circles...............

Asked by kirtichawla94 | 20th Nov, 2009, 07:08: PM

Expert Answer:

Let OP meets the circle at a point Q.

It  is given that OP = 2r

So, OP OP = QP + OP

QP = r.

Therfore, Q is the mid point of OP.

Now AP is tnagent at A and OA is radius

OA is perpendicular to AP.

Therefore, Triangle OAP is right triangle at A . Q is the mid point of hypotnuese OP.

Therfore, AQ = OQ = OA

i.e Triangle OAQ is equilateral.

So, AOQ = 60o

Therefore, APO = 180o - 90o- 60= 30o

APB = 2APO  = 60o

Also, PA = PB

PAB = PBA

Now in triangle APB, we have

PAB  + PBA + APB = 180o 

PAB = 60o

So, triangle APB is an equilateral triangle.

Answered by  | 24th Nov, 2009, 10:36: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.