Question

Two tangents TP and TQ are drawn to a circle with Centre O from an external point T. Prove that angle PTQ =2 angle OPQ ​

Answer 1

Solution:-

We know that lengths of tangent drawn from an external point to a Circle are equal.

∴ TP = TQ

Now, In ∆TPQ

⇒TP = TQ

∠TPQ = ∠TQP -------------eq(1)

( In ∆ Opposite sides have opposite angles equal to the)

NOW,

∠TPQ + ∠TQP + ∠PTQ = 180° ---( Angle sum property of Triangle)

2∠TPQ + PTQ -----( From eqn 1)

PTQ = 180 - 2TPQ ----( From eqn 2)

A tangent to a Circle is perpendicular to the radius through the point of contact

=> OP | PT

∠OPT = 90°

∠TPQ + ∠OPQ = 90°

∠OPQ = 90- ∠TPQ

=> 2∠OPQ =2(90 - ∠TPQ) = 180 - 2TPQ -----eqn(2)

From eqn (1) & (2) we get

∠PTQ = 2∠OPQ Hence Proved!

Answer 2

Answer:

Answer in attachment...(4 attachment)

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