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العثور على جميع زوايا مثلث معين

بالنظر إلى إحداثيات القمم الثلاثة للمثلث في المستوى ثنائي الأبعاد، تكون المهمة هي إيجاد الزوايا الثلاث.
مثال:  
 

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


 


لحل هذه المشكلة نستخدم أدناه قانون جيب التمام
 



جميع زوايا مثلث معين' title=

جافا شار إلى السلسلة


 

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


بعد إعادة الترتيب 
 

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


في علم المثلثات، قانون جيب التمام (المعروف أيضًا باسم صيغة جيب التمام أو قاعدة جيب التمام) يربط أطوال جوانب المثلث بجيب تمام إحدى زواياه.
 

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.


وفيما يلي تنفيذ الخطوات المذكورة أعلاه. 
 

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) 

الإخراج:  
 

alpha : 90 beta : 45 gamma : 45

تعقيد الوقت: O(log(n)) منذ استخدام وظائف sqrt المدمجة

المساحة المساعدة: يا(1)

مرجع
https://en.wikipedia.org/wiki/Law_of_cosines
 

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