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: Find the inscribed angle from the central angle.
By the central angle theorem, the angle at $S$ on the circle subtended by arc $PR$ is half of $\angle POR$: $\angle PST = \frac{65^\circ}{2} = 32.5^\circ$. Step 2: Note that SPT is a right triangle.
Since $SP$ is a diameter and $PT$ is tangent at $P$, the radius (and hence the diameter $SP$) is perpendicular to the tangent, so $\angle SPT = 90^\circ$. Step 3: Use the complementary angle property instead of the full angle sum.
In a right triangle, the two non-right angles add to $90^\circ$, so $\angle PST + \angle PTR = 90^\circ$.
\[ \angle PTR = 90^\circ - 32.5^\circ = 57.5^\circ \]
\[ \boxed{\angle PTR = 57.5^\circ} \]
