 Problem of the Week

Problem E and Solution

Angled III

Problem

In the circle with centre $$R$$ below, $$PQ$$ is a diameter. Point $$S$$ is a point on the circumference of the circle other than $$P$$ or $$Q$$.
Determine the measure of $$\angle PSQ$$. Solution

Join $$S$$ to the centre $$R$$. Since $$RP$$, $$RQ$$ and $$RS$$ are radii of the circle, $$RP=RQ=RS$$.
Since $$RP=RS$$, $$\triangle PRS$$ is isosceles and $$\angle RPS = \angle RSP = x^{\circ}$$.
Since $$RQ=RS$$, $$\triangle QRS$$ is isosceles and $$\angle RQS = \angle RSQ = y^{\circ}$$.
This new information is marked on the following diagram. The angles in a triangle add to $$180^{\circ}$$, so in $$\triangle PQS$$ \begin{aligned} \large \angle PSQ + \angle QPS + \angle PQS &=&180^{\circ}\\ (x^{\circ}+y^{\circ})+x^{\circ}+y^{\circ}&=&180^{\circ}\\ 2(x^{\circ}+y^{\circ})&=&180^{\circ}\\ x^{\circ}+y^{\circ}&=&90^{\circ}\\\end{aligned}

But $$\angle PSQ=x^{\circ}+y^{\circ}$$, so $$\angle PSQ=90^{\circ}$$.
This result is often expressed as a theorem for circles:
An angle($$\angle PSQ$$) inscribed in a circle by a diameter ($$PQ$$) of the circle is $$90^{\circ}$$.