In the diagram above, O is the centre of the circle, POM is a diameter, and \(\angle\) MNQ = 42º. Calculate \(\angle\) QMP.
POM is a diameter, so arc MP (lower semicircle via Q) = 180°.
∠MNQ = 42° is an inscribed angle subtending arc MQ.
Therefore, arc MQ = 2 × 42° = 84°
Arc MQ + arc QP = lower semicircle:
arc QP = 180° − 84° = 96°
- ∠QMP is an inscribed angle subtending arc QP.
Therefore, ∠QMP = \(\frac{1}{2}\) × 96° = 48°.
(Alternatively: in right triangle PQM, ∠PQM = 90° (angle in a semicircle), so the angles at M and P sum to 90°; the arc method directly gives the result.)
There is an explanation video available below.
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