Question

5. If PT is a tangent to a circle with centre o
and PQ is a chord of the circle such that
OPT = 70°, then find the measure of angle POQ.
​

Answer 1

Answer:

We know that the radius and tangent are perpendicular at their point of contact.

∴ ∠OPT = 90°

Now, ∠OPQ = ∠OPT - ∠TPQ = 90°

-70° = 20°

Since, OP = OQ as both are radius

∴ ∠OPQ = ∠OQP = 20° (Angles opposite to equal sides are equal)

Now, In isosceles Δ POQ

∠POQ + ∠OPQ + ∠OQP = 180°

(Angle sum property of a triangle)

⇒ ∠POQ =180° - 20° = 140°