PQ is tangent to a circle with centre O. If \(\angle POR = 65^\circ\), then m\(\angle PTR\) is
Show Hint
In any right-angled triangle like \(\Delta SPT\), the two acute angles must add up to \(90^\circ\).
Once you find the inscribed angle \(\angle PST = 32.5^\circ\), you can simply subtract it from \(90^\circ\) to find the answer:
\[ \angle PTR = 90^\circ - 32.5^\circ = 57.5^\circ \]
This shortcut avoids working with \(180^\circ\) and saves valuable time!
Step 1: Use the inscribed angle theorem first. The angle at the circumference is half the central angle for the same arc: $\angle PSR=\frac12\angle POR=\frac{65^\circ}{2}=32.5^\circ$. Step 2: Use the tangent-radius right angle as a shortcut. Since the diameter through $P$ meets tangent $PT$ at $90^\circ$, triangle $SPT$ is right-angled at $P$, so its two non-right angles add to $90^\circ$. Step 3: Subtract instead of using the full angle-sum equation. $\angle PTR=90^\circ-\angle PST=90^\circ-32.5^\circ=57.5^\circ$, matching option (C). \[ \boxed{57.5^\circ} \]
