使用边长和顶角的菱形对角线长度

原文:https://www . geesforgeks . org/菱形对角线长度-使用边长和顶角/

给定两个整数 A 和θ，分别表示一个菱形的边长和顶角，任务是求菱形对角线的长度。

示例:

输入: A = 10，θ= 30 T3】输出: 19.32 5.18

输入: A = 6，θ= 45 T3】输出: 11.09 4.59

方法: 这个问题可以用余弦定律来解决。使用由菱形的对角线和边形成的三角形上的余弦定律给出以下关系来计算对角线的长度:

Diagonal(p)=a\sqrt{2+2cos\theta} Diagonal(q)=a\sqrt{2-2cos\theta}

下面是上述方法的实现:

C++

// 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 语言(一种计算机语言，尤用于创建网站)

// 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 3

# 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

// 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

java 描述语言

<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>

Output: 

11.09 4.59

时间复杂度:O(1) T5辅助空间:** O(1)

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