8. As shown in Figure 5, in isosceles $\triangle A B C$, the base angle $\angle B=$ $15^{\circ}$, and the length of the leg $A B=10$. Then the area of this triangle is $\qquad$
8. 25 .

Draw the altitude $C D$ from $C$ to $B A$ extended, meeting at point $D$. In the right triangle $\triangle A C D$,
$$
\angle C A D=\angle B+\angle C=2 \angle B=30^{\circ} \text {. }
$$

Then $C D=\frac{1}{2} A C=\frac{1}{2} \times 10=5$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot C D=\frac{1}{2} \times 10 \times 5=25$.