距原点垂直距离为 D 且距原点的垂线与 x 轴的夹角为 A 的直线方程

给定两个整数D和A，分别代表从原点到直线的垂直距离和垂线与x轴正方向的夹角，任务是找到直线的方程。

例子：

Input: D = 10, A = 30 degrees
Output: 0.87x +0.50y = 10

Input: D = 12, A = 45 degrees
Output: 0.71x +0.71y = 12

方法：根据以下观察可以解决给定的问题：

图1

• 设垂直距离为(p) ，垂直线与正 x 轴之间的角度为(α) 度。
• 考虑在所需线上具有坐标(x, y)的点P。
• 从P绘制一条垂线，在L处与 x 轴相交。
• 从L ，在OQ 上在M处画一条垂线。
• 现在，从P画一条垂线，在N处与ML相交。

图2

Now consider right triangle OLM
$cos α = OM/OL$
$OM = OL cos α = x cos α$ — (1)

Now consider right triangle PNL
$cos (90 - α) = PN/PL$
$sin α = PN/PL$
$PN = PL sin α = y sin α$  
$MQ = PN = y sin α$ — (2)

Now $p = OQ = OM + MQ$
Using equations (1) and (2)
$p = x cos α + y sin α$ which is the equation of the required line

下面是上述方法的实现：

C++

// C++ program for the approach
#include 
using namespace std;
 
// Function to find equation of a line whose
// distance from origin and angle made by the
// perpendicular from origin with x-axis is given
void findLine(int distance, float degree)
{
    // Convert angle from degree to radian
    float x = degree * 3.14159 / 180;
 
    // Handle the special case
    if (degree > 90) {
        cout << "Not Possible";
        return;
    }
 
    // Calculate the sin and cos of angle
    float result_1 = sin(x);
    float result_2 = cos(x);
 
    // Print the equation of the line
    cout << fixed << setprecision(2)
         << result_2 << "x +"
         << result_1 << "y = " << distance;
}
 
// Driver Code
int main()
{
    // Given Input
    int D = 10;
    float A = 30;
 
    // Function Call
    findLine(D, A);
 
    return 0;
}

Java

// Java program for the approach
 
class GFG{
 
// Function to find equation of a line whose
// distance from origin and angle made by the
// perpendicular from origin with x-axis is given
static void findLine(int distance, float degree)
{
    // Convert angle from degree to radian
    float x = (float) (degree * 3.14159 / 180);
 
    // Handle the special case
    if (degree > 90) {
        System.out.print("Not Possible");
        return;
    }
 
    // Calculate the sin and cos of angle
    float result_1 = (float) Math.sin(x);
    float result_2 = (float) Math.cos(x);
 
    // Print the equation of the line
    System.out.print(String.format("%.2f",result_2)+ "x +"
         + String.format("%.2f",result_1)+ "y = " +  distance);
}
 
// Driver Code
public static void main(String[] args)
{
    // Given Input
    int D = 10;
    float A = 30;
 
    // Function Call
    findLine(D, A);
 
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the approach
import math
 
# Function to find equation of a line whose
# distance from origin and angle made by the
# perpendicular from origin with x-axis is given
def findLine(distance, degree):
 
    # Convert angle from degree to radian
    x = degree * 3.14159 / 180
  
    # Handle the special case
    if (degree > 90):
        print("Not Possible")
        return
  
    # Calculate the sin and cos of angle
    result_1 = math.sin(x)
    result_2 = math.cos(x)
  
    # Print the equation of the line
    print('%.2f' % result_2,
          "x +", '%.2f' % result_1,
          "y = ", distance, sep = "")
 
# Driver code
 
# Given Input
D = 10
A = 30
 
# Function Call
findLine(D, A)
 
# This code is contributed by mukesh07

C#

// C# program for the approach
using System;
class GFG
{
     
    // Function to find equation of a line whose
    // distance from origin and angle made by the
    // perpendicular from origin with x-axis is given
    static void findLine(int distance, float degree)
    {
        // Convert angle from degree to radian
        float x = (float)(degree * 3.14159 / 180);
      
        // Handle the special case
        if (degree > 90) {
            Console.WriteLine("Not Possible");
            return;
        }
      
        // Calculate the sin and cos of angle
        float result_1 = (float)(Math.Sin(x));
        float result_2 = (float)(Math.Cos(x));
      
        // Print the equation of the line
        Console.WriteLine(result_2.ToString("0.00") + "x +"
             + result_1.ToString("0.00") + "y = " + distance);
    }
 
  static void Main ()
  {
    // Given Input
    int D = 10;
    float A = 30;
  
    // Function Call
    findLine(D, A);
  }
}
 
// This code is contributed by suresh07.

Javascript

  
        // JavaScript program for the above approach
 
        // Function to find equation of a line whose
        // distance from origin and angle made by the
        // perpendicular from origin with x-axis is given
        function findLine(distance, degree) {
            // Convert angle from degree to radian
            let x = degree * 3.14159 / 180;
 
            // Handle the special case
            if (degree > 90) {
                document.write("Not Possible");
                return;
            }
 
            // Calculate the sin and cos of angle
            let result_1 = Math.sin(x);
            let result_2 = Math.cos(x);
 
            // Print the equation of the line
            document.write(result_2.toPrecision(2) + "x + "
             + result_1.toPrecision(2) + "y = " + distance);
        }
 
        // Driver Code
 
        // Given Input
        let D = 10;
        let A = 30;
 
        // Function Call
        findLine(D, A);
 
        // This code is contributed by Hritik
         

输出：

0.87x +0.50y = 10

时间复杂度： O(1)
辅助空间： O(1)

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