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Length of the chord of the circle whose radius and the angle subtended at the center by the chord is given

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Given a circle whose radius and the angle subtended at the centre by its chord is given. The task is to find the length of the chord.
Examples: 
 

Input: r = 4, x = 63 
Output: 4.17809

Input:: r = 9, x = 71
Output:: 10.448

 

Approach
 

  1. Let the circle has center at O and has radius r, and it’s chord be AB.
  2. length of the chord be 2d, and the angle subtended by it on the center be 2x degrees.
  3. As the perpendicular dropped at the chord bisects the chord so, the perpendicular also equally divides the subtended angle 2x in x degrees.
  4. So, from the diagram, 
    d/r = sin(x*π/180)(here x deg is converted in radians)
  5. So, d = rsin(x*π/180) 
    therefore, 2d = 2rsin(x*π/180) 
     
  6. So, length of the chord = 2 * radius * sin(angle/2)

Below is the implementation of the above approach: 
 

C++




// C++ program to find the length chord
// of the circle whose radius
// and the angle subtended at the centre
// is also given
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of the chord
void length_of_chord(double r, double x)
{
    cout << "The length of the chord"
         << " of the circle is "
         << 2 * r * sin(x * (3.14 / 180))
         << endl;
}
 
// Driver code
int main()
{
    double r = 4, x = 63;
    length_of_chord(r, x);
    return 0;
}


Java




// Java program to find the length chord
// of the circle whose radius
// and the angle subtended at the centre
// is also given
 
class GFG
{
 
// Function to find the length of the chord
static void length_of_chord(double r, double x)
{
    System.out.println("The length of the chord"
        + " of the circle is "
        + 2 * r * Math.sin(x * (3.14 / 180)));
}
 
// Driver code
public static void main(String[] args)
{
    double r = 4, x = 63;
    length_of_chord(r, x);
}
}
 
// This code contributed by Rajput-Ji


Python3




# Python3 program to find the length chord
# of the circle whose radius
# and the angle subtended at the centre
# is also given
 
import math as mt
 
# Function to find the length of the chord
def length_of_chord(r, x):
 
    print("The length of the chord"
        ," of the circle is "
        ,2 * r * mt.sin(x * (3.14 / 180)))
 
 
# Driver code
r = 4
x = 63;
length_of_chord(r, x)
 
# This code is contributed by mohit kumar


C#




// C# program to find the length chord
// of the circle whose radius
// and the angle subtended at the centre
// is also given
using System;
 
class GFG
{
     
    // Function to find the length of the chord
    static void length_of_chord(double r, double x)
    {
        Console.WriteLine("The length of the chord" +
                        " of the circle is " +
                        2 * r * Math.Sin(x * (3.14 / 180)));
                         
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        double r = 4, x = 63;
        length_of_chord(r, x);
    }
}
 
// This code is Contributed by Naman_Garg


PHP




<?php
// PHP program to find the length chord
// of the circle whose radius and the
// angle subtended at the centre
// is also given
 
// Function to find the length of the chord
function length_of_chord($r, $x)
{
    echo "The length of the chord",
    " of the circle is "
    ,2 * $r * sin($x * (3.14 / 180)) ;
     
}
    // Driver code
    $r = 4; $x = 63;
    length_of_chord($r, $x);
 
    // This code is contributed by Ryuga
 
?>


Javascript




<script>
 
// JavaScript program to find the length chord
// of the circle whose radius
// and the angle subtended at the centre
// is also given
 
 
// Function to find the length of the chord
function length_of_chord(r, x)
{
    document.write("The length of the chord"
        + " of the circle is "
        + 2 * r * Math.sin(x * (3.14 / 180))
        + "<br>");
}
 
// Driver code
 
    let r = 4, x = 63;
    length_of_chord(r, x);
 
// This code is contributed by Surbhi Tyagi.
 
</script>


Output: 

The length of the chord of the circle is 7.12603

 

Time Complexity: O(1)

Auxiliary Space: O(1)


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Last Updated : 07 Jun, 2022
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