Question

PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120° , then ∠OPQ is 

A. 60° 

B. 45° 

C. 30° 

D. 90°

Answer 1

Answer C. 30°

Given: 

∠POR = 120°

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

Property 2: Sum of all angles of a straight line = 180°. 

Property 3: Sum of all angles of a triangle = 180°. 

By property 1, 

∠PQO = 90° 

By property 2, 

POQ + ∠POR = 180° 

⇒ ∠POQ + 120° = 180° 

⇒ ∠POQ = 180° - 120°

⇒ ∠POQ = 60° 

Now by property 3 in ∆OPQ, 

∠POQ + ∠PQO + ∠OPQ = 180° 

⇒ ∠OPQ = 180° - ∠POQ + ∠PQO 

⇒ ∠OPQ = 180° - (60° + 90°) 

⇒ ∠OPQ = 180° - 150° 

⇒ ∠OPQ = 30° 

Hence, ∠OPQ = 30°