CBSE Class 10 Answered
circles
Asked by deepitapai | 19 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 | 20 Jan, 2009, 11:06: AM
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