Hey people. Here's the question.

Find the missing parts of the triangle by using the laws of cosines.

What's known is, B=27, a=11, c=22

Solving for b,

$\displaystyle b^2=11^2+22^2-2(11)(22)(cosB)$

$\displaystyle b^2=121+484-2(242)(.891)$

$\displaystyle b^2=605-432$

$\displaystyle b^2=\sqrt{173}$

$\displaystyle b=13.2$

Solving for A

$\displaystyle \frac{11}{sinA}=\frac{13.2}{sin27}$

Cross multiply.

$\displaystyle \frac {sin27\cdot 11}{13.2}=.3783(sin^-^1)=~22$

Solving for C

$\displaystyle \frac{22\cdot sin27}{13.2}=.7567(sin^-^1)=~49$

180-27-49-22=82? Where did I go so wrong. This formula has always seemed to work for me.