1. The expression $1000 \sin 10^{\circ} \cos 20^{\circ} \cos 30^{\circ} \cos 40^{\circ}$ can be simplified as $a \sin b^{\circ}$, where $a$ and $b$ are positive integers with $0<b<90$. Find the value of $100 a+b$.
1. Answer: 12560
$$
\begin{array}{c}
\sin 10^{\circ} \cos 20^{\circ} \cos 30^{\circ} \cos 40^{\circ}=\frac{2 \sin 10^{\circ} \cos 10^{\circ} \cos 20^{\circ} \cos 30^{\circ} \cos 40^{\circ}}{2 \cos 10^{\circ}} \\
=\frac{\sin 20^{\circ} \cos 20^{\circ} \cos 30^{\circ} \cos 40^{\circ}}{2 \cos 10^{\circ}} \\
=\frac{\sin 40^{\circ} \cos 40^{\circ} \cos 30^{\circ}}{4 \cos 10^{\circ}} \\
=\frac{\sin 80^{\circ} \cos 30^{\circ}}{8 \cos 10^{\circ}} \\
=\frac{\cos 10^{\circ} \sin 60^{\circ}}{8 \cos 10^{\circ}}
\end{array}
$$

Hence $1000 \sin 10^{\circ} \cos 20^{\circ} \cos 30^{\circ} \cos 40^{\circ}=125 \sin 60^{\circ}$, showing that $a=125$ and $b=60$. So, $100 a+b=12560$.