Trigonometry inside a trapezium 
I have the following image, and it's asked to find the values of $X$ and $Y$.
I've managed to find it using the this idea: Divide the image in two right triangles and let's call the height of the trapezium $H$. 
The opposite cathetus of the left triangle has a length of $\frac{H}{\tan{60^{\circ}}}$ and the opposite cathetus of the right triangle is $\frac{H}{\tan{30^{\circ}}}$.
The sum of this two catheti has to be equals to $12$, in this sum, we can assume that the height $H = 3\sqrt{3}$.
Applying trigonometrical functions in both triangles, I managed to find that $X = 6, Y = 6\sqrt{3}$
But a friend of mine has found $X = 8, Y = \frac{16\sqrt{3}}{3}$, and he did it in a completely different manner from mine.
Which one is right ?
 A: After the construction of the red parallel line, the problem becomes more easy to solve.
  
A: Call the left right-angled triangle's lower left $\;x\;$,  so that $\;X=2x\;,\;\;H=\sqrt3x\;$, and thus in the right right-angled triangle we have that the lower leg is $\;3x\;$, and its hypotenuse is $\;2\sqrt3 x\;$.
Adding both lower legs above we get
$$x+3x=4x=12\implies x=3\implies \begin{cases}X=6\\{}\\Y=2\sqrt3\cdot3=6\sqrt3\end{cases}$$ 
and the above only uses basic Euclidean geometry.
A: The easiest way to solve this problem is to create three right triangles: two of the inside the trapezium, and one outside.
the two inside will be created by drawing two lines down from and perpendicular to the top 8 cm leg. This will give you two right triangles with a rectangle between them.
A: You can do without finding H. Let $B$ and $b$ be the two bases of the trapezium. Then $$\frac 1 2 X+\frac {\sqrt 3}{2}Y=B-b \\\frac{\sqrt 3} 2 X = \frac 1{2}Y.$$ Surely you can solve this system.
A: If you just cut out eight centimeters from the top and from the bottom, then you have a $30^\circ{-}60^\circ{-}90^\circ$ triangle whose hypotenuse has length 12 centimeters.  Its shorter leg must therefore have length 6 centimeters, and then the other leg can be found via Pythagoras.
