Question

In figure TP and TQ are tangents drawn from an external point T to a circle with Centre O. If QTO = 25°, the find measure of POQ?​

Answer 1

Step-by-step explanation:

We know that length of taughts drawn from an external point to a circle are equal

∴ TP=TQ−−−(1)

4∴ ∠TQP=∠TPQ (angles of equal sides are equal)−−−(2)

Now, PT is tangent and OP is radius.

∴ OP⊥TP (Tangent at any point pf circle is perpendicular to the radius through point of cant act)

∴ ∠OPT=90

o

or, ∠OPQ+∠TPQ=90

o

or, ∠TPQ=90

o

−∠OPQ−−−(3)

In △PTQ

∠TPQ+∠PQT+∠QTP=180

o

(∴ Sum of angles triangle is 180

o

)

or, 90

o

−∠OPQ+∠TPQ+∠QTP=180

o

or, 2(90

o

−∠OPQ)+∠QTP=180

o

[from (2) and (3)]

or, 180

o

−2∠OPQ+∠PTQ=180

o

∴ 2∠OPQ=∠PTQ−−−− proved