CEMC Gauss (Grade 7), 2014

Solution 1

Since $PQR$ is a straight line segment, then $\angle PQR=180\degree$.

Since $\angle SQP+ \angle SQR=180\degree$, then $\angle SQR=180\degree -\angle SQP=180\degree-75\degree=105\degree$.

The three angles in a triangle add to $180\degree$, so $\angle QSR+\angle SQR+\angle QRS=180\degree$, or
$\angle QSR=180\degree-\angle SQR-\angle QRS=180\degree-105\degree-30\degree=45\degree$.

Solution 2

The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles of the triangle.

Since $\angle SQP$ is an exterior angle of $\triangle SQR$, and the two opposite interior angles are $\angle QSR$ and $\angle QRS$, then $\angle SQP=\angle QSR+\angle QRS$.
Thus, $75\degree=\angle QSR + 30\degree$ or $\angle QSR=75\degree-30\degree=45\degree$.