There are $12$ similar triangles with angles $30°, 60°$ and $90°$. In the diagram below you can see how the first three triangles are arranged. They are arranged so all of them have one of their points at the point $O$. One of the sides of the $(n+1)$th triangle is the same length as the $n$th triangle’s hypotenuse, and they are arranged so those two sides line up.
If the side $OX$ has length $1$, then what is the area of all of the triangles added together? Below is the image: http://www.ucl.ac.uk/clie/placement-tests/UPC/Maths-2/images/question21a.jpg
I only labeled the angles, other than that I have no clue as to how to solve this problem.
Hint
By the picture:
$$\cos 30º=\frac{l_n}{l_{n+1}}=\frac{\sqrt{3}}{2} \rightarrow l_{n+1}=\frac{2\sqrt{3}\cdot l_n}{3}$$
So the sides are in a geometric sequence.
We also have that the area of the $n-th$ triangle will be, as a function of $l_n$:
$$A_n=\frac{h_n\cdot l_n}{2}=\frac{l_n\cdot \tan 30º\cdot l_n}{2}=\frac{(l_n)^2\cdot \tan 30º}{2}$$
Can you finish?