12.077. The sides of the base of a right parallelepiped are in the ratio $1: 2$, and the acute angle in the base is $\alpha$. Find the angle between the smaller diagonal of the parallelepiped and the base plane, if the height of the parallelepiped is equal to the larger diagonal of the base.
Solution.

Let the sides of the base of the parallelepiped be denoted by $a$ and $b$. Then $a: b=1: 2$. The larger diagonal of the base

$$
d_{2}=\sqrt{a^{2}+b^{2}-2 a b \cos \left(180^{\circ}-\alpha\right)}=\sqrt{a^{2}+b^{2}+2 a b \cos \alpha} \text {. Considering that }
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-772.jpg?height=500&width=584&top_left_y=92&top_left_x=108)

Fig. 12.78

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-772.jpg?height=474&width=422&top_left_y=115&top_left_x=761)

Fig. 12.79

$b=2 a, d_{2}=\sqrt{5 a^{2}+4 a^{2} \cos \alpha}$. The smaller diagonal of the base $d_{1}=\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}=\sqrt{5 a^{2}-4 a^{2} \cos \alpha}$. Since the height $h=d_{2}=\sqrt{5 a^{2}+4 a^{2} \cos \alpha}$, then $\operatorname{tg} \beta$ (Fig. 12.79) can be found as $\operatorname{tg} \beta=\frac{h}{d_{1}}=\sqrt{\frac{5+4 \cos \alpha}{5-4 \cos \alpha}}$.

Answer: $\operatorname{arctg} \sqrt{\frac{5+4 \cos \alpha}{5-4 \cos \alpha}}$.