<s>[INST] Solve The Triangle I am having a tough time trying to solve this problem. 
I have utilized the 30, 60, 90 triangle measures for the length of sides. However, I am stuck since the side that would be √3 has 100 as its length. How do I solve then? 
This is what I have done so far: 

 [/INST] To find $x$ here, you will need to look at this as 2 separate triangles, and find the base of each separate triangle, then add the bases up to get $x$.
For the triangle on the left hand side, we can use the $\tan(x)$ function to relate the $60^{\circ}$ angle with the opposite side of length 100, and our adjacent side, which we are looking for (let's call it $y$).
So, we know $\tan(x) = \dfrac{\text{opp}}{\text{adj}}$, and this gives $\tan(60^{\circ}) = \dfrac{100}{y}$.  But $\tan(60^{\circ}) = \sqrt{3}$, so we have:
$\sqrt{3} = \dfrac{100}{y}$, and solving for $y$ gives $y = \dfrac{100}{\sqrt{3}}$.
Now, we need to solve for the adjacent side of the triangle on the right-hand side using the angle $30^{\circ}$.  Again, we want to relate the opposite side of $30^{\circ}$ to its adjacent side (the unknown -- we can call it $z$).  We will have to use $\tan(x)$ again.  So we get:
$\tan(30^{\circ}) = \dfrac{\text{opp}}{\text{adj}} = \dfrac{100}{z}$.  But, $\tan(30^{\circ}) = \dfrac{1}{\sqrt{3}}$, so we have the equation:
$\dfrac{1}{\sqrt{3}} = \dfrac{100}{z}$ and solving for $z$ gives:
$z = \dfrac{100}{\frac{1}{\sqrt{3}}} = 100\sqrt{3}$.
Finally, $x = y + z$, since $x$ is the length of both adjacent sides put together, so:
$x = \dfrac{100}{\sqrt{3}} + 100\sqrt{3} = \dfrac{100}{\sqrt{3}} + \dfrac{100\sqrt{3}}{1}*\dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{100 + 100*3}{\sqrt{3}} = \dfrac{400}{\sqrt{3}}$, which is your final answer.
Here is the picture to help you see what I am doing:

</s>