Length of diagonal of a parallelogram using adjacent sides and angle between them

Namachila November 19, 2025

0 0 3 minutes read

Given two integers a and b where a and b represents the length of adjacent sides of a parallelogram and an angle 0 between them, the task is to find the length of diagonal of the parallelogram.

Examples:

Input: a = 6, b = 10, 0=30
Output: 6.14

Input: a = 3, b = 5, 0=45
Output: 3.58

Approach: Consider a parallelogram ABCD with sides a and b, now apply cosine rule at angle A in the triangle ABD to find the length of diagonal p, similarly find diagonal q from triangle ABC.

Therefore the diagonals is given by:

$Diagonal (P)=\sqrt{a^2+b^2-2ab.cos(\theta)}$

C++

// C++ program to find length
// Of diagonal of a parallelogram
// Using sides and angle between them.
#include <bits/stdc++.h>
using namespace std;
#define PI 3.147
 
// Function to return the length
// Of diagonal of a parallelogram
// using sides and angle between them.
double Length_Diagonal(int a, int b, double theta)
{
    double diagonal = sqrt((pow(a, 2) + pow(b, 2)) - 
                      2 * a * b * cos(theta * (PI / 180)));
 
    return diagonal;
}
 
// Driver Code
int main()
{
 
    // Given sides
    int a = 3;
    int b = 5;
 
    // Given angle
    double theta = 45;
 
    // Function call
    double ans = Length_Diagonal(a, b, theta);
 
    // Print the final answer
    printf("%.2f", ans);
}
 
// This code is contributed by Amit Katiyar

Java

// Java program to find length 
// Of diagonal of a parallelogram 
// Using sides and angle between them.
class GFG{
 
// Function to return the length
// Of diagonal of a parallelogram
// using sides and angle between them.
static double Length_Diagonal(int a, int b,
                              double theta)
{
    double diagonal = Math.sqrt((Math.pow(a, 2) + 
                                 Math.pow(b, 2)) - 
                                 2 * a * b * 
                                 Math.cos(theta * 
                                 (Math.PI / 180)));
 
    return diagonal;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given sides
    int a = 3;
    int b = 5;
 
    // Given angle
    double theta = 45;
 
    // Function call
    double ans = Length_Diagonal(a, b, theta);
 
    // Print the final answer
    System.out.printf("%.2f", ans);
}
}
 
// This code is contributed by amal kumar choubey 

Python3

# Python3 Program to find length 
# Of diagonal of a parallelogram 
# Using sides and angle between them.
 
import math  
   
# Function to return the length
# Of diagonal of a parallelogram 
# using sides and angle between them.  
def Length_Diagonal(a, b, theta):  
   
    diagonal = math.sqrt( ((a**2) + (b**2))
    - 2 * a*b * math.cos(math.radians(theta)))
     
    return diagonal  
   
# Driver Code
 
# Given Sides
a = 3
b = 5
 
# Given Angle
theta = 45
   
# Function Call  
ans = Length_Diagonal(a, b, theta)  
   
# Print the final answer
print(round(ans, 2))

C#

// C# program to find length 
// Of diagonal of a parallelogram 
// Using sides and angle between them.
using System;
 
class GFG{
 
// Function to return the length
// Of diagonal of a parallelogram
// using sides and angle between them.
static double Length_Diagonal(int a, int b,
                              double theta)
{
    double diagonal = Math.Sqrt((Math.Pow(a, 2) + 
                                 Math.Pow(b, 2)) - 
                                 2 * a * b * 
                                 Math.Cos(theta * 
                                (Math.PI / 180)));
 
    return diagonal;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given sides
    int a = 3;
    int b = 5;
 
    // Given angle
    double theta = 45;
 
    // Function call
    double ans = Length_Diagonal(a, b, theta);
 
    // Print the readonly answer
    Console.Write("{0:F2}", ans);
}
}
 
// This code is contributed by amal kumar choubey

Javascript

<script>
 
// javascript program to find length 
// Of diagonal of a parallelogram 
// Using sides and angle between them.
 
// Function to return the length
// Of diagonal of a parallelogram
// using sides and angle between them.
function Length_Diagonal(a , b,theta)
{
    var diagonal = Math.sqrt((Math.pow(a, 2) + 
                                 Math.pow(b, 2)) - 
                                 2 * a * b * 
                                 Math.cos(theta * 
                                 (Math.PI / 180)));
 
    return diagonal;
}
 
// Driver Code
 
// Given sides
var a = 3;
var b = 5;
 
// Given angle
var theta = 45;
 
// Function call
var ans = Length_Diagonal(a, b, theta);
 
// Print the final answer
document.write(ans.toFixed(2));
 
// This code is contributed by 29AjayKumar 
 
</script>

Output:

3.58

Time Complexity: O(logn) as it is using inbuilt sqrt function
Auxiliary Space: O(1)

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