Question

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Answer 1

Step-by-step explanation:

Given, a circle with center O, from an external point T two tangents TP and TQ are drawn to the circle, where P and Q are the points of contact.

Prove:  ∠PTQ = 2∠OPQ

Proof:

TP = TQ {∴ Tangents from an external point are equal}

So, TPQ is an isosceles triangle,

∴  ∠TPQ = ∠TQP

⇒ ∠TPQ + ∠TQP + ∠PTQ = 180°

⇒ 2∠TPQ = 180° - ∠PTQ

⇒ ∠TPQ = 90° - (1/2)∠PTQ

⇒ (1/2)∠PTQ = 90° - ∠TPQ     ------- (i)

Also, OPT = 90° {Radius and Tangent are perpendicular}

⇒ ∠OPQ + ∠TPQ = 90°

⇒ ∠OPQ = 90° - ∠TPQ    ------- (ii)

From (i) & (ii), we get

⇒ (1/2)∠PTQ = ∠OPQ

⇒ ∠PTQ = 2∠OPQ.

Hope it helps!

Answer 2

TP = TQ {∴ Tangents from an external point are equal}

So, TPQ is an isosceles triangle,

∴  ∠TPQ = ∠TQP

⇒ ∠TPQ + ∠TQP + ∠PTQ = 180°

⇒ 2∠TPQ = 180° - ∠PTQ

⇒ ∠TPQ = 90° - (1/2)∠PTQ

⇒ (1/2)∠PTQ = 90° - ∠TPQ     ------- (i)

Also, OPT = 90° {Radius and Tangent are perpendicular}

⇒ ∠OPQ + ∠TPQ = 90°

⇒ ∠OPQ = 90° - ∠TPQ    ------- (ii)

From (i) & (ii), we get

⇒ (1/2)∠PTQ = ∠OPQ

⇒ ∠PTQ = 2∠OPQ.