Trigonometry inside a trapezium !enter image description here 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 ?

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.