12.426 The axial section of the cylinder is a square. The segment $A B$, connecting point $\boldsymbol{A}$ on the circumference of the upper base with point $B$ on the circumference of the lower base of the cylinder, is equal to $a$ and is at a distance $b$ from the axis of the cylinder. Find the angle between the line $\boldsymbol{A}$ and the plane of the base of the cylinder.
## Solution.

A

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-303.jpg?height=328&width=302&top_left_y=761&top_left_x=175)

B
Let $OP$ be the axis of the given cylinder, $M$ and $N$ be the midpoints of $OP$ and $AB$, and $AC$ be the generatrix of the cylinder. In $\triangle ABC$, draw the line $NK$. Then $PK \perp BC$ (by the property of a chord and a radius passing through its midpoint) $\Rightarrow PK \perp AB$ (Theorem 12). Additionally, $NK = \frac{1}{2} AC = \frac{1}{2} OP = MP$ and $NK \| MP \Rightarrow MNKP$ is a rectangle, i.e., $MN$ is the common perpendicular of the lines $AB$ and $OP \Rightarrow MN = PK = b$.

By the condition, $PC = \frac{1}{2} AC = \frac{a}{2} \cdot \sin \angle ABC ; KC = \frac{1}{2} BC = \frac{a}{2} \cdot \cos \angle ABC \Rightarrow$ $\Rightarrow b^2 = PK^2 = PC^2 - KC^2 = -\frac{a^2}{4} \cos (2 \angle ABC) \Rightarrow \cos (2 \angle ABC) = -\frac{4 b^2}{a^2}$.

Answer: $\angle ABC = \frac{1}{2} \arccos \left(-\frac{4 b^2}{a^2}\right)$.