Question

PQ is a tangent to a circle with centre O at point P. If ∆OPQ is an

isosceles triangle with P as vertex, then find ∠OQP​

Answer 1

Step-by-step explanation:

OPQ=90°( tanget is perpendicular to line joining by centre)

if PQO is isoscles triangle then

POQ+OQP+OPQ=180

x+x+90°=180°. (POQ=OQP because given

2x=90 ∆ le is isoscles)

x=45°

OQP=45°

Answer 2

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pq is tangent and Oq is radius

we know that tangent is perpendiculatr to radius

Angle Opq = 90°

Angle OPQ is isoceles

• OP = PQ

Angle OQP = Angle POQ

SO,

angle Oqp = 1/2 X 90°

= 45°

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