Length of diagonals of a Rhombus using length of Side and vertex Angle

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Given two integers A and theta, denoting the length of a side of a rhombus and the vertex angle respectively, the task is to find the length of the diagonals of the rhombus.

Examples:

Input: A = 10, theta = 30
Output: 19.32 5.18

Input: A = 6, theta = 45
Output: 11.09 4.59

Approach:
The problem can be solved using the law of cosines. Using the law of cosines on triangles formed by the diagonals and sides of the rhombus gives the following relation to calculate the length of diagonals:

$Diagonal(p)=a\sqrt{2+2cos\theta}$
$Diagonal(q)=a\sqrt{2-2cos\theta}$

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the length
// of diagonals of a rhombus using
// length of sides and vertex angle
double Length_Diagonals(int a, double theta)
{
    double p = a * sqrt(2 + (2 * cos(
           theta * (3.141 / 180))));
    double q = a * sqrt(2 - (2 * cos(
           theta * (3.141 / 180))));
            
    cout << fixed << setprecision(2) << p
         << " " << q; 
}
 
// Driver Code 
int main() 
{ 
    int a = 6;
    int theta = 45;
   
    // Function Call 
    Length_Diagonals(a, theta); 
   
    return 0; 
} 
 
// This code is contributed by Virusbuddah_

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to calculate the length
// of diagonals of a rhombus using
// length of sides and vertex angle
static double[] Length_Diagonals(int a, double theta)
{
    double p = a * Math.sqrt(2 + (2 *
                   Math.cos(theta * (Math.PI / 180))));
 
    double q = a * Math.sqrt(2 - (2 *
                   Math.cos(theta * (Math.PI / 180))));
 
    return new double[]{ p, q };
}
 
// Driver Code
public static void main(String[] args) 
{
    int A = 6;
    double theta = 45;
     
    double[] ans = Length_Diagonals(A, theta);
 
    System.out.printf("%.2f" + " " + "%.2f",
                      ans[0], ans[1]);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python Program to implement
# the above approach
import math
 
# Function to calculate the length
# of diagonals of a rhombus using
# length of sides and vertex angle
def Length_Diagonals(a, theta):
 
    p = a * math.sqrt(2 + (2 * \
            math.cos(math.radians(theta))))
             
    q = a * math.sqrt(2 - (2 * \
            math.cos(math.radians(theta))))
 
    return [p, q]
 
 
# Driver Code
A = 6
theta = 45
 
ans = Length_Diagonals(A, theta)
 
print(round(ans[0], 2), round(ans[1], 2))

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to calculate the length
// of diagonals of a rhombus using
// length of sides and vertex angle
static double[] Length_Diagonals(int a, double theta)
{
    double p = a * Math.Sqrt(2 + (2 *
                   Math.Cos(theta * (Math.PI / 180))));
 
    double q = a * Math.Sqrt(2 - (2 *
                   Math.Cos(theta * (Math.PI / 180))));
 
    return new double[]{ p, q };
}
 
// Driver Code
public static void Main(String[] args) 
{
    int A = 6;
    double theta = 45;
     
    double[] ans = Length_Diagonals(A, theta);
 
    Console.Write("{0:F2}" + " " + "{1:F2}",
                            ans[0], ans[1]);
}
}
 
// This code is contributed by gauravrajput1

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function to calculate the length
// of diagonals of a rhombus using
// length of sides and vertex angle
function Length_Diagonals(a, theta)
{
    let p = a * Math.sqrt(2 + (2 *
                Math.cos(theta * (Math.PI / 180))));
   
    let q = a * Math.sqrt(2 - (2 *
                Math.cos(theta * (Math.PI / 180))));
   
    return [ p, q ];
}
     
// Driver Code
     
      let A = 6;
    let theta = 45;
       
    let ans = Length_Diagonals(A, theta);
   
    document.write(ans[0].toFixed(2) + " "
    + ans[1].toFixed(2));
      
</script>

Output:

11.09 4.59

Time Complexity: O(1)
Auxiliary Space: O(1)

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