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

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:

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)