Find the ordered pair $(a,b)$ of integers such that
\[\sqrt{9 - 8 \sin 50^\circ} = a + b \csc 50^\circ.\]
We write
\[9 - 8 \sin 50^\circ = \frac{9 \sin^2 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ} = \frac{9 \sin^2 50^\circ - 6 \sin 50^\circ + 6 \sin 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ}.\]By the triple angle identity,
\begin{align*}
6 \sin 50^\circ - 8 \sin^3 50^\circ &= 2 \sin (3 \cdot 50^\circ) \\
&= 2 \sin 150^\circ \\
&= 1,
\end{align*}so
\[9 - 8 \sin 50^\circ = \frac{9 \sin^2 50^\circ - 6 \sin 50^\circ + 1}{\sin^2 50^\circ} = \left( \frac{3 \sin 50^\circ - 1}{\sin 50^\circ} \right)^2.\]Since $3 \sin 50^\circ > 3 \sin 30^\circ = \frac{3}{2} > 1,$ $3 \sin 50^\circ - 1 > 0.$  Therefore,
\[\sqrt{9 - 8 \sin 50^\circ} = \frac{3 \sin 50^\circ - 1}{\sin 50^\circ} = 3 - \csc 50^\circ,\]so $(a,b) = \boxed{(3,-1)}.$
We write
\[9 - 8 \sin 50^\circ = \frac{9 \sin^2 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ} = \frac{9 \sin^2 50^\circ - 6 \sin 50^\circ + 6 \sin 50^\circ - 8 \sin^3 50^\circ}{\sin^2 50^\circ}.\]By the triple angle identity,
\begin{align*}
6 \sin 50^\circ - 8 \sin^3 50^\circ &= 2 \sin (3 \cdot 50^\circ) \\
&= 2 \sin 150^\circ \\
&= 1,
\end{align*}so
\[9 - 8 \sin 50^\circ = \frac{9 \sin^2 50^\circ - 6 \sin 50^\circ + 1}{\sin^2 50^\circ} = \left( \frac{3 \sin 50^\circ - 1}{\sin 50^\circ} \right)^2.\]Since $3 \sin 50^\circ > 3 \sin 30^\circ = \frac{3}{2} > 1,$ $3 \sin 50^\circ - 1 > 0.$  Therefore,
\[\sqrt{9 - 8 \sin 50^\circ} = \frac{3 \sin 50^\circ - 1}{\sin 50^\circ} = 3 - \csc 50^\circ,\]so $(a,b) = \boxed{(3,-1)}.$
The final answer is $(3,-1)$. I hope it is correct.