#### Answer

Refer to the triangle below.

#### Work Step by Step

Recall:
For an angle $\theta$ whose terminal side contains the point $(x, y)$, we have:
$\sin{\theta}=\dfrac{y}{r}$ where $r^2=x^2+y^2$
Thus, with $\sin{\theta}=\frac{3}{4}$, we can assume that $y=3$ and $r=4$.
Solve for $x$ using the formula above to obtain:
\begin{align*}
r^2&=x^2+y^2\\
4^2&=x^2+3^2\\
4^2-3^2&=x^2\\
7&=x^2\\
\pm \sqrt{7}&=x
\end{align*}
Hence, one triangle whose solution is found from $\sin{\theta}=\dfrac{3}{4}$ is one where $x=\sqrt7$ and $y=3$.
Refer to the triangle above.