Question

two tangents TP and TQ arw drawn to a circle with centre O from the external point T.Prove that angle PTQ=2angleOPQ

Answer 1

Hi ,

First draw a rough diagram from the information given in the problem. And join P and Q.

We are given a circle with centre O,

an external point T and two tangents

PT and PQ to the circle , where P , Q are the points of contact.

We need to prove that

< PTR = 2 ×
Let
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By the theorem,

The lengths of tangents drawn from an external point to a circle are equal.

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TP = TQ,

So triangle TPQ is an isosceles

triangle.

Therefore,


(sum of the three angles in a triangle )

< TPQ = < TQP = 1/ 2 ( 180 - θ )

= 90 - θ /2

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By the theorem

The tangent at any point of a circle is

perpendicular to the radius through

the point of contact.

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< OPQ = < OPT - < TPQ

= 90 - ( 90 - θ / 2 ) = θ / 2

= 1/2
This gives

< OPQ = 1/ 2 < PTQ

Therefore,


Similarly

< PTQ = 2 < OQP

Hence proved.

I hope this will useful to u.

Answer 2

Answer:

Step-by-step explanation: