acos(x) * 180) /          3.141592), " degrees")         # Stores the numerator and the    # denominator to find angle B    numerator = a * a + b * b - c * c - d * d    x = numerator / denominator    print("B: ", '%.2f' % ((math.acos(x) * 180) /          3.141592), " degrees")         # Stores the numerator and the    # denominator to find angle C:    numerator = c * c + b * b - a * a - d * d    x = numerator / denominator    print("C: ", '%.2f' % ((math.acos(x) * 180) /          3.141592), " degrees")         # Stores the numerator and the    # denominator to find angle D:    numerator = d * d + c * c - a * a - b * b    x = numerator / denominator    print("D: ", '%.2f' % ((math.acos(x) * 180) /          3.141592), " degrees")     # Driver Codeif __name__ == "__main__":         A = 10    B = 15    C = 20    D = 25         findAngles(A, B, C, D) # This code is contributed by ukasp

## C#

 // C# program for the above approachusing System; class GFG{     // Function to find the interior angles// of the cyclic quadrilateralstatic void findAngles(double a, double b,                       double c, double d){         // Stores the numerator and the    // denominator to find angle A    double numerator = a * a + d * d -                       b * b - c * c;     double denominator = 2 * (a * b + c * d);     double x = numerator / denominator;     Console.WriteLine("A: " +       Math.Round(((Math.Acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");     // Stores the numerator and the    // denominator to find angle B    numerator = a * a + b * b - c * c - d * d;     x = numerator / denominator;     Console.WriteLine("B: " +       Math.Round(((Math.Acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");     // Stores the numerator and the    // denominator to find angle C:    numerator = c * c + b * b -                a * a - d * d;     x = numerator / denominator;     Console.WriteLine("C: " +       Math.Round(((Math.Acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");     // Stores the numerator and the    // denominator to find angle D:    numerator = d * d + c * c -                a * a - b * b;     x = numerator / denominator;     Console.WriteLine("D: " +       Math.Round(((Math.Acos(x) * 180) /                       3.141592) * 100.0) /                       100.0 + " degrees");} // Driver Codepublic static void Main(string[] args){    double A = 10, B = 15, C = 20, D = 25;         findAngles(A, B, C, D);}} // This code is contributed by AnkThon

## Javascript

 
Output:
A: 85.59 degrees
B: 122.58 degrees
C: 94.41 degrees
D: 57.42 degrees

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

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