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Check if right triangle possible from given area and hypotenuse
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Find all angles of a given triangle

Last Updated : 16 Feb, 2023
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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 


Next Article
Check if right triangle possible from given area and hypotenuse

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Pratik Chhajer
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Article Tags :
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  • DSA
Practice Tags :
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    Problems based on 3D Objects

    Find the perimeter of a cylinder
    Given diameter and height, find the perimeter of a cylinder.Perimeter is the length of the outline of a two - dimensional shape. A cylinder is a three - dimensional shape. So, technically we cannot find the perimeter of a cylinder but we can find the perimeter of the cross-section of the cylinder. T
    3 min read
    Find the Surface area of a 3D figure
    Given a N*M matrix A[][] representing a 3D figure. The height of the building at (i, j) is A[i][j] . Find the surface area of the figure. Examples : Input : N = 1, M = 1 A[][] = { {1} } Output : 6 Explanation : The total surface area is 6 i.e 6 side of the figure and each are of height 1. Input : N
    11 min read
    Calculate Volume of Dodecahedron
    Given the edge of the dodecahedron calculate its Volume. Volume is the amount of the space which the shapes takes up. A dodecahedron is a 3-dimensional figure made up of 12 faces, or flat sides. All of the faces are pentagons of the same size.The word 'dodecahedron' comes from the Greek words dodeca
    3 min read
    Program to calculate volume of Octahedron
    Given the side of the Octahedron then calculate the volume of Octahedron. Examples: Input : 3 Output : 12.7279 Input : 7 Output : 161.692 Approach: The volume of an octahedron is given by the formula: Volume = √2/3 × a3 where a is the side of Octahedron Below is the implementation to calculate volum
    2 min read
    Program to calculate Volume and Surface area of Hemisphere
    Calculate volume and surface area of a hemisphere. Hemisphere : In geometry, it is an exact half of a sphere. We can find many of the real life examples of the hemispheres such as our planet Earth can be divided into two hemisphere the southern & northern hemispheres.   Volume of Hemisphere : Vo
    4 min read
    Program for volume of Pyramid
    A pyramid is a 3-dimensional geometric shape formed by connecting all the corners of a polygon to a central apex. There are many types of pyramids. Most often, they are named after the type of base they have. Let's look at some common types of pyramids below. Volume of a square pyramid [base of Pyra
    7 min read
    Program to calculate volume of Ellipsoid
    Ellipsoid, closed surface of which all plane cross sections are either ellipses or circles. An ellipsoid is symmetrical about three mutually perpendicular axes that intersect at the center. It is a three-dimensional, closed geometric shape, all planar sections of which are ellipses or circles. An el
    4 min read
    Program for Volume and Surface Area of Cube
    Cube is a 3-dimensional box-like figure represented in the 3-dimensional plane. Cube has 6 squared-shape equal faces. Each face meet another face at 90 degree each. Three sides of cube meet at same vertex. Examples: Input : Side of a cube = 2 Output : Area = 8 Total surface area = 24 Input : Side of
    3 min read

    Problems based on Quadrilateral

    Number of parallelograms when n horizontal parallel lines intersect m vertical parallel lines
    Given two positive integers n and m. The task is to count number of parallelogram that can be formed of any size when n horizontal parallel lines intersect with m vertical parallel lines. Examples: Input : n = 3, m = 2Output : 32 parallelograms of size 1x1 and 1 parallelogram of size 2x1.Input : n =
    8 min read
    Program for Circumference of a Parallelogram
    Given the sides of Parallelogram then calculate the circumference.Examples : Input: a = 10, b = 8 Output: 36.00 Input: a = 25.12, b = 20.4 Output: 91.04 The opposite sides of a parallelogram are of equal length and parallel. The angles are pairwise equal but not necessarily 90 degree. The circumfere
    3 min read
    Program to calculate area and perimeter of Trapezium
    A trapezium is a quadrilateral with at least one pair of parallel sides, other two sides may not be parallel. The parallel sides are called the bases of the trapezium and the other two sides are called it's legs. The perpendicular distance between parallel sides is called height of trapezium. Formul
    5 min read
    Program to find area of a Trapezoid
    Definition of Trapezoid : A Trapezoid is a convex quadrilateral with at least one pair of parallel sides. The parallel sides are called the bases of the trapezoid and the other two sides which are not parallel are referred to as the legs. There can also be two pairs of bases. In the above figure CD
    4 min read
    Find all possible coordinates of parallelogram
    Find all the possible coordinates from the three coordinates to make a parallelogram of a non-zero area.Let's call A, B, and C are the three given points. We can have only the three possible situations:  (1) AB and AC are sides, and BC a diagonal(2) AB and BC are sides, and AC a diagonal (3) BC and
    5 min read
    Maximum area of quadrilateral
    Given four sides of quadrilateral a, b, c, d, find the maximum area of the quadrilateral possible from the given sides .Examples:  Input : 1 2 1 2Output : 2.00It is optimal to construct a rectangle for maximum area .  According to Bretschneider's formula, the area of a general quadrilateral is given
    5 min read
    Check whether four points make a parallelogram
    Given four points in a 2-dimensional space we need to find out whether they make a parallelogram or not. A parallelogram has four sides. Two opposite sides are parallel and are of same lengths. Examples:Points = [(0, 0), (4, 0), (1, 3), (5, 3)]Above points make a parallelogram.Points = [(0, 0), (2,
    15+ min read
    Find the Missing Point of Parallelogram
    Given three coordinate points A, B and C, find the missing point D such that ABCD can be a parallelogram.Examples : Input : A = (1, 0) B = (1, 1) C = (0, 1) Output : 0, 0 Explanation: The three input points form a unit square with the point (0, 0) Input : A = (5, 0) B = (1, 1) C = (2, 5) Output : 6,
    13 min read

