circles

Asked by deepitapai | 19th Jan, 2009, 07:26: PM

Expert Answer:

Consider Quadrilateral OAPB,

Angle A =90 degrees

Angle B=90 degrees (as they are angle made by the radius thru' the point of contact with the tangent0

So, Angle APB+ Angle AOB=180 ( as sum of the angles of a quadrilateral is 360, sum of two of them is 180 , so sum of the remaining two is also 180)

Now,

 consider triangle AOB

AO=BO(radii)

So,

Angle OAB= Angle OBA

Let each of them = x ( say)

So Angle AOB=180-2x (as sum of the angles of a triangle is 180 degrees)

But we have already proved that Angle AOB+ Angle APB=180 degres

So Angle APB=180-Angle AOB=180-(180-2x)=2x=2(Angle OAB)

Hence proved.

Answered by  | 20th Jan, 2009, 11:06: AM

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