1. If $a, b, c$ are all integers, and satisfy
$$
\begin{array}{l}
\sqrt{9-8 \sin 50^{\circ}}=a+b \sin C^{\circ} . \\
\text { then } \frac{a+b}{c}=
\end{array}
$$
$$
\text { II. 1. } \frac{1}{2} \text {. }
$$

Since $9-8 \sin 50^{\circ}$
$$
\begin{aligned}
= & 9+8 \cos 80^{\circ} \\
& -8 \cos 80^{\circ}-8 \cos 40^{\circ} \\
= & 9+8 \sin 10^{\circ} \\
& -16 \cos 60^{\circ} \cos 20^{\circ} \\
= & 9+8 \sin 10^{\circ} \\
& -8 \cos 20^{\circ} \\
= & 1+8 \sin 10^{\circ} \\
& +8\left(1-\cos 20^{\circ}\right) \\
= & 1+8 \sin 10^{\circ}+16 \sin ^{2} 10^{\circ} \\
= & \left(1+4 \sin 10^{\circ}\right)^{2},
\end{aligned}
$$

Comparing, we get $a=1, b=4, c=10$.
$$
\therefore \frac{a+b}{c}=\frac{1}{2} \text {. }
$$