Since AB and AC are tangents from external point A, triangle ABC is isosceles with AB = AC, so ∠ACB = ∠ABC = 42°.
∠BAC = 180° − 42° − 42° = 96°. → Option A
Equal tangents from an external point (Theorem 10.2) make △ABC isosceles, so base angles ∠ABC = ∠ACB = 42°. Apply angle sum: 180° − 84° = 96°. Watch for this standard isosceles-tangent triangle setup in MCQs.