带推导、示例和实现的正弦规则

给定角度（以度为单位） A 、 C和边c ，对应于下图，任务是找到剩余的两条边a和b 。

例子：

Input: A = 45, C = 35, c = 23
Output:
28.35 
39.49
Explanation: 
a is 28.35 and b is 39.49

Input: A = 45, C = 45, c = 10
Output:
10
14.14

方法：想法是使用正弦规则。它指出任何三角形的边都与它们对角的正弦成正比。 a / Sin(A) = b / Sin(B) = c / Sin(C) 。推导描述如下：

As is evident from the figure above:

A perpendicular of length h has been drawn on BC from A. From General trigonometric rules:

SinB=h/c——–(1)

SinC=h/b——–(2)

From the above two equations, we get:

c x SinB=b x SinC

Or b/SinB=c/SinC—–(3)

Similarly, if a perpendicular is drawn from B to AC, we can get:

a/SinA=c/SinC——-(4)

From Equations (3) and (4), we get:

a/SinA=b/SinB=c/SinC  

请按照以下步骤解决问题：

• 将角度A和C从度数更改为弧度，以便能够在内置函数中使用。
• 使用三角形的角之和为 180 度的观察结果计算角度B。
• 使用正弦规则计算边a和b 。

下面是上述方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate remaining two sides
void findSides(double A, double C, double c)
{
    // Calculate angle B
    double B = 180 - (A + C);
 
    // Convert angles to their respective radians for
    // using trigonometric functions
    A = A * (3.14159 / 180);
    C = C * (3.14159 / 180);
    B = B * (3.14159 / 180);
 
    // Sine rule
    double a = (c / sin(C)) * sin(A);
    double b = (c / sin(C)) * sin(B);
 
    // Precision of 2 decimal spaces
    cout << fixed << setprecision(2);
 
    // Print the answer
    cout << a << endl;
    cout << b << endl;
}
 
// Driver Code
int main()
{
    // Input
    double A = 45.0;
    double C = 35.0;
    double c = 23;
 
    // Function Call
    findSides(A, C, c);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to calculate remaining two sides
static void findSides(double A, double C,
                      double c)
{
     
    // Calculate angle B
    double B = 180 - (A + C);
 
    // Convert angles to their respective
    // radians for using trigonometric functions
    A = A * (3.14159 / 180);
    C = C * (3.14159 / 180);
    B = B * (3.14159 / 180);
 
    // Sine rule
    double a = (c / Math.sin(C)) * Math.sin(A);
    double b = (c / Math.sin(C)) * Math.sin(B);
 
    // Print the answer
    System.out.println(String.format("%.2f", a));
    System.out.println(String.format("%.2f", b));
}
 
// Driver code
public static void main(String[] args)
{
     
    // Input
    double A = 45.0;
    double C = 35.0;
    double c = 23;
 
    // Function Call
    findSides(A, C, c);
}
}
 
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
import math
 
# Function to calculate remaining two sides
def findSides(A, C, c):
     
    # Calculate angle B
    B = 180 - (A + C)
 
    # Convert angles to their respective radians
    # for using trigonometric functions
    A = A * (3.14159 / 180)
    C = C * (3.14159 / 180)
    B = B * (3.14159 / 180)
 
    # Sine rule
    a = (c / math.sin(C)) * math.sin(A)
    b = (c / math.sin(C)) * math.sin(B)
 
    # Precision of 2 decimal spaces
 
    # Print the answer
    print("{0:.2f}".format(a))
    print("{0:.2f}".format(b))
 
# Driver Code
 
# Input
A = 45.0
C = 35.0
c = 23
 
# Function Call
findSides(A, C, c)
 
# This code is contributed by target_2

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to calculate remaining two sides
static void findSides(double A, double C,
                      double c)
{
     
    // Calculate angle B
    double B = 180 - (A + C);
 
    // Convert angles to their respective
    // radians for using trigonometric functions
    A = A * (3.14159 / 180);
    C = C * (3.14159 / 180);
    B = B * (3.14159 / 180);
 
    // Sine rule
    double a = (c / Math.Sin(C)) * Math.Sin(A);
    double b = (c / Math.Sin(C)) * Math.Sin(B);
 
    // Print the answer
    Console.WriteLine("{0:F2}",a);
    Console.WriteLine("{0:F2}",b);
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Input
    double A = 45.0;
    double C = 35.0;
    double c = 23;
 
    // Function Call
    findSides(A, C, c);
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

       // JavaScript program for the above approach
 
 
       // Function to calculate remaining two sides
       function findSides(A, C, c)
       {
        
           // Calculate angle B
           let B = 180 - (A + C);
 
           // Convert angles to their respective radians for
           // using trigonometric functions
           A = A * (3.14159 / 180);
           C = C * (3.14159 / 180);
           B = B * (3.14159 / 180);
 
           // Sine rule
           let a = (c / Math.sin(C)) * Math.sin(A);
           let b = (c / Math.sin(C)) * Math.sin(B);
 
           // Precision of 2 decimal spaces
 
           // Print the answer
           document.write(a.toPrecision(4) + "<br>");
           document.write(b.toPrecision(4) + "<br>");
       }
 
       // Driver Code
 
       // Input
       let A = 45.0;
       let C = 35.0;
       let c = 23;
 
       // Function Call
       findSides(A, C, c);
 
 // This code is contributed by Potta Lokesh
   

输出

28.35
39.49

时间复杂度： O(1)
辅助空间： O(1)