Question

Tangents PA and PB are drawn to a circle with centre O from a point P outside the circle. Prove that angle OPA is equal to angle OAM

Answer 1

Answer:

In ∆POA & ∆POB

PA = PB ( tangents on circle from external point are equal)

OA=OB(Radius of circle)

OP=OP(common side)

So both are congruent .

So

$\angle \: opb \: = \angle \: opa$