Question

Determine the missing parts of triangle $ABC$ from the given information.
$b=29,c=17,A={103}^o$

Answer 1

We know $\cos A=\frac{b^2+c^2-a^2}{2bc}$
So we get $a^2=1130-986\cos \left(103\right)=1345$

$\Rightarrow a=\sqrt{1345}=36.67$
By sine rule, we get $\frac{a}{\sin A}=\frac{b}{\sin B}$

$\Rightarrow \sin B=\frac{b}{a}\sin A=0.771$

$\Rightarrow B=arc\sin \left(0.771\right)=50.22$ degrees
$\Rightarrow C=180-103-50.44=26.78$ degrees