Finding angles of triangle by given sides in C programming - c

Question

I run it in my code::blocks to show 3 angles of triangle where 3 sides are given by the user. But if I run and give 3 sides such as 3,4,5 output will be -1.#J -1.#J -1.#J. What's wrong with my code?
#include <stdio.h>
#include <math.h>
int main()
{
float a, b, c, A, B, C, R, s, pi, area;
pi = acos(-1);
scanf("%f %f %f", &a, &b, &c);
s = (a+b+c)/2;
area = sqrt(s*(s-a)*(s-b)*(s-c));
R = (a*b*c)/(4*area);
A = (180/pi)*asin(a/2*R);
B = (180/pi)*asin(b/2*R);
C = (180/pi)*asin(c/2*R);
printf("%.2f %.2f %.2f", A, B, C);
return 0;
}

Answer 1

You need more parentheses. You have:
A = (180/pi)*asin(a/2*R);
You need:
A = (180/pi)*asin(a/(2*R));
What you wrote is equivalent to:
A = (180 / pi) * asin((R * a) / 2);
You should also check your inputs so that a 'triangle' with sides 1, 1, 3 is rejected, for example. Negative and zero lengths should probably be rejected too.
Revised code
#include <stdio.h>
#include <math.h>
int main(void)
{
float a, b, c, A, B, C, R, s, pi, area;
pi = acos(-1);
if (scanf("%f %f %f", &a, &b, &c) != 3)
{
fprintf(stderr, "Failed to read 3 numbers\n");
return 1;
}
if (a <= 0 || b <= 0 || c <= 0)
{
fprintf(stderr, "Sides must be strictly positive\n");
return 1;
}
s = (a + b + c) / 2;
if (a > s || b > s || c > s)
{
fprintf(stderr, "The three sides %.2f, %.2f, %.2f do not form a triangle\n",
a, b, c);
return 1;
}
area = sqrt(s * (s - a) * (s - b) * (s - c));
R = (a * b * c) / (4 * area);
A = (180 / pi) * asin(a / (2 * R));
B = (180 / pi) * asin(b / (2 * R));
C = (180 / pi) * asin(c / (2 * R));
printf("Sides: %6.2f %6.2f %6.2f\n", a, b, c);
printf("Angles: %6.2f %6.2f %6.2f\n", A, B, C);
return 0;
}
Note that error messages are reported on standard error. All lines of output end with a newline in the format. The input data is echoed. All of these are good practices.
Example runs
$ triangle <<< "0 1 3"
Sides must be strictly positive
$ triangle <<< "-1 1 3"
Sides must be strictly positive
$ triangle <<< "1 1 3"
The three sides 1.00, 1.00, 3.00 do not form a triangle
$ triangle <<< "3 4 5"
Sides: 3.00 4.00 5.00
Angles: 36.87 53.13 90.00
$ triangle <<< "3 3 3"
Sides: 3.00 3.00 3.00
Angles: 60.00 60.00 60.00
$ triangle <<< "1 1.4141 1"
Sides: 1.00 1.41 1.00
Angles: 45.00 nan 45.00
$ triangle <<< "1 1 1.4141"
Sides: 1.00 1.00 1.41
Angles: 45.00 45.00 nan
$ triangle <<< "1 1 1.414"
Sides: 1.00 1.00 1.41
Angles: 45.01 45.01 89.98
$ triangle <<< "1 1 1.41421356237309504880"
Sides: 1.00 1.00 1.41
Angles: 45.00 45.00 nan
$
I'm a little puzzled about the nan values. However, changing the data type from float to double and adjusting the scanf() formats appropriately (%lf instead of %f) seems to 'resolve' (maybe 'evade' is a better word) that problem.

Answer 2

It should be
A = (180/pi)*asin(a/(2*R));
rather than
A = (180/pi)*asin(a/2*R);
Insufficient parentheses.

Answer 3

Additional detail:
asin(x) is only defined for values [-1.0 ... 1.0]. With select input, corrected code may get a a/(2*R) that is just a little bit greater than 1.0 due to calculation issues with floating point.
s = (a+b+c)/2;
// Jonathan Leffler tests
area = sqrt(s*(s-a)*(s-b)*(s-c));
R = (a*b*c)/(4*area);
// add test
double R2 = R*2.0;
A = (a>R2) ? 180/pi : (180/pi)*asin(a/R2);
...

Answer 4

I could not get your posted code to work, even given all the comments and prior answers.
However, this works very well.
Note: I did not include code to error check the call to scanf() to keep the presentation simple
#include <stdio.h>
#include <math.h>
#define RAD_2_DEG (57.2958)
#define TOTAL_DEG (180.0)
int main( void )
{
int a, b, c; // user entered lengths of sides of triangle
double A, B, C; // calculated angle values in degrees
scanf( "%d %d %d", &a, &b, &c );
// note: acos returns the angle in radians
// and one radian is (approx) 57.2958 degrees
A = RAD_2_DEG * acos((double)(b*b + c*c - a*a)/(2.0*b*c));
B = RAD_2_DEG * acos((double)(c*c + a*a - b*b)/(2.0*a*c));
// third angle done this way to absorb fractional degree errors
C = TOTAL_DEG -(A + B);
printf("%.2f %.2f %.2f", A, B, C);
return 0;
}