Question

QP is a tangent to a circle with centre O at a point P on the circle. If ΔOPQ is isosceles, then ∠OQP equals.

Options

• 30°

• 45°

• 60°

• 90°

Answer 1

45°

Explanation:

We know that, the radius and tangent are perpendicular at their point of contact.

Now, in isosceles triangle POQ

∠POQ + ∠OPQ + ∠OQP = 180°

⇒ 2∠OQP + 90° = 180°   ...(Equal sides subtend equal angles)

⇒ ∠OQP = 45°