# Thread: [SOLVED] law of cosines issue

1. ## [SOLVED] law of cosines issue

I have an issue solving the following problem. This starts using the law of cosines. Anyway, given the following, solve the triangle

side a = 12
b = 5
c = 13

law of cosines is $b^2 = a^2 + c^2 - 2accosB$ which yields me an angle B of 23 degrees, so far so good.

so next you're supposed to use the law of sines to solve for the next angle. I'm solving for A for the hell of it, so I get
$\frac{sin23^\circ}{5} = \frac{sinA}{12}$
or
$\frac{12sin23^\circ}{5} = sinA$
that gives me A = 69degrees...but according to the book, it's 67 degrees.

and if I solve for C instead, I get
$\frac{sin23^\circ}{5} = \frac{sinC}{13}$
or
$\frac{13sin23^\circ}{5} = sin C$
and that doesn't even work, since the left hand of the equation is great than 1, it's out of domain or whatever.

So...what am I doing wrong?

2. Do not round (or round to 4 decimal places) until you get to the end. This is where your error is coming from.

3. Originally Posted by ANDS!
Do not round (or round to 4 decimal places) until you get to the end. This is where your error is coming from.
Thanks a lot, I appreciate it. I'm thrilled it's that simple.