3-0. In an isosceles triangle, the base is twice the diameter of the inscribed circle. Find the sine of the largest angle of the triangle.
Answer: 0.96

Solution. Let $AC$ be the base of the given triangle, $B$ its vertex, $I$ the center of the inscribed circle, $H$ the midpoint of side $AC$, $r$ the radius of the inscribed circle, $\angle A=\alpha, \angle B=\beta$. From the condition $AH=2r$. In triangle $AIH$, $AI=\sqrt{r^{2}+(2r)^{2}}=r\sqrt{5}$. Therefore, $\sin \frac{\alpha}{2}=\frac{1}{\sqrt{5}}, \cos \frac{\alpha}{2}=\frac{2}{\sqrt{5}}$, then $\sin \alpha=2 \cdot \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}}=\frac{4}{5}$. From the isosceles property of triangle $ABC$, we have $\frac{\beta}{2}=90^{\circ}-\alpha$, and then $\cos \frac{\beta}{2}=\frac{4}{5}, \sin \frac{\beta}{2}=\frac{3}{5}$, and then $\sin \beta=2 \cdot \frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}$. Clearly, $\beta$ is the largest angle (since $\cos \alpha=\frac{3}{5}>\frac{1}{2}$, and therefore, $\alpha<60^{\circ}$).