How to solve this problem

Dear Student,

$$\begin{gathered} △ABC is an equilateral triangle \\ So, \angle ABC= \angle BCA =\angle BAC= 60° \{ all angles of an equilateral △are of 60°\} \end{gathered}$$
$$\begin{gathered} Now, \sin ce PAB\hspace{0.17em}is a straight line, \\ \angle ABP+\angle ABC= 180° \{ sum of angles on a straight line is 180°\} \\ \angle ABP+ 60°= 180° \{ \angle ABC=60° as it is an angle of equilateral△\} \\ \angle ABP= 120° ........eqs\left(1\right) \\ Now, in △ABP, \\ \angle APB +\angle ABP+ \angle PAB=180° \{ sum of angles of a △is 180°\} \\ 40° +120°+x=180° \{ from eqs 1 and figure\} \\ 160°+x=180° \\ x=20° \end{gathered}$$
$$\begin{gathered} \angle ACB=60° \{ angle of an equilateral △ABC\} \\ BCQ is a straight line \\ \angle ACB+\angle ACQ= 180° \{sum of angles on a straight line is 180°\} \\ 60°+ \angle ACQ=180° \\ \angle ACQ=120° ......eqs\left(2\right) \\ In △ACQ, \\ \angle ACQ+\angle CQA+\angle QAC=180° \\ 120°+30°+y=180° \{ from eqs 2 and figure\} \\ 150°+y=180° \\ y=30° \end{gathered}$$
$$\begin{gathered} In △PAQ, \\ \angle PAQ+\angle APQ+\angle PQA=180° \{sum of angles of a △is 180°\} \\ \angle PAQ+40°+30°=180° \{from figure\} \\ \angle PAQ+70°=180° \\ \angle PAQ= 180°-70°=110° \end{gathered}$$
Therefore, x=20$°$, y=30$°$, $\angle PAQ=110°$  Ans. 

Regards