11.195. The radius of the base of the cone is $R$, and the angle of the sector of its lateral surface is $90^{\circ}$. Determine the volume of the cone.
## Solution.

Let $l$ be the slant height of the cone. Since the length of the arc of the lateral surface development of the cone is equal to the circumference of the base, we have $2 \pi R = \frac{2 \pi l}{4}$ (by the condition that the development is a quarter circle), from which $l = 4R$.

Next, we find the height of the cone: $h = \sqrt{l^2 - R^2} = \sqrt{16R^2 - R^2} = R \sqrt{15}$. Then, $V = \frac{1}{3} \pi R^2 \cdot R \sqrt{15} = \frac{\pi R^3 \sqrt{15}}{3}$.

Answer: $\frac{\pi R^3 \sqrt{15}}{3}$.

## Solutions to Chapter 12

## GEOMETRY PROBLEMS WITH THE APPLICATION OF TRIGONOMETRY

## SOME RELATIONS BETWEEN ELEMENTS OF FIGURES

1. The area of the parallelogram $ABCD$ (Fig. 12.1) can be calculated using the following formulas:

$$
\begin{gathered}
S = \frac{AC^2 - BD^2}{4} \tan A \\
S = \frac{AB^2 - AD^2}{2} \tan \angle AOD
\end{gathered}
$$

where $O$ is the point of intersection of the diagonals $AC$ and $BD$.

2. Let the lengths $b$ and $c$ of two sides of triangle $ABC$ and the angle $A$ between them (Fig. 12.2) be known. Then the length of the angle bisector $AD$ of the triangle, drawn from the vertex of this angle, is given by the formula

$$
l_a = \frac{2bc \cos(A/2)}{b+c}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0892.jpg?height=290&width=537&top_left_y=1371&top_left_x=109)

Fig. 12.1

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0892.jpg?height=371&width=487&top_left_y=1290&top_left_x=677)

Fig. 12.2

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0893.jpg?height=716&width=507&top_left_y=76&top_left_x=94)

Fig. 12.3

3. The following relations hold between the elements of a sphere and a cone inscribed in it:

$$
\begin{gathered}
l = 2R \sin \alpha \\
l^2 = 2RH
\end{gathered}
$$

where $R$ is the radius of the sphere, $l$ is the slant height of the cone, $H$ is its height, and $\alpha$ is the angle between the slant height and the plane of the base.

These relations are also valid for a pyramid inscribed in a sphere, where the lateral edges have length $l$ and form an angle $\alpha$ with the plane of the base.

4. Let $A_1B_1$ be a lateral edge of a pyramid or prism, $A_1O$ its projection on the base plane, $\angle B_1A_1O = \alpha, \angle OA_1A_2 = \beta, \angle B_1A_1A_2 = \gamma$ (Fig. 12.3). Then the equality $\cos \gamma = \cos \alpha \cos \beta$ holds.

Proofs of these relations can be found in any edition of this problem book from recent years.