Discussion

Explanation:

In quadrilateral PQOT, ∠QOT = 360 − ∠QPT − ∠PQO − ∠PTO 

Since PQ and PT are tangents to the circle at Q and T, ∠PQO = ∠PTO = 90°

∴ ∠QOT = 360 − 55 − 90 − 90 = 125° 

∴ ∠QOS + ∠SOT  = ∠QOT = 125° 

Now, when two tangents are drawn from a point to the circle, the line joining the external point and the centre of the circle is the angle bisector of the central angle subtended by the two tangential points at the centre.   
 
For instance, when PQ and PT are tangents from P to the circle with centre O, then line PO is the angle bisector of ∠QOT. 

Similarly, AQ and AS are tangents from A to the circle with centre O. Hence, AO is the angle bisector of ∠QOS. 

∴ ∠QOA = ∠AOS = (∠QOS)/2

Similarly, for quadrilateral BSOT, ∠TOB = ∠BOS = (∠SOT)/2 

Now, ∠AOB = ∠AOS + ∠BOS 

= (∠QOS)/2 + (∠SOT)/2 = (∠QOS + ∠SOT)/2 

= 125/2 = 62.5°

Hence, option (b).

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