3. A right cylinder and a cone have a common base, and the vertex of the cone is located at the center of the other base of the cylinder. Determine the angle between the slant height of the cone and the axis of the cylinder if the ratio of the surface areas of the cylinder and the cone is 7:4.
Solution. Let $r$ and $h$ be the radius of the base and the height of the cylinder, respectively. Then the areas of the cylinder and the cone are:

$$
P_{1}=2 \pi r h+2 r^{2} \pi, \quad P_{2}=r \pi \sqrt{r^{2}+h^{2}}+r^{2} \pi
$$

From the condition $\frac{P_{1}}{P_{2}}=\frac{7}{4}$, we sequentially obtain:

![](https://cdn.mathpix.com/cropped/2024_06_05_53a7c72cd2352523c04ag-1.jpg?height=548&width=320&top_left_y=1685&top_left_x=1128)

$$
\begin{aligned}
& 8 h+8 r=7 \sqrt{r^{2}+h^{2}}+7 r \\
& 48 r^{2}-16 r h-15 h^{2}=0 \\
& 48\left(\frac{r}{h}\right)^{2}-16 \frac{r}{h}-15=0
\end{aligned}
$$

and finally $\frac{r}{h}=\frac{3}{4}$. If we denote the required angle by $\varphi$, then $\operatorname{tg} \varphi=\frac{r}{h}=\frac{3}{4}$, so $\varphi=\operatorname{arctg} \frac{3}{4}$.