14. It is known that
$$
\sqrt{9-8 \sin 50^{\circ}}=a+b \sin c^{\circ}
$$
for exactly one set of positive integers $(a, b, c)$, where $0<c<90$. Find the value of $\frac{b+c}{a}$.
14. Answer: 14
We have
$$
\begin{aligned}
9-8 \sin 50^{\circ} & =8+8 \sin 10^{\circ}-8 \sin 10^{\circ}-8 \sin 50^{\circ} \\
& =9+8 \sin 10^{\circ}-8\left(2 \sin 30^{\circ} \cos 20^{\circ}\right) \\
& =9+8 \sin 10^{\circ}-8\left(1-2 \sin ^{2} 10^{\circ}\right) \\
& =16 \sin ^{2} 10^{\circ}+8 \sin 10^{\circ}+1 \\
& =\left(1+4 \sin 10^{\circ}\right)^{2} \\
\therefore \sqrt{9-8 \sin 50^{\circ}} & =1+4 \sin 10^{\circ} .
\end{aligned}
$$

Thus, $a=1, b=4$ and $c=10$, and hence $\frac{b+c}{a}=14$.