# Find all angles of a given triangle

• Difficulty Level : Medium
• Last Updated : 28 Jul, 2022

Given coordinates of all three vertices of the triangle in the 2D plane, the task is to find all three angles.
Example:

Input : A = (0, 0),
B = (0, 1),
C = (1, 0)
Output : 90, 45, 45

To solve this problem we use below Law of cosines

c^2 = a^2 + b^2 - 2(a)(b)(cos beta)

After re-arranging

beta = acos( ( a^2 + b^2 - c^2 ) / (2ab) )

In trigonometry, the law of cosines (also known as the cosine formula or cosine rule) relates the lengths of the sides of a triangle to the cosine of one of its angles.

First, calculate the length of all the sides.
Then apply above formula to get all angles in
degrees.

Below is implementation of above steps.

## C++

 // Code to find all three angles // of a triangle given coordinate // of all three vertices #include #include // for pair #include // for math functions using namespace std;   #define PI 3.1415926535   // returns square of distance b/w two points int lengthSquare(pair X, pair Y) {     int xDiff = X.first - Y.first;     int yDiff = X.second - Y.second;     return xDiff*xDiff + yDiff*yDiff; }   void printAngle(pair A, pair B,                 pair C) {     // Square of lengths be a2, b2, c2     int a2 = lengthSquare(B,C);     int b2 = lengthSquare(A,C);     int c2 = lengthSquare(A,B);       // length of sides be a, b, c     float a = sqrt(a2);     float b = sqrt(b2);     float c = sqrt(c2);       // From Cosine law     float alpha = acos((b2 + c2 - a2)/(2*b*c));     float beta = acos((a2 + c2 - b2)/(2*a*c));     float gamma = acos((a2 + b2 - c2)/(2*a*b));       // Converting to degree     alpha = alpha * 180 / PI;     beta = beta * 180 / PI;     gamma = gamma * 180 / PI;       // printing all the angles     cout << "alpha : " << alpha << endl;     cout << "beta : " << beta << endl;     cout << "gamma : " << gamma << endl; }   // Driver code int main() {     pair A = make_pair(0,0);     pair B = make_pair(0,1);     pair C = make_pair(1,0);       printAngle(A,B,C);       return 0; }

## Java

 // Java Code to find all three angles // of a triangle given coordinate // of all three vertices   import java.awt.Point; import static java.lang.Math.PI; import static java.lang.Math.sqrt; import static java.lang.Math.acos;   class Test {     // returns square of distance b/w two points     static int lengthSquare(Point p1, Point p2)     {         int xDiff = p1.x- p2.x;         int yDiff = p1.y- p2.y;         return xDiff*xDiff + yDiff*yDiff;     }           static void printAngle(Point A, Point B,             Point C)     {     // Square of lengths be a2, b2, c2     int a2 = lengthSquare(B,C);     int b2 = lengthSquare(A,C);     int c2 = lengthSquare(A,B);           // length of sides be a, b, c     float a = (float)sqrt(a2);     float b = (float)sqrt(b2);     float c = (float)sqrt(c2);           // From Cosine law     float alpha = (float) acos((b2 + c2 - a2)/(2*b*c));     float betta = (float) acos((a2 + c2 - b2)/(2*a*c));     float gamma = (float) acos((a2 + b2 - c2)/(2*a*b));           // Converting to degree     alpha = (float) (alpha * 180 / PI);     betta = (float) (betta * 180 / PI);     gamma = (float) (gamma * 180 / PI);           // printing all the angles     System.out.println("alpha : " + alpha);     System.out.println("betta : " + betta);     System.out.println("gamma : " + gamma);     }           // Driver method     public static void main(String[] args)     {         Point A = new Point(0,0);         Point B = new Point(0,1);         Point C = new Point(1,0);                printAngle(A,B,C);     } }

