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指定された三角形のすべての角度を求める

2D 平面内の三角形の 3 つの頂点すべての座標が与えられた場合、タスクは 3 つの角度すべてを見つけることです。
例:  
 

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


 


この問題を解決するには、以下を使用します 余弦の法則 。 
 



与えられた三角形のすべての角度' title=

ウェブブラウザの設定


 

scan.next Java
c^2 = a^2 + b^2 - 2(a)(b)(cos beta)


整理し直してから 
 

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


三角法では、余弦の法則 (余弦公式または余弦定理とも呼ばれます) は、三角形の辺の長さをその角度の 1 つの余弦に関連付けます。
 

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.


以下は上記の手順の実装です。 
 

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) 

出力:  
 

alpha : 90 beta : 45 gamma : 45

時間計算量: 組み込みの sqrt 関数を使用しているため、O(log(n))

補助スペース: ○(1)

jvm

参照
https://en.wikipedia.org/wiki/余弦の法則
 

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