Answer
$\angle A \approx 13.0^{\circ}$
Work Step by Step
In this exercise, we are given the measure of a non-included angle and two sides. We can use the law of sines to find the measure of another angle in the triangle. The law of sines states the following:
$\frac{sin A}{\overline{BC}}$ = $\frac{sin B}{\overline{AC}}$ = $\frac{sin C}{\overline{AB}}$, where $A$, $B$, and $C$ are the measures of the angles in the triangle and $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$ are the measures of the sides opposite those angles, respectively.
Let's plug in what we know into the formula:
$\frac{sin 72}{38}$ = $\frac{sin \angle A}{9}$
Multiply each side by $9$:
$\frac{sin 72}{38}(9)$ = sin $\angle A$
Take sin$^{-1}$ of both sides to solve for $\angle A$:
$\angle A \approx 13.0^{\circ}$
