Question

In the adjoining figure, AB is a diameter of the circle
with centre 0. AP, BQ and PRO are tangents. Prove that
angle POQ = 90°.

I need it quick, answer it fast.
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Answer 1

$\huge{\mathcal {\purple {A}\green {n}\pink {s}\blue {w}\purple{e}\green {r}}}$

☆qp is a tangent

and oR intersects pq

so if a line interested any angest it forms a right angle.

then while it cuts the other 2 sides are 45° each

and as we know the sum of a triangle is 180°

POQ=180°-90°

POQ=90°

$\huge\colorbox{orange}{çhű࿐ ❤}$

Answer 2

Answer:

☆qp is a tangent

and oR intersects pq

so if a line interested any angest it forms a right angle.

then while it cuts the other 2 sides are 45° each

and as we know the sum of a triangle is 180°

POQ=180°-90°

POQ=90°