12.154. In an acute isosceles triangle, the radius of the inscribed circle is 4 times smaller than the radius of the circumscribed circle. Find the angles of the triangle.
Solution.

Let in $\triangle A B C(A B=B C) \quad A C=a, \angle B A C=\alpha, B D$ - altitude, $O-$ center of the inscribed circle, $R$ and $r$ - radii of the circumscribed and inscribed circles (Fig. 12.27)

In $\triangle A D O\left(\angle A D O=90^{\circ}\right):$
$r=O D=A D \operatorname{tg} \angle O A D=\frac{a}{2} \operatorname{tg} \frac{\alpha}{2} \cdot R=\frac{A C}{2 \sin \angle A B C}=\frac{a}{2 \sin 2 \alpha}$.

Since $R=4 r$, then $\frac{a}{2 \sin \alpha}=4 \frac{a}{2} \operatorname{tg} \frac{\alpha}{2} ; \Leftrightarrow 4 \sin 2 \alpha \operatorname{tg} \frac{\alpha}{2}=1 \Leftrightarrow$

$16 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos \alpha \operatorname{tg} \frac{\alpha}{2}=1 \Leftrightarrow 16 \sin ^{2} \frac{\alpha}{2} \cos \alpha=1 \Leftrightarrow 8(1-\cos \alpha) \cos \alpha=1 \Leftrightarrow$

$$
\begin{aligned}
& \Leftrightarrow 8 \cos ^{2} \alpha-8 \cos \alpha+1=0 ; \quad \cos \alpha=\frac{2 \pm \sqrt{2}}{4} \\
& \cos \angle A B C=\cos \left(180^{\circ}-2 \alpha\right)=-\cos 2 \alpha=1-2 \cos ^{2} \alpha
\end{aligned}
$$

If $\cos \alpha=\frac{2+\sqrt{2}}{4}$, then $\cos \angle A B C=1-2\left(\frac{2+\sqrt{2}}{4}\right)^{2}=\frac{1-2 \sqrt{2}}{4}<0$, which contradicts the condition, as the given triangle is acute-angled. Thus, $\cos \alpha=\frac{2-\sqrt{2}}{4}, \cos \angle A B C=1-2\left(\frac{2-\sqrt{2}}{4}\right)^{2}=\frac{1+2 \sqrt{2}}{4}$. Answer: $\arccos \frac{2-\sqrt{2}}{4} ; \arccos \frac{2 \sqrt{2}+1}{4}$.