Question

note: figure not drawn to scale. given rectangle nopq and triangle prq share side pq. solve for twice the m∠psq. (a) 240° (b) 200° (c) 150° (d) 120°

Explanation:

Step1: Recall properties of rectangles

In rectangle NOPQ, diagonals bisect each other at point S. So, $\triangle PSQ$ is isosceles. Also, $\angle PSQ + \angle OSP+\angle OSQ = 360^{\circ}$. Since $\angle OSP=\angle OSQ = 90^{\circ}$ (diagonals of rectangle are perpendicular bisectors of each other), and $\angle PSQ = 120^{\circ}$ (by the fact that the sum of angles around a point is $360^{\circ}$ and the right - angled nature of the rectangle's diagonal intersection).

Step2: Calculate twice the angle

We need to find $2\times m\angle PSQ$. If $m\angle PSQ = 120^{\circ}$, then $2\times m\angle PSQ=2\times120^{\circ} = 240^{\circ}$.

Answer:

A. $240^{\circ}$