3.69. The base of the triangle is equal to \(a\), and the angles adjacent to it contain \(45^{\circ}\) and \(15^{\circ}\). A circle with a radius equal to the height dropped to this base is drawn from the vertex opposite the base. Find the area of the part of the corresponding circle that is enclosed within the triangle.
3.69. By the condition, $\angle A C B=$ $=45^{\circ}, \angle A B C=15^{\circ}, B C=a$, $A M \perp B C, A M$ - radius of the circle, $E$ and $F$ - points of intersection of the circle with sides $A B$ and $A C$; it is required to find $S_{A F M E}$ (Fig. 3.70). From $\triangle A M C$ and $\triangle A M B$ we find $M C=A M \operatorname{ctg} 45^{\circ}=A M, M B=$ $=A M \operatorname{ctg} 15^{\circ}$. Then $a=C M+$

$+M B=A M\left(1+\operatorname{ctg} 15^{\circ}\right)$, i.e., $A M=\frac{a}{1+\operatorname{ctg} 15^{\circ}}$. Note that $\angle F A B=$ $=180^{\circ}-\left(45^{\circ}+15^{\circ}\right)=120^{\circ}$. Consequently,

$$
\begin{gathered}
S_{A F M E}=\frac{1}{3} \pi \cdot \frac{a^{2}}{\left(1+\operatorname{ctg} 15^{\circ}\right)^{2}}=\frac{\pi a^{2}}{3\left(\operatorname{ctg} 45^{\circ}+\operatorname{ctg} 15^{\circ}\right)^{2}}= \\
\frac{\pi a^{2} \sin ^{2} 45^{\circ} \sin ^{2} 15^{\circ}}{3 \sin ^{2} 60^{\circ}}=\frac{\pi a^{2}\left(1-\cos 30^{\circ}\right)}{9}=\frac{\pi a^{2}(2-\sqrt{3})}{18}
\end{gathered}
$$