We know that MN = MQ, which makes triangle MNQ an isosceles triangle. By definition, then, ∠MQR = ∠MNR and the height from the vertex of the triangle (point M) to the base (QN) is perpendicular and bisects ∠QMN. In other words, ∠QMR = ∠NMR = 36°.
Now, since QMR is a triangle, all three angles add up to 180, so we can find the angle MQR:
∠MQR = 180 - 36 - 90 = 54°
Now, since PQ = NO, we can say that PQNO is an isosceles trapezoid, so by definition, ∠PQN = ∠ONQ. And, since ∠MQR = ∠MNR = 54° and ∠MNO = 125, we can find ∠PQN = ∠ONQ:
∠PQN = ∠ONQ = 125 - 54 = 71°
Since PQRS is a quadrilateral, by definition, all angles will add up to 360. So, we have:
∠PQR + ∠QRS + ∠RSP + ∠SPQ = 360
71 + 90 + 90 + ∠SPQ = 360
∠SPQ = 109
Since ∠SPQ is the same angle as ∠OPQ, the answer is A.
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