Question

AP and AQ are two tangents of a circle with centre ‘O’ and ∠POQ = 60°, then ∠OPQ = 

(A) 30° 

(B) 45° 

(C) 60° 

(D) 90°

Answer 1

Correct option is: (C) 60°

In \(\triangle\) OPQ, we have

OP = OQ = radii of circle

\(\therefore\) \(\angle \)OQP = \(\angle \)....(1)

(Angles opposite to equal sides are equal in measure)

Also, \(\angle \) POQ + \(\angle \) OQP + \(\angle \) OPQ = 180°

= 2 \(\angle \) OPQ + 60° = 180° (\(\because\)\(\angle \) OPQ = 60° and from (1))

= 2 \(\angle \) OPQ = 180° - 60°  = 120°

= \(\angle \) OPQ = \(\frac {120^\circ}2 = 60^\circ\).

Answer 2

Correct option is: (C) 60°