You're on a roll — a) Find $\angle PSO$. Step 1: Identify the properties of triangle POS. OP and OS are both radii of the circle, so $OP = OS$. This means triangle POS is an isosceles triangle. Step 2: Use the property of isosceles triangles. In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, $\angle PSO = \angle OPS$. From the diagram, we are given $\angle OPS = 25^\circ$ (assuming the typo $\angle OPC$ refers to $\angle OPS$). $$ \angle PSO = 25^\circ $$ $$\boxed{\angle PSO = 25^\circ}$$ b) Find $\angle PST$. Step 1: Use the property of a tangent and radius. QRST is a tangent to the circle at point S. OS is the radius to the point of tangency. The radius to the point of tangency is perpendicular to the tangent. Therefore, $\angle OST = 90^\circ$. Step 2: Calculate $\angle PST$. We know $\angle OST = 90^\circ$ and from part (a), $\angle PSO = 25^\circ$. $$ \angle PST = \angle OST - \angle PSO $$ $$ \angle PST = 90^\circ - 25^\circ $$ $$ \angle PST = 65^\circ $$ $$\boxed{\angle PST = 65^\circ}$$ c) Find $\angle PRQ$. Step 1: Find $\angle POS$. In triangle POS, the sum of angles is $180^\circ$. We know $\angle OPS = 25^\circ$ and $\angle PSO = 25^\circ$. $$ \angle POS = 180^\circ - (\angle OPS + \angle PSO) $$ $$ \angle POS = 180^\circ - (25^\circ + 25^\circ) $$ $$ \angle POS = 180^\circ - 50^\circ $$ $$ \angle POS = 130^\circ $$ Step 2: Find $\angle ROS$. POR is a straight line, so $\angle POS$ and $\angle ROS$ are angles on a straight line. $$ \angle ROS = 180^\circ - \angle POS $$ $$ \angle ROS = 180^\circ - 130^\circ $$ $$ \angle ROS = 50^\circ $$ Step 3: Find $\angle PRQ$ (which is $\angle ORS$). Consider triangle ORS. We know $\angle OSR = 90^\circ$ (radius OS perpendicular to tangent QRST) and $\angle ROS = 50^\circ$. The sum of angles in triangle ORS is $180^\circ$. $$ \angle ORS + \angle OSR + \angle ROS = 180^\circ $$ $$ \angle ORS + 90^\circ + 50^\circ = 180^\circ $$ $$ \angle ORS + 140^\circ = 180^\circ $$ $$ \angle ORS = 180^\circ - 140^\circ $$ $$ \angle ORS = 40^\circ $$ Since $\angle PRQ$ is the same as $\angle ORS$: $$ \angle PRQ = 40^\circ $$ $$\boxed{\angle PRQ = 40^\circ}$$ What's next?