## Python3

 # Python3 code to find all three angles # of a triangle given coordinate # of all three vertices import math   # returns square of distance b/w two points def lengthSquare(X, Y):     xDiff = X[0] - Y[0]     yDiff = X[1] - Y[1]     return xDiff * xDiff + yDiff * yDiff       def printAngle(A, B, C):           # Square of lengths be a2, b2, c2     a2 = lengthSquare(B, C)     b2 = lengthSquare(A, C)     c2 = lengthSquare(A, B)       # length of sides be a, b, c     a = math.sqrt(a2);     b = math.sqrt(b2);     c = math.sqrt(c2);       # From Cosine law     alpha = math.acos((b2 + c2 - a2) /                          (2 * b * c));     betta = math.acos((a2 + c2 - b2) /                          (2 * a * c));     gamma = math.acos((a2 + b2 - c2) /                          (2 * a * b));       # Converting to degree     alpha = alpha * 180 / math.pi;     betta = betta * 180 / math.pi;     gamma = gamma * 180 / math.pi;       # printing all the angles     print("alpha : %f" %(alpha))     print("betta : %f" %(betta))     print("gamma : %f" %(gamma))           # Driver code A = (0, 0) B = (0, 1) C = (1, 0)   printAngle(A, B, C);   # This code is contributed # by ApurvaRaj

## C#

 // C# Code to find all three angles // of a triangle given coordinate // of all three vertices using System;       class GFG {     class Point     {         public int x, y;         public Point(int x, int y)         {             this.x = x;             this.y = y;         }     }           // returns square of distance b/w two points     static int lengthSquare(Point p1, Point p2)     {         int xDiff = p1.x - p2.x;         int yDiff = p1.y - p2.y;         return xDiff * xDiff + yDiff * yDiff;     }           static void printAngle(Point A, Point B, Point C)     {         // Square of lengths be a2, b2, c2         int a2 = lengthSquare(B, C);         int b2 = lengthSquare(A, C);         int c2 = lengthSquare(A, B);                   // length of sides be a, b, c         float a = (float)Math.Sqrt(a2);         float b = (float)Math.Sqrt(b2);         float c = (float)Math.Sqrt(c2);                   // From Cosine law         float alpha = (float) Math.Acos((b2 + c2 - a2) /                                            (2 * b * c));         float betta = (float) Math.Acos((a2 + c2 - b2) /                                            (2 * a * c));         float gamma = (float) Math.Acos((a2 + b2 - c2) /                                            (2 * a * b));                   // Converting to degree         alpha = (float) (alpha * 180 / Math.PI);         betta = (float) (betta * 180 / Math.PI);         gamma = (float) (gamma * 180 / Math.PI);                   // printing all the angles         Console.WriteLine("alpha : " + alpha);         Console.WriteLine("betta : " + betta);         Console.WriteLine("gamma : " + gamma);     }           // Driver Code     public static void Main(String[] args)     {         Point A = new Point(0, 0);         Point B = new Point(0, 1);         Point C = new Point(1, 0);               printAngle(A, B, C);     } }   // This code is contributed by Rajput-Ji

## Javascript

 // JavaScript program // Code to find all three angles // of a triangle given coordinate // of all three vertices   // returns square of distance b/w two points function lengthSquare(X, Y){     let xDiff = X[0] - Y[0];     let yDiff = X[1] - Y[1];     return xDiff*xDiff + yDiff*yDiff; }   function printAngle(A, B, C){           // Square of lengths be a2, b2, c2     let a2 = lengthSquare(B,C);     let b2 = lengthSquare(A,C);     let c2 = lengthSquare(A,B);       // length of sides be a, b, c     let a = Math.sqrt(a2);     let b = Math.sqrt(b2);     let c = Math.sqrt(c2);       // From Cosine law     let alpha = Math.acos((b2 + c2 - a2)/(2*b*c));     let beta = Math.acos((a2 + c2 - b2)/(2*a*c));     let gamma = Math.acos((a2 + b2 - c2)/(2*a*b));       // Converting to degree     alpha = alpha * 180 / Math.PI;     beta = beta * 180 / Math.PI;     gamma = gamma * 180 / Math.PI;       // printing all the angles     console.log("alpha : ", alpha);     console.log("beta : ", beta);     console.log("gamma : ", gamma); }   // Driver code let A = [0, 0]; let B = [0, 1]; let C = [1, 0];   printAngle(A,B,C);   // The code is contributed by Gautam goel (guatamgoel962)

Output:

alpha : 90
beta : 45
gamma : 45

Time Complexity: O(log(n)) since using inbuilt sqrt functions

Auxiliary Space: O(1)

Reference
https://en.wikipedia.org/wiki/Law_of_cosines
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