Question

Studying for a test, need some help solving this question

Answer 1

Solution

Part A

To find C

$_{\angle\text{ C+}\angle A+\angle B=180^0(Sum\text{ of angles in a triangle\rparen}}$$\begin{gathered} \angle C+33.84+19.32=180 \\ \angle C+53.16=180 \\ \angle C=180-53.16 \\ \angle C=126.84^0 \\ \end{gathered}$

Part B

To find a

Using sine rule

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$$\begin{gathered} \frac{sin33.84}{a}=\frac{sin126.84}{13.95} \\ cross\text{ multiply} \\ asin126.84=13.95sin33.84 \\ divide\text{ both side by sin126.84} \\ \\ a=9.70672 \end{gathered}$

Part C

To find b

Using sine rule

$\begin{gathered} \frac{sin33.84}{9.70672}=\frac{sin19.32}{b} \\ \\ \\ b=5.76683 \end{gathered}$