[Geometry]
Problem: If $0^{\circ}<x<90^{\circ}$ and $3 \sin (x)-\cos \left(15^{\circ}\right)=0$, what is the value of $x$ to the nearest tenth of a degree?

Solution: Rearranging the equation,

$$
\begin{aligned}
3 \sin (x) & =\cos \left(15^{\circ}\right) \\
\sin (x) & =\frac{1}{3} \cos \left(15^{\circ}\right) \\
\sin (x) & \approx 0.3220
\end{aligned}
$$

Using a calculator, $x \approx 18.78^{\circ}$. To the nearest tenth of a degree, $x=18.8^{\circ}$.

Final answer: $18.8^{\circ}$