Answer
$x = \frac{\sqrt{45}}{7}$
Work Step by Step
$\theta = arcsin~\frac{2}{7}$
$sin~\theta = \frac{2}{7} = \frac{opposite}{hypotenuse}$
Note that angle $\theta$ is in quadrant I. We can find the length of the adjacent side.
$adjacent = \sqrt{7^2-2^2} = \sqrt{45}$
We can find the value of $cos~\theta$:
$cos~\theta = \frac{adjacent}{hypotenuse}$
$cos~\theta = \frac{\sqrt{45}}{7}$
$\theta = arccos~\frac{\sqrt{45}}{7}$
Therefore, $arccos~\frac{\sqrt{45}}{7} = arcsin~\frac{2}{7}$ and thus $x = \frac{\sqrt{45}}{7}$
