POM is a diameter, so arc MP (lower semicircle via Q) = 180^\circ. ∠MNQ = 42^\circ is an inscribed angle subtending arc MQ. Therefore, arc MQ = 2 \times 42^\circ = 84^\circ Arc MQ + arc QP = lower semicircle: arc QP = 180^\circ − 84^\circ = 96^\circ - ∠QMP is an inscribed angle subtending arc QP. Therefore, ∠QMP = \(\frac{1}{2}\) \times 96^\circ = 48^\circ. (Alternatively: in right triangle PQM, ∠PQM = 90^\circ (angle in a semicircle), so the angles at M and P sum to 90^\circ; the arc method directly gives the result.)