Question

我正在编写一个程序，读取三角形的每一边，计算其度数并确定它是直角三角形，斜角三角形还是三角形。

我计算余弦然后从余弦中找出度数：the cos equation

以下是代码：

#include <stdio.h>
#include <math.h>
#define PI 3.14159265

int a,b,c;
double val, cA, cB, cC, angleA, angleB, angleC;

int main()
{
    printf("Input the length of each side!");
    printf("\na: ");
    scanf("%d", &a);
    printf("\nb: ");
    scanf("%d", &b);
    printf("\nc: ");
    scanf("%d", &c);

    //calculating the cosine
    cA = ((b*b)+(c*c)-(a*a))/(2*b*c);
    cB = ((a*a)+(c*c)-(b*b))/(2*a*c);
    cC = ((a*a)+(b*b)-(c*c))/(2*a*b);

    val = 180.0/PI;

    if(cA>=0&&cA<=1){
        angleA = acos(cA)*val;
    } else{
        angleA = 0;
    }

    if(cB>=0&&cB<=1){
        angleB = acos(cB)*val;
    } else{
        angleB = 0;
    }

    if(cC>=0&&cC<=1){
        angleC = acos(cC)*val;
    } else{
        angleC = 0;
    }

    printf("\ncos A= %lf", cA);
    printf("\ncos B= %lf", cB);
    printf("\ncos C= %lf", cC);     

    printf("\nAngle A= %lf", angleA);
    printf("\nAngle B= %lf", angleB);
    printf("\nAngle C= %lf", angleC);
}

我还在尝试计算余弦，但只要我输入三角形的边值，cA，cB，cC只给出零。

cos A= 0.000000
cos B= 0.000000
cos C= 0.000000

如何使cos方程返回正弦的正确值？

Answer 1

在以下行中：

cA = ((b^2)+(c^2)-(a^2))/(2*b*c);

您在int上进行计算并将结果保存到double。这意味着您最终可能会得到8 / 10之类的中间结果，它只是0而不是分数。

但更大的问题是^是XOR操作，而不是您的想法。使用b*b或pow(b, 2)。

Answer 2

所以我改变了一点来解决除以零结果：

#include <stdio.h>
#include <math.h>
#define PI 3.14159265

double a,b,c;
double val, cA, cB, cC, angleA, angleB, angleC;

int main()
{
    printf("Input the length of each side!");
    printf("\na: ");
    scanf("%lf", &a);
    printf("\nb: ");
    scanf("%lf", &b);
    printf("\nc: ");
    scanf("%lf", &c);

    //calculating the cosine
    cA = ((b*b)+(c*c)-(a*a))/(2*b*c);
    cB = ((a*a)+(c*c)-(b*b))/(2*a*c);
    cC = ((a*a)+(b*b)-(c*c))/(2*a*b);

    val = 180.0/PI;

    if(cA>=0&&cA<=1){
        angleA = acos(cA)*val;
    } else{
        angleA = 0;
    }

    if(cB>=0&&cB<=1){
        angleB = acos(cB)*val;
    } else{
        angleB = 0;
    }

    if(cC>=0&&cC<=1){
        angleC = acos(cC)*val;
    } else{
        angleC = 0;
    }

    printf("\ncos A= %lf", cA);
    printf("\ncos B= %lf", cB);
    printf("\ncos C= %lf", cC);     

    printf("\nAngle A= %lf", angleA);
    printf("\nAngle B= %lf", angleB);
    printf("\nAngle C= %lf", angleC);       
}

我猜整数输入不起作用，所以我把它改成了双倍。

C中cos的计算给出零值

2 个答案: