Основанием четырехугольной пирамиды является ромб, у которого косинус угла равен $\frac{7}{8}$ и длина стороны равна $8$ . Все боковые грани пирамиды наклонены к плоскости ее основания под углом альфа, а высота пирамиды равна $18$ . Найдите значение выражения $2\sqrt{15} \, \text{tg} \, альфа$ .

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Q. Основанием четырехугольной пирамиды является ромб, у которого косинус угла равен

\frac{7}{8}

и длина стороны равна

8

. Все боковые грани пирамиды наклонены к плоскости ее основания под углом альфа, а высота пирамиды равна

18

. Найдите значение выражения

2\sqrt{15} \, \text{tg} \, альфа

Find Tangent of Angle: To find the tangent of the angle $\alpha$ ( $\text{tg} \alpha$ ), we need to relate the height of the pyramid to the slant height of the lateral faces. The slant height can be found using the Pythagorean theorem on a right triangle formed by the height of the pyramid, half of the diagonal of the rhombus base, and the slant height.
Find Diagonal Length: First, we need to find the length of the diagonal of the rhombus. Since the cosine of the angle is $\frac{7}{8}$ and the side length is $8$ , we can use the formula for the cosine of an angle in a rhombus, which is $\cos(\theta) = \frac{e}{2a}$ , where $e$ is the length of the short diagonal, and $a$ is the side length. So, $e = 2a \cdot \cos(\theta) = 2 \cdot 8 \cdot \left(\frac{7}{8}\right) = 14$ .
Calculate Long Diagonal: Now we have the length of the short diagonal $e = 14$ . The length of the long diagonal $f$ can be found using the Pythagorean theorem since the diagonals of a rhombus are perpendicular bisectors of each other. So, $f = \sqrt{4a^2 - e^2} = \sqrt{4 \times 8^2 - 14^2} = \sqrt{256 - 196} = \sqrt{60} = 2\sqrt{15}$ .
Find Slant Height: With the long diagonal $f = 2\sqrt{15}$ , we can find the slant height $l$ of the pyramid using the Pythagorean theorem again, where $l$ is the hypotenuse, the height of the pyramid is one leg, and half of the long diagonal is the other leg. So, $l = \sqrt{h^2 + (\frac{f}{2})^2} = \sqrt{18^2 + (\frac{\sqrt{60}}{2})^2} = \sqrt{324 + 15} = \sqrt{339}$ .
Calculate Tangent: Now that we have the slant height $l = \sqrt{339}$ , we can find $\text{tg } \alpha$ , which is the ratio of the opposite side (height of the pyramid) to the adjacent side (half of the long diagonal of the base). So, $\text{tg } \alpha = \frac{h}{(f/2)} = \frac{18}{(\sqrt{60}/2)} = \frac{18}{\sqrt{15}} = \frac{18\sqrt{15}}{15}$ .
Final Expression Calculation: Finally, we can find the value of the expression $2\sqrt{15} \, \text{tg} \, \alpha = 2\sqrt{15} \times \left(\frac{18\sqrt{15}}{15}\right) = \frac{2 \times 18 \times 15}{15} = 2 \times 18 = 36$ .