(C) 96°
Since RJ = RL (tangents from external point), △RJL is isosceles, so ∠RLJ = ∠RJL = 42°. Thus ∠JRL = 180° − 84° = 96°. Since ∠JOL + ∠JRL = 180° (tangent ⊥ radius makes ROJL a cyclic quadrilateral with two right angles), ∠JOL = 96°.
In quadrilateral RJOL: ∠OJR = ∠OLR = 90° (tangent ⊥ radius). So ∠JOL + ∠JRL = 180°. Find ∠JRL using the isosceles triangle property (RJ = RL), then subtract from 180°. Examiners expect you to recall that the angle at centre and angle at external point are supplementary.