利用相邻边和它们之间的角度的平行四边形的对角线长度
原文:https://www . geesforgeks . org/平行四边形对角线长度-使用相邻边和它们之间的角度/
给定两个整数 a 和 b ，其中 a 和 b 代表平行四边形相邻边的长度和它们之间的角度 ~~0~~ ，任务是求平行四边形对角线的长度。
示例:
输入: a = 6，b = 10，~~0~~= 30 T5】输出: 6.14
输入: a = 3，b = 5，~~0~~= 45 T5】输出: 3.58
逼近:考虑一个平行四边形 ABCD 有边 a 和 b ，现在在三角形 ABD 中以角度 A 应用余弦法则求对角线 p 的长度，同样从三角形 ABC 中求对角线 q 。
因此对角线由下式给出:
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<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
时间复杂度:O(1) T3】辅助空间: O(1)
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