Solve The Triangle

Let's name the triangle as $\triangle ABC$ which is right angled at B so that $x$ is hypotenuse. And this is a 30-60-90 triangle.

Let the perpendicular from B, drop on the Hypotenuse at D giving $\triangle ABD$ and $\triangle BDC$, and both of these small triangle are 30-60-90 triangle. So, apply the trignometry identities.

The length $x$ will be $$=100\cdot tan 30^\circ+100 \cdot tan 60^\circ$$

Or, $$100 \cdot {\frac{4}{\sqrt 3}}$$

A little hint, remembering sin, cos and tan identities and values is lot simple and elegant than remembering thew cosec, sec and cot values.

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:

First, lets take a generic hypotenuse-opposite-adjacent right triangle and divide each side by the length of the opposite side (opposite to angle $t$). This gives us a similar triangle with sides $\csc t$, $1$, and $\cot t$. Next, multiply each side by 100 (see below) to get another similar triangle with a height of $100$, hypotenuse of $100\csc t$, and a horizontal base of $100\cot t$.

Now we want to apply our result to a $60$ degreed, and a $30$ degreed right triangle, and add the horizontal bases together.

Our result becomes $$100\cot 60^\circ + 100\cot 30^\circ=100(\sqrt{3}+1/\sqrt{3})=400/\sqrt{3}$$

Both of the current answers utilize trig functions. I thought I would add an answer that doesn't use trig functions, but rather uses scaling of two $(30,60,90)$ triangles.

We are used to seeing the sides of a $(30, 60, 90)$ triangle being written as $(1, \sqrt{3}, 2)$ (where $1$ is the length opposite to the $30$ degree angle, $\sqrt 3$ is the length opposite to the $60$ degree angle, etc). We can then scale the side lengths by whatever multiplier we wish. The diagram below shows how to scale the side lengths appropriately for this problem.

For the left-hand triangle (top), we choose the multiplier $\frac{100}{\sqrt 3}$ (because the side opposite the $60$ degree angle is $100$ units long). For the right-hand triangle (bottom), we choose the multiplier $100$ (because the side opposite the $30$ degree angle is $100$ units long).

Thus, the length $x$ can be found easily: $$x = 100\sqrt{3} + \frac{100}{\sqrt 3} = \frac{400\sqrt{3}}{3}$$