15. (2002 Shanghai College Entrance Exam) In the right triangular prism $A B O-A_{1} B_{1} O_{1}$, $O O_{1}=4, O A=4, O B=3$, $\angle A O B=90^{\circ}$, $D$ is the midpoint of the line segment $A_{1} B_{1}$, $P$ is a point on the lateral edge $B B_{1}$, if $O P \perp B D$, find the size of the angle formed by $O P$ and the base $A O B$ (the result should be expressed using the inverse trigonometric function).
15. According to the problem, establish a spatial rectangular coordinate system $O-xyz$, then $B(3,0,0), D\left(\frac{3}{2}\right.$, $2,4), \overrightarrow{B D}=\left(-\frac{3}{2}, 2,4\right)$, let $P=(3,0, z)$, since $B D \perp O P$, hence $\overrightarrow{B D} \cdot \overrightarrow{O P}=$ $\left(-\frac{9}{2}\right)+4 z=0$, so $z=\frac{9}{8}, \overrightarrow{O P}=\left(3,0, \frac{9}{8}\right)$, it is easy to know that $\vec{k}=(0,0,1)$ is the normal vector of plane $A O B$, let the required angle be $\theta$, then $\sin \theta=|\vec{k} \cdot \overrightarrow{O P}| /|\vec{k}||\overrightarrow{O P}|=3 \sqrt{73} / 73$, so $\theta=\operatorname{arc} \sin 3 \sqrt{73} / 73$.