26. Suppose $A, B, C$ are three angles such that $A \geq B \geq C \geq \frac{\pi}{8}$ and $A+B+C=\frac{\pi}{2}$. Find the largest possible value of the product $720 \times(\sin A) \times(\cos B) \times(\sin C)$.
26. Answer: 180
Note that
$$
A=\frac{\pi}{2}-B-C \leq \frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4},
$$
and $\sin (B-C) \geq 0$. Therefore
$$
\begin{aligned}
\sin A \cos B \sin C & =\frac{1}{2} \cdot \sin A \cdot[\sin (B+C)-\sin (B-C)] \\
& \leq \frac{1}{2} \sin A \sin (B+C) \\
& =\frac{1}{2} \sin A \cos A \\
& =\frac{1}{4} \sin 2 A \leq \frac{1}{4} \sin \frac{\pi}{2}=\frac{1}{4} .
\end{aligned}
$$

When $A=\frac{\pi}{4}$ and $B=C=\frac{\pi}{5}$, we have
$$
\sin A \cos B \sin C=\sin \frac{\pi}{4} \cos \frac{\pi}{8} \sin \frac{\pi}{8}=\sin \frac{\pi}{4} \cdot \frac{1}{2} \sin \frac{\pi}{4}=\frac{1}{4} .
$$

Hence the largest possible value of $720(\sin A)(\cos B)(\sin C)$ is $720 \times \frac{1}{4}=180$.