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:

OP is the radius and PT is tangent to the circle.

So, ∠OPT=90

∘

But ∠QPT=70

∘

∠OPQ=90

∘

−70

∘

=20

∘

From right ΔOPQ

OP=OQ (radius of circle)

∠OQP=∠OPQ=20

∘

and ∠POQ=180

∘

−(∠OPQ+∠OQP)

=180

∘

−(20

∘

+20

∘

)

=180

∘

−40

∘

=140

∘

Measure of ∠POQ is 140 degrees

Answer 2

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°

hope it helps...