Answer:
From the given figure in the textbook, join OC. The diagram will now be as-
![](https://www.learnatnoon.com/s//wp-content/uploads/2021/09/k-1.png)
Now using the SSS congruency the triangles △OPA and △OCA are similar:
(i) OP = OC since they are the same circle’s radii
(ii) AO = AO as it is the common side
(iii) AP = AC as these are the tangents from point A
As a result, △OPA ≅ △OCA
In the same way,
△OQB ≅ △OCB
As a result,
∠POA = ∠COA … (Eq. (i))
And, ∠QOB = ∠COB … (Eq. (ii))
Because the line POQ is a straight line, it can be regarded as a circle’s diameter.
As a result, ∠POA +∠COA +∠COB +∠QOB = 180°
Now, from eq. (i) and eq. (ii) we get,
2∠COA+2∠COB = 180°
∠COA+∠COB = 90°
Therefore, ∠AOB = 90°