坐标 C 的三角形角度

Question

i have to find angle ∠BAC in triangle ABC with given coordinates. I'm trying to find cos, then use function acos(cos) and get degrees(not radian) anwser. The input is Ax,Ay,Bx,By,Cx,Cy coordinates

#include <math.h>

int main(){
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    int n;
    int ax,ay,bx,by,cx,cy;
    scanf("%d",&n);
    for(int i = 0; i<n; i ++){
        scanf("%d %d %d %d %d %d",&ax,&ay,&bx,&by,&cx,&cy);
        double cosab = (ax*bx+ay*by)/ (sqrt(ax*ax+ay*ay)* sqrt(bx*bx+by*by));
        cosab =180* acos(cosab)/M_PI;
        printf("cosab is %0.20g\n",cosab);
    }
    return 0;
}

With input

8
2 1 2 3 5 5
2 1 4 3 2 3
3 1 3 5 2 3
0 0 1 0 10 0
0 0 1 0 -10 1
7 4 3 3 3 3
0 0 1 0 1 1e-5
0 0 1 0 1e-5 1

I get

cosab is 29.744881296942211
cosab is 10.304846468766044
cosab is 40.601294645004479
cosab is -1.#IND
cosab is -1.#IND
cosab is 15.255118703057764
cosab is -1.#IND
cosab is -1.#IND

But the anwser should be

36.869897645844019962
45
26.565051177077990019
0
174.28940686250035697
0
0.00057295779511172474814
89.999427042204885652

What is "-1.#IND" and what's wrong with my code?

Answer 1

As others have pointed out, the input coordinates are floating point, so let us read them into double s:

        double ax,ay,bx,by,cx,cy;
        /* Read coordinates of A, B, and C. */
        scanf("%lf%lf%lf%lf%lf%lf",&ax,&ay,&bx,&by,&cx,&cy);

∠BAC is the angle between lines AB and AC. Let us consider those as vectors AB and AC represented by their x and y components:

        /* Convert to vectors AB and AC (as x and y components). */
        double abx = bx - ax;   /* x component of vector AB */
        double aby = by - ay;   /* y component of vector AB */
        double acx = cx - ax;   /* x component of vector AC */
        double acy = cy - ay;   /* y component of vector AC */

Now there is a nice formula using ATAN2 on the vector components to get the angle of rotation from one 2D vector to another ¹ :

        /*
         * Get angle of rotation from vector AB to vector AC
         * in the range [-M_PI, +M_PI].
         */
        double rot_ab_ac = atan2(acy*abx - acx*aby, acx*abx + acy*aby);

We want the absolute angle, not a signed angle:

        /* Convert to absolute angle BAC. */
        double bac = fabs(rot_ab_ac);

Finally, we want the angle in degrees, not radians:

        /* Convert to degrees. */
        double bac_deg = bac * 180.0 / M_PI;

Note: M_PI is not defined in the C standard, but can be easily defined by a macro if necessary:

#ifndef M_PI
#define M_PI 3.14159265358979323846
#endif

---

¹ Angle Between Two Vectors 2D Formula

Answer 2

Input is not only integers

1e-5 is like 0.00001 . Read that into a double .

// int ax,ay,bx,by,cx,cy;
double ax,ay,bx,by,cx,cy;
...
    //scanf("%d %d %d %d %d %d",&ax,&ay,&bx,&by,&cx,&cy);
    scanf("%lf %lf %lf %lf %lf %lf",&ax,&ay,&bx,&by,&cx,&cy);

Be careful with acos()

The |x| in acos(x) may be just a tad larger than 1.0 due to computationally issues. Best to check

// add
if (cosab > 1.0) cosab = 1.0;
else if (cosab < -1.0) cosab = -1.0;

cosab = 180 * acos(cosab) / M_PI;

Perhaps better precision

// sqrt(ax*ax+ay*ay)
hypot(ax, ay)

Other issues may exist too

Worth reviewing if proper formula is used: Law of cosines