The diagonals of a rectangle PQRS intersects at O. If ∠QOR = 44°, ∠OPS =?

The correct option is B. 68°

$\angle SOP=\angle QOR={44}^{\circ }$ (Vertically opposite angles)

In $\Delta SOP$

$\angle OPS+\angle OSP+\angle SOP={180}^{\circ }$ [Angle Sum Property]

$2\angle OPS={136}^{\circ }$ [As $SO=OP$, diagonals of a rectangle are equal]

$\therefore \angle OPS={68}^{\circ }$