Question

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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

Explanation:

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

hope this will help you

Answer 2

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

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