3.19. In the rhombus $A B C D$ and the triangle $A B C$, which contains its larger diagonal, circles are inscribed. Find the ratio of the radii of these circles if the acute angle of the rhombus is $\alpha$.
3.19. Given that $A B C D-$ is a rhombus (Fig. 3.21), $\angle B A D=$ $=\alpha\left(\alpha<90^{\circ}\right), r_{1}-$ is the radius of the circle inscribed in the rhombus, $r_{2}$ is the radius of the circle inscribed in $\triangle A B C$. Let $A B=a$. Draw $B E \perp A D$. In $\triangle B E A$ we have $B E=$ $=2 r_{1}=a \sin \alpha$, hence $r_{1}=\frac{a \sin \alpha}{2}$. From $\triangle A B C$ using the cosine rule, we find $A C=\sqrt{2 a^{2}-2 a^{2} \cos \left(180^{\circ}-\alpha\right)}=2 a \cos \frac{\alpha}{2}$. Therefore,

$$
\begin{gathered}
P_{\triangle A B C}=2 a+2 a \cos \frac{\alpha}{2}=4 a \cos ^{2} \frac{\alpha}{4} \\
S_{\triangle A B C}=\frac{1}{2} a^{2} \sin \left(180^{\circ}-\alpha\right)=\frac{1}{2} a^{2} \sin \alpha
\end{gathered}
$$

from which

$$
r_{2}=\frac{2 S_{\triangle A B C}}{P_{\triangle A B C}}=\frac{a^{2} \sin \alpha}{4 a \cos ^{2} \frac{\alpha}{4}}=\frac{a \sin \alpha}{4 \cos ^{2} \frac{\alpha}{4}}
$$

Thus,

$$
\frac{r_{1}}{r_{2}}=\frac{a \sin \alpha \cdot 4 \cos ^{2} \frac{\alpha}{4}}{2 a \sin \alpha}=2 \cos ^{2} \frac{\alpha}{4}
$$