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Наћи све углове датог троугла

Дате координате сва три врха троугла у 2Д равни задатак је пронаћи сва три угла.
Пример:  
 

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


 


Да бисмо решили овај проблем, користимо се у наставку Закон косинуса
 



сви углови датог троугла' title=


 

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


После преуређења 
 

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


У тригонометрији закон косинуса (такође познат као косинусна формула или косинусно правило) повезује дужине страница троугла са косинусом једног од његових углова.
 

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.


Испод је имплементација горњих корака. 
 

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<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) 

Излаз:  
 

alpha : 90 beta : 45 gamma : 45

Временска сложеност: О(лог(н)) пошто се користе уграђене скрт функције

Помоћни простор: О(1)

Референце
хттпс://ен.википедиа.орг/вики/Лав_оф_цосинес
 

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