Question

Two tangents tp and tq are drawn to a circle with centre o from an external point t prove that angle ptq equal to 2 opq

Answer 1

there are two methods.....

whichever u feel easy u can use...

both are right....

Answer 2

Answer:

Step-by-step explanation:

Given a circle with centre O ,

an external point T and two tangents

TP and TQto the circle ,

Where P , Q are the points of contact.

We need to prove that ,

<PTQ = 2<OPQ

Let <PTQ = x°

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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, TPQ is an Isosceles triangle .

Therefore ,

<TPQ + <TQPn+ <PTQ = 180°

[ Angle sum property ]

=> <TPQ = <TQP = ( 180° - x )/2

= 90° - x/2

[ By , theorem

The tangent at any point of a circle is

perpendicular to the radius through

the point of contact ]

<OPT = 90°

<OPQ = <OPT - <TPQ

= 90° - ( 90 - x/2 )

= x/2

= <PTQ/2

This gives ,

<OPQ = <PTQ/2

Therefore ,

<PTQ = 2<OPQ

Similarly

<PTQ = 2<OQP