Trigonometry confusion with triangle in weird question I was wondering how do you get x from the triangle below: 
 A:  First method:
$$
\frac n {\sqrt 3} = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sqrt{1-\cos^2\theta}}{\cos\theta} = \frac{\sqrt{1-(1/3)^2}}{1/3}
$$
Second method:
$$
\frac 1 3 = \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt 3}{\text{hypotenuse}}
$$
Given the above, you can find the length of the hypontenuse.  Then you can use the Pythagorean theorem to find $n$.
A: Using the top piece of information $\cos\theta = \frac{1}{3}$, we can set up the proportion $\frac{1}{3} = \frac{\sqrt3}{x}$ to solve for the hypotenuse $x=3\sqrt3$, then use the Pythagorean Theorem to solve for the missing side $n$: $\sqrt3^2 + n^2 = (3\sqrt3)^2$, getting $n=2\sqrt6$.
The second piece of information, $n=\sqrt3 \sin\theta$ is confusing me because using the definition of $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, that would make the hypotenuse and a leg of the triangle equal to $\sqrt3$ which is impossible. Is there anything I'm missing?
Using Mr. Hardy's methods above I'm still getting two different values for $n$, fyi.
