In the given figure, what is the measure of ‘q’?

Clearly $\mathrm{p}={360}^{\circ }-{270}^{\circ }={90}^{\circ }$  (Angles at a point)
Through C, draw a line $\mathrm{\iota }$ parallel to AB and DE
 

$\begin{array}{l}\therefore {42}^{\circ }+\mathrm{x}={180}^{\circ }\text{ and }\mathrm{q}+\mathrm{y}={180}^{\circ }\\ \Rightarrow \mathrm{x}={180}^{\circ }-{42}^{\circ }={138}^{\circ }\\ \therefore \mathrm{y}={270}^{\circ }-{138}^{\circ }={132}^{\circ }\\ \therefore \mathrm{q}={180}^{\circ }-{132}^{\circ }={48}^{\circ }.\end{array}$