Question

✒Answer please

Best answer will be marked as brainlist..........

Two tangents TP and TQ are drawn to a circle with centre O from an external point T.

Prove. that <PTQ = 2<OPQ.​

Answer 1

Answer:

We know that, the lengths of tangents drawn from an external point to a circle are equal.

∴ TP = TQ

In ΔTPQ,

TP = TQ

⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)

∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)

∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))

⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)

We know that, a tangent to a circle is perpendicular to the radius through the point of contact.

OP ⊥ PT,

∴ ∠OPT = 90º

⇒ ∠OPQ + ∠TPQ = 90º

⇒ ∠OPQ = 90º – ∠TPQ

⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)

From (1) and (2), we get

∠PTQ = 2∠OPQ

Answer 2

Step-by-step explanation:

Given:

TP and TQ are tangents to the circle with centre O.

To Prove:

<PTQ = 2<OPQ

Let <PTQ be A

TP = TQ ............(th. 10.2)

So,TPQ is an isosceles ∆

Therefore, <TPQ = <TQP = 1/2 (180°-A)

=90° - 1/2 A

<OPT = 90°............(th. 10.1)

So, <OPQ = <OPT - <TPQ = 90° - (90° - 1/2 A)

=1/2 A

=1/2<PTQ

Therefore, <PTQ = 2<OPQ

Hence Proved