From a point T, a tangent TP at point P is drawn to a circle with centre O. A secant TQR (point Q is near to point T) is drawn from point T. ∆PQR is inscribed into the circle by joining the points P, Q, and R. Draw lines OQ and OR. If ∠PTQ = 27° and ∠TPQ = 55°, what is the measure of ∠ROQ in degrees?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 86°.
Here, ∠PTQ = 27° and ∠TPQ = 55°
An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
So, ∠PQR = ∠PTQ + ∠TPQ = 27° + 55° = 82°
∠PRQ = ∠TPQ = 55° (By alternate segment theorem)
So, ∠RPQ = 180° - 55° - 82° = 180° - 137° = 43°
The angle at the center of a circle is twice the angle at the circumference.
∠ROQ = 2∠RPQ = 2 × 43° = 86°