Question

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

angle PTO = angle OPQ
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Answer 1

Let ∠PTQ = x

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a triangle = 180°.

By property 1,

TP = TQ (tangent from T)

Therefore, ∠TPQ = ∠TQP

Now,

By property 3 in ∆PAB,

∠TPQ + ∠TQP + ∠PTQ = 180°

⇒ ∠TPQ + ∠TQP = 180° – ∠PTQ

⇒ ∠TPQ + ∠TQP = 180° – x

By property 2,

∠TPO = 90°

Now,

∠ TPO = ∠ TPQ + ∠OPQ

⇒ ∠OPQ = ∠ TPO – ∠ TPQ

⇒ ∠PTQ = 2∠OPQ

Hence, Proved

Step-by-step explanation:

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Answer 2

Answer:

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