The length of the drawn through lines are given. Is there a way to determine $h_b$ ? The angle $\alpha$ is unknown, but can be assumed to be small.
My approach so far was to to consider $\sin(\alpha) \approx \tan(\alpha) \Rightarrow \frac{h_b-x}{a}= \frac{h_b}{b}$, and $h_b-x=h_a$, leading to $h_b= \frac{x}{1-\frac{a}{b}}$. How to continue? Is there an elementary way to solve this?
$hyp_a = \sqrt{a^2 + h_a^2}$
Then $\frac{b}{\sqrt{a^2 + h_a^2}}=\frac{h_b}{h_a}$
Hence $h_b = \frac{bh_a}{\sqrt{a^2+h_a^2}}$