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. Since \(\angle POR = 65^\circ\) is the angle at the centre on arc PR, the angle at any point on the major arc, such as S, subtending the same arc PR is half of this. \[ \angle PST = \angle PSR = \frac{65^\circ}{2} = 32.5^\circ \] Step 2: Recognize triangle SPT is right angled at P. Since SP is a diameter and PQ is tangent at P, the radius (and hence diameter SP) meets the tangent at a right angle, so \(\angle SPT = 90^\circ\). Step 3: Use the complementary angle property of a right triangle. In a right triangle, the two non-right angles always add up to \(90^\circ\), so: \[ \angle PTR = 90^\circ - \angle PST \] Step 4: Substitute and simplify. \[ \angle PTR = 90^\circ - 32.5^\circ = 57.5^\circ \] This matches option (C). \[ \boxed{\angle PTR = 57.5^\circ} \]
