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 Codeint 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 Codepublic 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 Sidesa = 3b = 5# Given Angletheta = 45 # Function Call ans = Length_Diagonal(a, b, theta) # Print the final answerprint(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 Codepublic 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 sidesvar a = 3;var b = 5;// Given anglevar theta = 45;// Function callvar ans = Length_Diagonal(a, b, theta);// Print the final answerdocument.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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