5. In tetrahedron $ABCD$, the sums of three plane angles at each vertex are $180^{\circ}$. Find the volume of the tetrahedron if $BC=4, \cos \angle BAC=3/4, \sin \angle CBD=5\sqrt{7}/16$. (20 points)
Solution. Let's make a development of the tetrahedron on the plane $ABC$ relative to the sides of the triangle $ABC$. Since the sums of the three planar angles at each vertex are $180^{\circ}$, we will get a triangle $D_{1} D_{2} D_{3}$, for which the segments $AB, AC$, and $BC$ are the midlines. From this, we obtain that the sides of the triangle $\mathrm{ABC}$ are equal to the opposite edges of the tetrahedron. Therefore, the tetrahedron $ABCD$ is equifacial. Let's denote

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-06.jpg?height=545&width=877&top_left_y=1561&top_left_x=1109)

$BC=a, \angle BAC=\alpha, \angle CBD=\beta, AC=b, AB=$

c. Note that $\angle ACB=\beta$. By the Law of Sines for triangle $ABC$, we have $\frac{a}{\sin \alpha}=\frac{c}{\sin \beta}$. Since $a=4$, $\cos \alpha=3 / 4, \sin \angle \beta=5 \sqrt{7} / 16$, then

$$
\frac{16}{\sqrt{7}}=\frac{16 c}{5 \sqrt{7}}, \quad c=5
$$

By the Law of Cosines, we find $b$:

$$
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-06.jpg?height=542&width=557&top_left_y=2202&top_left_x=1315)
$16=b^{2}+25-7.5 b, b^{2}-7.5 b+9=0, b=6$. The second root $b=1.5$ is not suitable, as in this case the angle $ACB$ is obtuse.

To find the volume of the tetrahedron $ABCD$, we will describe a rectangular parallelepiped around it such that the edges of the tetrahedron are the diagonals of the faces of the parallelepiped. Let the sides of the parallelepiped be $x, y, z$. Then the volume of the tetrahedron $V=(1-4 / 6) x y z=$ $x y z / 3$.

We have $x^{2}+z^{2}=a^{2}, x^{2}+y^{2}=b^{2}, z^{2}+y^{2}=c^{2}, \Rightarrow$

$$
x^{2}+y^{2}+z^{2}=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right), \quad x^{2}=\frac{1}{2}\left(a^{2}+b^{2}-c^{2}\right)
$$

$$
\begin{aligned}
& y^{2}=\frac{1}{2}\left(-a^{2}+b^{2}+c^{2}\right), z^{2}=\frac{1}{2}\left(a^{2}-b^{2}+c^{2}\right) \\
& V=\frac{1}{12} \sqrt{2\left(a^{2}+b^{2}-c^{2}\right)\left(-a^{2}+b^{2}+c^{2}\right)\left(a^{2}-b^{2}+c^{2}\right)} \\
& V=\frac{1}{12} \sqrt{2\left(4^{2}+6^{2}-5^{2}\right)\left(-4^{2}+6^{2}+5^{2}\right)\left(4^{2}-6^{2}+5^{2}\right)} \\
& \quad=\frac{15 \sqrt{6}}{4} .
\end{aligned}
$$

Answer: $\frac{15 \sqrt{6}}{4}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-07.jpg?height=1102&width=1202&top_left_y=661&top_left_x=628)