Question

(Ptq) (D*2-pq+q*2) is equal to​

Answer 1

Answer:

Answer

Refer image

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=90o

or, ∠OPQ+∠TPQ=90o

or, ∠TPQ=90o−∠OPQ−−−(3)

In △PTQ

∠TPQ+∠PQT+∠QTP=180o (∴  Sum of angles triangle is 180o)

or, 90o−∠OPQ+∠TPQ+∠QTP=180o

or, 2(90o−∠OPQ)+∠QTP=180o [from (2) and (3)]

or, 180o−2∠OPQ+∠PTQ=180o

∴ 2∠OPQ=∠PTQ−−−− proved