In figure 3.91, line PR touches the circle at point Q. Answer the following questions with the help of the figure.

(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)]