How to find angle in between two right triangles when one side is twice of other? The problem is as follows:

$\begin{array}{ll}
1.&10^\circ\\
2.&12^\circ\\
3.&15^\circ\\
4.&16^\circ\\
\end{array}$
I'm not sure which sort of construction can be used here to solve this problem?.
I've attempted to draw a perpendicular line from $B$ to segment $AC$. But this did not yielded good results.
Then I've attempted tracing a perpendicular segment to $BC$ intersecting $AC$, from looking on this possibility. I got still stuck.
I don't know how to use the given angles, they suspiciously add up to $3\alpha$.
I still don't know how to use $QC=2HC$. Can someone give me some ideas on what to do?. Should congruence be used here?.
I'm assuming that the intended approach is try to spot triangles such as:
$3-4-5$ or $1-\sqrt{3}-2$ or something along those special right triangles.
I do hope someone could help me on how to solve this problem relying on euclidean geometry can this be done?.


Construct isosceles triangle $CQD$ and connect $PD$. Note that
$$\angle HPD = \angle HPB + \angle BPD =(90^\circ +\angle HCP) + (90^\circ -\angle BDP)=180^\circ
$$
Thus, $H$, $P$ and $D$ are colinear and
$$\cos\angle HCD = \cos5\alpha = \frac{HC}{DC} = \frac12$$
which yields
$$\alpha = 12^\circ$$
