Question

\In Fig. 10.13, XY and XY' are two parallel tangents to a circle with centre 0 and
another tangent AB with point of contact Cintersecting XY at A and X'Y' at B. Prove
that <AOB = 90°.
​

Answer 1

Answer:

Hope above attachment help you

Answer 2

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First, draw a circle with centre O. Choose an external point P and draw two tangents PA and PB at point A and point B respectively. Now, join A and B to make AB in a way that it subtends ∠AOB at the center of the circle. The diagram is as follows:

Ncert solutions class 10 chapter 10-12

From the above diagram, it is seen that the line segments OA and PA are perpendicular.

So, ∠OAP = 90°

In a similar way, the line segments OB ⊥ PB and so, ∠OBP = 90°

Now, in the quadrilateral OAPB,

∴∠APB+∠OAP +∠PBO +∠BOA = 360° (since the sum of all interior angles will be 360°)

By putting the values we get,

∠APB + 180° + ∠BOA = 360°

So, ∠APB + ∠BOA = 180° (Hence proved).