Correct answer is (b) 50°
From △OPA and △OBP
OA = OB (Radii of the same circle)
OP (Common side)
PA = PB (Since tangents drawn from an external point to a circle are equal)
\(\therefore \triangle OPA\cong \triangle OPB\) (SSS rule)
\(\therefore \triangle APO=\triangle BPO\)
\(\therefore \angle APO=\frac{1}{2}\angle APB=40^\circ\)
And ∠OAP = 90°
(Since tangents drawn from an external point are perpendicular to the radius at point of contact)
