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  • In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that angle AOB = 90°.

9.  In Fig. 10.13, XY and X'Y'are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that \angle AOB = 90°.

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To prove- \angleAOB = 90^0
Proof-
In \DeltaAOP and \DeltaAOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence,  \DeltaAOP \cong \DeltaAOC
and by CPCT,  \angle PAO = \angle OAC
\Rightarrow \angle PAC = 2\angle OAC..................(i)

Similarly, from \DeltaOBC and \DeltaOBQ, we get;
\angleQBC = 2.\angleOBC.............(ii)

Adding eq (1) and eq (2)

\anglePAC + \angle QBC = 180 
2(\angleOBC + \angleOAC) = 180
(\angleOBC + \angleOAC) = 90

Now,  in \Delta OAB,
Sum of interior angle is 180.
So, \angleOBC + \angleOAC + \angleAOB = 180
\therefore \angleAOB = 90
hence proved.

Posted by

manish

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