**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 \(90^{\circ}\).*