Longitud de las diagonales de un rombo usando la longitud del lado y el ángulo del vértice

Dados dos números enteros A y theta , que denotan la longitud de un lado de un rombo y el ángulo del vértice respectivamente, la tarea es encontrar la longitud de las diagonales del rombo.

Ejemplos:

Entrada: A = 10, theta = 30 
Salida: 19,32 5,18

Entrada: A = 6, theta = 45 
Salida: 11,09 4,59

Planteamiento: 
El problema se puede resolver usando la ley de los cosenos . Usando la ley de los cosenos en los triángulos formados por las diagonales y los lados del rombo se obtiene la siguiente relación para calcular la longitud de las diagonales:

A continuación se muestra la implementación del enfoque anterior:

// 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 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

# 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# 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

<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>

11.09 4.59

Tiempo Complejidad: O(1) 
Espacio Auxiliar: O(1)