(1)ATSQ is a cyclic quadrilateral.

TAQ + TSQ = 180˚ [Opposite angles of cyclic quadrilateral are supplementary]

(2)Given PR is the tangent.

Seg AQ is the secant.

AQP = ½ m(arc AQ) [Angle between tangent and secant]

ASQ = ½ m(arc AQ) [Inscribed angle]

AQP ASQ

(3)QTS = QAS [ Angles inscribed in same arc are equal]

∠QTS = ½ m(arc QS) [Inscribed angle theorem]

But, ∠SQR = ½ m(arc QS) [Theorem of angle between tangent and secant]

∴ ∠QTS ∠SQR

(4) ∠TQS = ∠TAS [Angles inscribed in the same arc]

∴ ∠TQS = 65°

Now, ∠TQS = ½ m(arc TS) [Inscribed angle theorem]

∴ 65°= ½ m(arcTS)

∴ m(arc TS) = 2×65°

∴ m(arc TS) = 130°

(5) ∠AQP + ∠AQS + ∠SQR = 180° [linear pair]

∴ 42° + ∠AQS + 58° = 180°

∴∠AQS + 100° = 180° ………… (i)

AQS = 180-100 =80˚

But, AQST is a cyclic quadrilateral.

∴ ∠AQS + ∠ATS = 180° ………… (ii) [Opposite angles of cyclic quadrilateral are supplementary]

∴ ∠ATS = 180-80 = 100° [From (i) and (ii)]