Question

Studying for a test, need some help solving this question

Answer 1

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

`(sinA)/(a)=(sinB)/(b)=(sinC)/(c)`
`\begin{gathered} (sin33.84)/(a)=(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} (sin33.84)/(9.70672)=(sin19.32)/(b) \\ \\ \\ b=5.76683 \end{gathered}`