CEMC Fermat, 2014

Solution 1

$\angle SRP$ is an exterior angle for $\triangle PQR$.

Therefore, $\angle SRP = \angle RPQ + \angle RQP$ or $(180-x)^\circ = 30^\circ + 2x^\circ$.

Thus, $180-x = 30+2x$ or $3x=150$ and so $x=50$.

Solution 2

Since $QRS$ is a straight line segment and $\angle SRP = (180-x)^\circ$, then $\angle PRQ$ is the supplement of $\angle SRP$ so $\angle PRQ = x^\circ$.

Since the angles in a triangle add to $180^\circ$, then $\angle PRQ + \angle PQR + \angle RPQ = 180^\circ$, and so $x^\circ + 2x^\circ + 30^\circ = 180^\circ$.

From this, we obtain $3x = 150$ and so $x=50$.