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Question:

Two tangents are drawn from a point P on the circle with centre at O, touching the circle at point Q and T respectively. Another tangent AB touches the circle at point S. If angle QPT=55°, find the angle AOB=?

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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).