In \(\triangle PQS\), \(R\) lies on \(PQ\) such that \(PR=RQ=RS\) and \(\angle QRS =z\degree\).
Determine the measure of \(\angle PSQ\).
Extension: Consider the circle with centre \(O\) and diameter \(AB\). Suppose point \(C\) is a point on the circumference of the circle other than \(A\) or \(B\). Determine the measure of \(\angle ACB\).
Let \(\angle RPS = x\degree\) and \(\angle RQS = y\degree\).
In \(\triangle PRS\), since \(PR=RS\), \(\triangle PRS\) is isosceles and \(\angle RSP =\angle RPS=x\degree\).
Similarly, in \(\triangle QRS\), since
\(RQ=RS\), \(\triangle QRS\) is isosceles and \(\angle RSQ=\angle RQS=y\degree\).
From here we proceed with two different solutions.
Solution 1
Since \(PRQ\) is a straight line, \(\angle PRS + \angle QRS=180\degree\). Since \(\angle QRS =z\degree\), we have \(\angle PRS =180 - z\degree\).
The angles in a triangle sum to \(180\degree\), so in \(\triangle PRS\) \[\begin{aligned} \angle RPS +\angle RSP + \angle PRS&=180\degree\\ x\degree + x\degree + 180\degree - z\degree&=180\degree\\ 2x&=z\\ x&=\frac{z}{2} \end{aligned}\] The angles in a triangle sum to \(180\degree\), so in \(\triangle QRS\) \[\begin{aligned} \angle RQS +\angle RSQ + \angle QRS&=180\degree\\ y\degree + y\degree + z\degree&=180\degree\\ 2y&=180-z\\ y&= \frac{180-z}{2} \end{aligned}\] Then \(\angle PSQ=\angle RSP + \angle RSQ=x\degree+y\degree=\frac{z}{2}\degree+\left( \frac{180-z}{2} \right)\degree=\left(\frac{180}{2}\right)\degree=90\degree\).
Therefore, the measure of \(\angle PSQ\) is \(90\degree\).
It turns out that it is not necessary to determine expressions for \(x\) and \(y\) in terms of \(z\) to solve this problem, as we’ll see in Solution 2.
Solution 2
The angles in a triangle sum to \(180\degree\), so in \(\triangle PQS\) \[\begin{aligned} \angle QPS +\angle PSQ + \angle PQS&=180\degree\\ x\degree + (x\degree+y\degree) + y\degree&=180\degree\\ (x\degree+ y\degree) + (x\degree+y\degree) &=180\degree\\ 2(x\degree+y\degree)&=180\degree\\ x\degree+y\degree&=90\degree \end{aligned}\] But \(\angle PSQ=\angle RSP + \angle RSQ=x\degree+y\degree=90\degree\).
Therefore, the measure of \(\angle PSQ\) is \(90\degree\).
Extension Solution:
Join \(C\) to the centre \(O\). Since \(OA\), \(OB\) and \(OC\) are radii of the circle, \(OA=OB=OC\). Let \(\angle CAO = x\degree\) and \(\angle CBO = y\degree\).
Since \(OA=OC\), \(\triangle AOC\) is isosceles and \(\angle ACO = \angle CAO = x\degree\).
Since \(OB=OC\), \(\triangle BOC\) is isosceles and \(\angle BCO = \angle CBO = y\degree\).
The angles in a triangle add to \(180\degree\), so in \(\triangle ABC\) \[\begin{aligned} \angle ACB + \angle CAB + \angle CBA &=180\degree\\ (x\degree+y\degree)+x\degree+y\degree&=180\degree\\ 2(x\degree+y\degree)&=180\degree\\ x\degree+y\degree&=90\degree \end{aligned}\] But \(\angle ACB=x\degree+y\degree\), so \(\angle ACB=90\degree\).
This result is often expressed as a theorem for circles: An angle inscribed in a circle by a diameter of the circle is \(90\degree\).