Finding Angle

$\angle BCA=30^o$ , hence $CP$ is the external angle bisector of $\angle BCA . \therefore \frac{BP}{PA}=\frac{BC}{CA}$

$AD$ is altitude and median (isosceles !) . $CD=BD$ .

By Menelaus theorem , $\displaystyle \frac{CD}{DB}.\frac{BP}{PA}.\frac{AQ}{QC}=1 \Rightarrow \frac{QC}{QA}=\frac{BP}{PA}=\frac{BC}{CA}=\frac{BC}{AB}$

$\displaystyle \therefore \frac{QC}{QA}=\frac{BC}{AB}$

$BQ$ bisects $\angle ABC$ , $\angle ABQ=\frac{1}{2}\angle ABC=15^o$

$\angle BQA=180^o-\angle BAQ-\angle ABQ=180^o-120^o-15^o=45^o$