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:

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

C++

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// 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

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Java

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// 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 

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Python3

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# 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))

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C#

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// 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

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Output: 

3.58

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

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