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?.
 A: 
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
$$\alpha = 12^\circ$$
A: Hint: Using trigonometry and $|QC|=2|HC|$, express the lengths of $BQ,BP,PC$, and $BC$ in terms of $|HC|$. Now observe that $B,P,C$ are collinear.
A: Since they are complementary to angles that subtend $3\alpha$,
$$
\overbrace{\ \angle HPC\ }^{\text{co-}HCP}=\frac\pi2-3\alpha=\overbrace{\ \angle QPB\ }^{\text{co-}PQB}\tag1
$$
Since $\angle BPC=\pi$, we get
$$
\angle QPH=6\alpha\tag2
$$

Extend $\overline{QB}$ by an equal length to $R$; that is, $\overline{BR}=\overline{QB}$. Since $\triangle QCR$ is isosceles,
$$
\angle BCR=\angle BCQ=2\alpha\tag3
$$
Likewise,
$$
\angle BPR=\angle BPQ=\frac\pi2-3\alpha\tag4
$$
Thus, $\angle RPH=\pi$. That is, $R$, $P$, and $H$ are colinear.
Furthermore,
$$
\overline{RC}=\overline{QC}\tag5
$$
Extend $\overline{HC}$ by an equal length to $S$; that is, $\overline{HS}=\overline{HC}$. Thus,
$$
2\overline{HC}=\overline{SC}\tag6
$$
and since $\triangle SRC$ is isosceles, we have
$$
\overline{RS}=\overline{RC}\tag7
$$
Now we use that $\color{#C00}{\overline{QC}=2\overline{HC}}$. $\color{#090}{(7)}$, $\color{#44F}{(5)}$, and $\color{#CA0}{(6)}$ say
$$
\overline{RS}\color{#090}{=}\overline{RC}\color{#44F}{=}\overline{QC}\color{#C00}{=}2\overline{HC}\color{#CA0}{=}\overline{SC}\tag8
$$
Thus, $\triangle RCS$ is equilateral, and therefore,
$$
5\alpha=\angle RCS=\frac\pi3=60^{\large\circ}\tag9
$$
Thus,
$$
\alpha=\frac\pi{15}=12^{\large\circ}\tag{10}
$$
