Please solve question no.b.

Dear Student
Let O be the center of the circle.

$$\begin{gathered} In △ACT \\ \angle ACT +\angle CTA +\angle TAC =180° (Angle sum property of trinagle) \\ 48° +36° + \angle TAC = 180° \\ \angle TAC = 96 ° \\ \angle CAB = \angle ACT +\angle CTA ( Exterior angle ) \\ \angle CAB = 48 + 36 =84° \\ OC \perp CT ( Tangent meet radius at 90°) \\ \Rightarrow \angle OCT =90° \\ Now, \\ \angle OCA = \angle OCT -\angle ACT \\ = 90 -48 \\ =42° \\ OC = OA (radius ) \\ \angle OCA\hspace{0.17em} =\angle OAC (angle opposite to eqal sides are equal (OC\hspace{0.17em}= OA\left)\right) \\ \angle OCA\hspace{0.17em} =\angle OAC =42° \\ Now, \\ \angle OAB = \angle CAB -\angle OAC \\ = 84 -42 \\ = 42° \\ In △OBA\hspace{0.17em} \\ OA\hspace{0.17em}= OB (Radius ) \\ \angle OAB\hspace{0.17em} =\angle OBA (angle opposite to eqal sides are equal (OB\hspace{0.17em}= OA\left)\right) \\ \angle OAB\hspace{0.17em} =\angle OBA =42° \\ In △AOB \\ \angle OBA +\angle BOA+\angle OAB =180° (Angle sum property) \\ 42+\angle BOA +42 =180 \\ \angle BOA = 180-42-42 \\ \angle BOA = 96° \\ The angle subtended by AB at the centre of the circle = \angle BOA =96° \end{gathered}$$
Regards