За даними координатами всіх трьох вершин трикутника на 2D площині завдання полягає в тому, щоб знайти всі три кути.
приклад:
Input : A = (0 0) B = (0 1) C = (1 0) Output : 90 45 45
Для вирішення цієї проблеми ми використовуємо нижче Закон косинусів .
значення xdxd

c^2 = a^2 + b^2 - 2(a)(b)(cos beta)
Після переаранжування
beta = acos( ( a^2 + b^2 - c^2 ) / (2ab) )
У тригонометрії закон косинусів (також відомий як формула косинусів або правило косинусів) пов’язує довжини сторін трикутника з косинусом одного з його кутів.
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.
Нижче наведено реалізацію вищевказаних кроків.
словник сортування pythonC++
// 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<intint> X pair<intint> Y) { int xDiff = X.first - Y.first; int yDiff = X.second - Y.second; return xDiff*xDiff + yDiff*yDiff; } void printAngle(pair<intint> A pair<intint> B pair<intint> C) { // Square of lengths be a2 b2 c2 int a2 = lengthSquare(BC); int b2 = lengthSquare(AC); int c2 = lengthSquare(AB); // 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<intint> A = make_pair(00); pair<intint> B = make_pair(01); pair<intint> C = make_pair(10); printAngle(ABC); 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(BC); int b2 = lengthSquare(AC); int c2 = lengthSquare(AB); // 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(00); Point B = new Point(01); Point C = new Point(10); printAngle(ABC); } }
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(BC); let b2 = lengthSquare(AC); let c2 = lengthSquare(AB); // 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(ABC); // The code is contributed by Gautam goel (guatamgoel962)
Вихід:
alpha : 90 beta : 45 gamma : 45
Часова складність: O(log(n)) з використанням вбудованих функцій sqrt
Допоміжний простір: О(1)
довідка :
https://en.wikipedia.org/wiki/Law_of_cosines
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