$\left[\begin{array}{l}{\left[\begin{array}{l}\text { Cone } \\ {[\text { Surface of revolution solids }]}\end{array}\right]}\end{array}\right]$

The angle between the lines, each containing one generatrix of the cone, is $45^{\circ}$. A line perpendicular to both these generatrices intersects the base plane of the cone at an angle of $\frac{\pi}{3}$. Find the angle of the lateral development of the cone, given that it is greater than $270^{\circ}$.
Let $A$ be the vertex of the cone, $AB$ and $AC$ be the generators of the cone, $AO$ be the height of the cone, $M$ be the midpoint of the chord $BC$ of the base of the cone, and $D$ be the point where the line passing through point $A$ intersects the plane of the base of the cone (we can assume that the line mentioned in the problem passes through point $A$). Denote $AB = AC = l$, $OB = OC = r$, $AO = h$. The line $AD$ is perpendicular to the plane $ABC$ because it is perpendicular to two intersecting lines $AB$ and $AC$ of this plane. Therefore, $AD \perp BC$ and $AD \perp AM$. Since $DO$ is the orthogonal projection of the line $AD$ onto the plane of the base of the cone, $DO \perp BC$ by the theorem of three perpendiculars, and since $OM \perp BC$, the points $D$, $O$, and $M$ lie on the same line. Consider the right triangle $MAD$. In it, it is known that $AO$ is the height drawn from the vertex of the right angle,

$$
\angle OAM = \angle ADM = \frac{\pi}{8}, \quad AO = h, \quad AM = \frac{AO}{\cos OAM} = \frac{h}{\cos \frac{\pi}{8}}.
$$

Assume that $\angle BAC = \frac{\pi}{4}$. From the isosceles triangle $ABC$, we find that

$$
AM = AB \cdot \cos \angle BAM = l \cos \frac{\pi}{8}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3af6d4d9a845a74ce61dg-19.jpg?height=94&width=698&top_left_y=1279&top_left_x=680)

$$
\frac{h}{l} = \cos 2 \frac{\pi}{8} = \frac{1 + \cos \frac{\pi}{4}}{2} = \frac{1}{2} \left(1 + \frac{\sqrt{2}}{2}\right)
$$

From the right triangle $AOB$, we find that

$$
r = OB = \sqrt{AB^2 - OA^2} = \sqrt{l^2 - h^2}
$$

Let $\gamma$ be the desired angle in the lateral surface development of the cone. Then

$$
\begin{aligned}
\gamma &= \frac{2 \pi r}{2 \pi l} \cdot 2 \pi = \frac{r}{l} \cdot 2 \pi = \frac{\sqrt{l^2 - h^2}}{l} \cdot 2 \pi = \sqrt{1 - \left(\frac{h}{l}\right)^2} \cdot 2 \pi = \\
&= \sqrt{1 - \left(\frac{1}{2} \left(1 + \frac{\sqrt{2}}{2}\right)\right)^2} \cdot 2 \pi = \pi \sqrt{\frac{5 - 2 \sqrt{2}}{2}} \cdot \frac{3}{2} \pi.
\end{aligned}
$$

## Answer

Two lines, each containing one generator of the cone, intersect at an angle of $30^\circ$. The angle between the plane of the base of the cone and the plane passing through these generators is $\frac{5 \pi}{12}$. Find the angle of the lateral surface development of the cone if it is greater than $180^\circ$.

## Answer

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