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Two tangents TP and TQ are drawn to a circle with center O from an external point T. Then ∠PTQ= ____ ∠OPQ.

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Detailed Solution

TP = TQ …(1)
⇒∠TQP=∠TPQ …(2)
OP is perpendicular to TP
⇒∠OPT=

90^{\circ}

⇒∠OPQ + ∠TPQ=

90^{\circ}

⇒∠TPQ=

90^{\circ}

−∠OPQ …(3)
In triangle PTQ,
⇒∠TPQ + ∠PQT + ∠QTP =

180^{\circ}

90^{\circ}

−∠OPQ +∠TPQ + ∠QTP =

180^{\circ}

⇒ 2(

90^{\circ}

−∠OPQ) + ∠QTP =

180^{\circ}

⇒

180^{\circ}

−2∠OPQ + ∠PTQ =

180^{\circ}

⇒ 2∠OPQ = ∠PTQ
⇒ 2(

90^{\circ}

− ∠OPQ) + ∠QTP =

180^{\circ}

⇒

180^{\circ}

− 2∠OPQ + ∠PTQ =

180^{\circ}

⇒ 2∠OPQ = ∠PTQ
Two tangents TP and TQ are drawn to a circle with center O from an external point T. Then ∠PTQ= 2∠OPQ.

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