In the plane $\alpha$, two perpendicular lines are drawn. The line $l$ forms angles of $45^{\circ}$ and $60^{\circ}$ with them. Find the angle between the line $l$ and the plane $\alpha$.
Let the line $l$ form angles of $60^{\circ}$ and $45^{\circ}$ with the lines $a$ and $b$ in the plane $\alpha$, respectively, and intersects this plane at point $P$. Take a point $M$ on the line $l$ such that $P M=1$. Let $O$ be the orthogonal projection of point $M$ onto the plane $\alpha$, and let $A$ and $B$ be the feet of the perpendiculars dropped from point $M$ to the lines $a 1$ and $b 1$, respectively, which are parallel to the lines $a$ and $b$ and pass through point $P$. Then, $\angle MRO$ is the angle between the line $l$ and the plane $\alpha$, $\angle MPA$ is the angle between the lines $l$ and $a 1$ (and thus between the lines $l$ and $a$), and $\angle MPB$ is the angle between the lines $l$ and $b 1$ (and thus between the lines $l$ and $b$). According to the problem, $\angle MPA=60^{\circ}$ and $\angle MPB=45^{\circ}$. From the right triangles MRA and MRB, we find that

$$
A P=M P \cos \angle M P A=1 \cdot \cos 60^{\circ}=\frac{1}{2}
$$

$$
B P=M P \cos \angle M P B=1 \cdot \cos 45^{\circ}=\frac{\sqrt{2}}{2}.
$$

By the theorem of three perpendiculars, $O A \perp a$ and $O B \perp b$. Therefore, $O A P B$ is a rectangle. Hence, $O A=B P=\frac{\sqrt{2}}{2}$. From the right triangle OAR, we find that

$$
O P=\sqrt{A P^{2}+O A^{2}}=\sqrt{\frac{1}{4}+\frac{1}{2}}=\frac{\sqrt{3}}{2}
$$

Therefore,

$$
\cos \angle M P O=\frac{O P}{M P}=\frac{\sqrt{3}}{1}=\frac{\sqrt{3}}{2}
$$

Thus, $\angle M P O=30^{\circ}$.

## Answer

$30^{\circ}$.