# Count of obtuse angles in a circle with ‘k’ equidistant points between 2 given points

A circle is given with k equidistant points on its circumference. 2 points A and B are given in the circle. Find the count of all obtuse angles (angles larger than 90 degree) formed from /_ACB, where C can be any point in circle other than A or B.

Note :

A and B are not equal.

A < B.

Points are between 1 and K(both inclusive).

Examples :

Input : K = 6, A = 1, B = 3. Output : 1 Explanation : In the circle with 6 equidistant points, when C = 2 i.e. /_123, we get obtuse angle. Input : K = 6, A = 1, B = 4. Output : 0 Explanation : In this circle, there is no such C that form an obtuse angle.

It can be observed that if A and B have equal elements in between them, there can’t be any C such that ACB is obtuse. Also, the number of possible obtuse angles are the smaller arc between A and B.

Below is the implementation :

## C++

`// C++ program to count number of obtuse ` `// angles for given two points. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `countObtuseAngles(` `int` `a, ` `int` `b, ` `int` `k) ` `{ ` ` ` `// There are two arcs connecting a ` ` ` `// and b. Let us count points on ` ` ` `// both arcs. ` ` ` `int` `c1 = (b - a) - 1; ` ` ` `int` `c2 = (k - b) + (a - 1); ` ` ` ` ` `// Both arcs have same number of ` ` ` `// points ` ` ` `if` `(c1 == c2) ` ` ` `return` `0; ` ` ` ` ` `// Points on smaller arc is answer ` ` ` `return` `min(c1, c2); ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `k = 6, a = 1, b = 3; ` ` ` `cout << countObtuseAngles(a, b, k); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to count number of obtuse ` `// angles for given two points ` `class` `GFG { ` ` ` ` ` `static` `int` `countObtuseAngles(` `int` `a, ` ` ` `int` `b, ` `int` `k) ` ` ` `{ ` ` ` ` ` `// There are two arcs connecting a ` ` ` `// and b. Let us count points on ` ` ` `// both arcs. ` ` ` `int` `c1 = (b - a) - ` `1` `; ` ` ` `int` `c2 = (k - b) + (a - ` `1` `); ` ` ` ` ` `// Both arcs have same number of ` ` ` `// points ` ` ` `if` `(c1 == c2) ` ` ` `return` `0` `; ` ` ` ` ` `// Points on smaller arc is answer ` ` ` `return` `min(c1, c2); ` ` ` `} ` ` ` ` ` `// Driver Program to test above function ` ` ` `public` `static` `void` `main(String arg[]) ` ` ` `{ ` ` ` ` ` `int` `k = ` `6` `, a = ` `1` `, b = ` `3` `; ` ` ` `System.out.print(countObtuseAngles(a, b, k)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Anant Agarwal. ` |

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## Python

`# C++ program to count number of obtuse ` `# angles for given two points. ` ` ` `def` `countObtuseAngles( a, b, k): ` ` ` `# There are two arcs connecting a ` ` ` `# and b. Let us count points on ` ` ` `# both arcs. ` ` ` `c1 ` `=` `(b ` `-` `a) ` `-` `1` ` ` `c2 ` `=` `(k ` `-` `b) ` `+` `(a ` `-` `1` `) ` ` ` ` ` `# Both arcs have same number of ` ` ` `# points ` ` ` `if` `(c1 ` `=` `=` `c2): ` ` ` `return` `0` ` ` ` ` `# Points on smaller arc is answer ` ` ` `return` `min` `(c1, c2) ` ` ` `# Driver code ` `k, a, b ` `=` `6` `, ` `1` `, ` `3` `print` `countObtuseAngles(a, b, k) ` ` ` `# This code is contributed by Sachin Bisht ` |

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## C#

`// C# program to count number of obtuse ` `// angles for given two points ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `countObtuseAngles(` `int` `a, ` ` ` `int` `b, ` `int` `k) ` ` ` `{ ` ` ` ` ` `// There are two arcs connecting ` ` ` `// a and b. Let us count points ` ` ` `// on both arcs. ` ` ` `int` `c1 = (b - a) - 1; ` ` ` `int` `c2 = (k - b) + (a - 1); ` ` ` ` ` `// Both arcs have same number ` ` ` `// of points ` ` ` `if` `(c1 == c2) ` ` ` `return` `0; ` ` ` ` ` `// Points on smaller arc is ` ` ` `// answer ` ` ` `return` `Math.Min(c1, c2); ` ` ` `} ` ` ` ` ` `// Driver Program to test above ` ` ` `// function ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` ` ` `int` `k = 6, a = 1, b = 3; ` ` ` ` ` `Console.WriteLine( ` ` ` `countObtuseAngles(a, b, k)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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## PHP

`<?php ` `// PHP program to count number ` `// of obtuse angles for given ` `// two points. ` ` ` `function` `countObtuseAngles(` `$a` `, ` `$b` `, ` `$k` `) ` `{ ` ` ` `// There are two arcs connecting a ` ` ` `// and b. Let us count points on ` ` ` `// both arcs. ` ` ` `$c1` `= (` `$b` `- ` `$a` `) - 1; ` ` ` `$c2` `= (` `$k` `- ` `$b` `) + (` `$a` `- 1); ` ` ` ` ` `// Both arcs have same number of ` ` ` `// points ` ` ` `if` `(` `$c1` `== ` `$c2` `) ` ` ` `return` `0; ` ` ` ` ` `// Points on smaller arc is answer ` ` ` `return` `min(` `$c1` `, ` `$c2` `); ` `} ` ` ` `// Driver code ` `$k` `= 6; ` `$a` `= 1; ` `$b` `= 3; ` `echo` `countObtuseAngles(` `$a` `, ` `$b` `, ` `$k` `); ` ` ` `// This code is contributed by aj_36 ` `?> ` |

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**Output :**

1

This article is contributed by **Rohit Thapliyal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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