    Problems based on Polygon and Convex Hull

    How to check if a given point lies inside or outside a polygon?
    Given a polygon and a point 'p', find if 'p' lies inside the polygon or not. The points lying on the border are considered inside. Examples: Recommended ProblemPlease solve it on PRACTICE first, before moving on to the solution Solve ProblemApproach: The idea to solve this problem is based on How to
    9 min read
    Area of a polygon with given n ordered vertices
    Given ordered coordinates of a polygon with n vertices. Find the area of the polygon. Here ordered means that the coordinates are given either in a clockwise manner or anticlockwise from the first vertex to last.Examples : Input : X[] = {0, 4, 4, 0}, Y[] = {0, 0, 4, 4}; Output : 16 Input : X[] = {0,
    6 min read
    Tangents between two Convex Polygons
    Given two convex polygons, we aim to identify the lower and upper tangents connecting them. As shown in the figure below, TRL and TLR represent the upper and lower tangents, respectively. Examples: Input: First Polygon : [[2, 2], [3, 3], [5, 2], [4, 0], [3, 1]] Second Polygon : [[-1, 0], [0, 1], [1,
    15 min read
    Find number of diagonals in n sided convex polygon
    Given n > 3, find number of diagonals in n sided convex polygon.According to Wikipedia, In geometry, a diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge. Informally, any sloping line is called diagonal.Examples : Input : 5Outp
    3 min read
    Convex Hull using Jarvis' Algorithm or Wrapping
    Given a set of points in the plane. the convex hull of the set is the smallest convex polygon that contains all the points of it.We strongly recommend to see the following post first. How to check if two given line segments intersect?The idea of Jarvis's Algorithm is simple, we start from the leftmo
    13 min read
    Convex Hull using Graham Scan
    A convex hull is the smallest convex polygon that contains a given set of points. It is a useful concept in computational geometry and has applications in various fields such as computer graphics, image processing, and collision detection.A convex polygon is a polygon in which all interior angles ar
    15+ min read
    Dynamic Convex hull | Adding Points to an Existing Convex Hull
    Given a convex hull, we need to add a given number of points to the convex hull and print the convex hull after every point addition. The points should be in anti-clockwise order after addition of every point. Examples: Input : Convex Hull : (0, 0), (3, -1), (4, 5), (-1, 4) Point to add : (100, 100)
    15 min read
    Deleting points from Convex Hull
    Given a fixed set of points. We need to find convex hull of given set. We also need to find convex hull when a point is removed from the set. Example: Initial Set of Points: (-2, 8) (-1, 2) (0, 1) (1, 0) (-3, 0) (-1, -9) (2, -6) (3, 0) (5, 3) (2, 5) Initial convex hull:- (-2, 8) (-3, 0) (-1, -9) (2,
    15+ min read
    Minimum area of a Polygon with three points given
    Given three points of a regular polygon(n > 3), find the minimum area of a regular polygon (all sides same) possible with the points given.Examples: Input : 0.00 0.00 1.00 1.00 0.00 1.00 Output : 1.00 By taking point (1.00, 0.00) square is formed of side 1.0 so area = 1.00 . One thing to note in
    13 min read
    Find Simple Closed Path for a given set of points
    Given a set of points, connect the dots without crossing. Example: Input: points[] = {(0, 3), (1, 1), (2, 2), (4, 4), (0, 0), (1, 2), (3, 1}, {3, 3}}; Output: Connecting points in following order would not cause any crossing {(0, 0), (3, 1), (1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (0, 3)} We strongl
    11 min read
    Minimum Distance between Two Points
    You are given an array arr[] of n distinct points in a 2D plane, where each point is represented by its (x, y) coordinates. Find the minimum Euclidean distance between two distinct points.Note: For two points A(px,qx) and B(py,qy) the distance Euclidean between them is:Distance = \sqrt{(p_{x}-q_{x})
    15+ min read
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