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Find all angles of a given triangle

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

all angles of a given triangle

 

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 
radian. Then convert angles from radian into 
degrees.

Below is implementation of above steps. 
 

C++




// Code to find all three angles
// of a triangle given coordinate
// of all three vertices
#include <iostream>
#include <utility> // for pair
#include <cmath> // for math functions
using namespace std;
 
#define PI 3.1415926535
 
// returns square of distance b/w two points
int lengthSquare(pair<int,int> X, pair<int,int> Y)
{
    int xDiff = X.first - Y.first;
    int yDiff = X.second - Y.second;
    return xDiff*xDiff + yDiff*yDiff;
}
 
void printAngle(pair<int,int> A, pair<int,int> B,
                pair<int,int> 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<int,int> A = make_pair(0,0);
    pair<int,int> B = make_pair(0,1);
    pair<int,int> 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
This article is contributed by Pratik Chhajer . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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