Here is our situation. We are given the following setup:
I have omitted the "top" of the triangle (vertex $A$). We know the following angles:
$$
\begin{align}
\angle ABC = \angle SBC &= u + v = 62^{\circ} \\
\angle ACB = \angle HCB &= s + t = 104^{\circ} \\
\angle HBC &= u = 50^{\circ} \\
\angle SCB &= t = 94^{\circ} \\
\end{align}
$$
We wish to find:
$$
\angle HSC = x = {?}^{\circ}
$$
We can solve for the unknown angles $w$, $y$, and $z$ by noticing two facts: the angles in a triangle add up to a constant ($\pi$) and the diagonals of the quadrilateral $HSBC$ create two pairs of triangles that share an angle. Therefore:
$$
\begin{align}
t + u &= x + y \\
s + z &= v + w \\
\end{align}
$$
To get a third equation, note that the total of the interior angles of any quadrilateral add up to $2\pi$:
$$
s + t + u + v + w + x + y + z = 2\pi
$$
Solving this system of three equations gives us:
$$
\begin{align}
w &= \pi - t - u - v \\
y &= t + u - x \\
z &= \pi - s - t - u
\end{align}
$$
Now, we can use the law of sines to write a second set of relations:
$$
\begin{align}
\frac{\overline{HS}}{\sin s} &= \frac{\overline{CH}}{\sin x} &
\frac{\overline{CH}}{\sin u} &= \frac{\overline{BC}}{\sin z} &
\frac{\overline{BC}}{\sin w} &= \frac{\overline{SB}}{\sin t} &
\frac{\overline{SB}}{\sin y} &= \frac{\overline{HS}}{\sin v}
\end{align}
$$
Multiplying them together, the lengths cancel:
$$
\sin s \sin u \sin w \sin y = \sin t \sin v \sin x \sin z
$$
Adding in our substitutions from before (and remembering $\sin(\pi-x)=\sin x$):
$$
\sin s \sin u \sin (t+u+v) \sin (t+u-x) = \sin t \sin v \sin x \sin (s+t+u)
$$
(This approach is adapted from the webpage Angular Angst that Fimpellizieri posted in this comment on the question.)
Now proving that the answer is $x=34^{\circ}$ (oops, did I say that out loud?) reduces to proving the following trigonometric identity:
$$
\sin 10^{\circ} \sin 50^{\circ} \sin 110^{\circ} \sin 156^{\circ} = \sin 12^{\circ} \sin 34^{\circ} \sin 94^{\circ} \sin 154^{\circ}
$$
or equivalently:
$$
\begin{multline}
\sin\left(\frac{\pi}{18}\right) \sin\left(\frac{5\pi}{18}\right) \sin\left(\frac{11\pi}{18}\right) \sin\left(\frac{13\pi}{15}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{17\pi}{90}\right) \sin\left(\frac{47\pi}{90}\right) \sin\left(\frac{77\pi}{90}\right)
\end{multline}
$$
We can expand some of the fractions into some interesting sums:
$$
\begin{multline}
\sin\left(\frac{\pi}{6}-\frac{\pi}{9}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{9}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{9}\right) \sin\left(\frac{13\pi}{15}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{45}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{45}\right) \sin\left(\frac{5\pi}{6}+\frac{\pi}{45}\right)
\end{multline}
$$
Remember that $\sin x=\sin(\pi-x)$; this allows us to put our identity into an even more intriguing form:
$$
\begin{multline}
\sin\left(\frac{2\pi}{15}\right) \sin\left(\frac{\pi}{6}-\frac{\pi}{9}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{9}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{9}\right) \\ = \sin\left(\frac{\pi}{15}\right) \sin\left(\frac{\pi}{6}+\frac{\pi}{45}\right) \sin\left(\frac{\pi}{6}-\frac{\pi}{45}\right) \sin\left(\frac{\pi}{2}+\frac{\pi}{45}\right)
\end{multline}
$$
Let's see if we can't make a useful identity with the pattern that we see:
$$
\sin\left(\frac{\pi}{6}-\alpha\right) \sin\left(\frac{\pi}{6}+\alpha\right) \sin\left(\frac{\pi}{2}+\alpha\right) \\
= \left(\frac{1}{2}\cos\alpha - \frac{\sqrt{3}}{2}\sin\alpha \right)\left(\frac{1}{2}\cos\alpha + \frac{\sqrt{3}}{2}\sin\alpha \right)\cos\alpha \\
= \frac{1}{4}\left(\cos^2\alpha - 3\sin^2\alpha\right)\cos\alpha \\
= \frac{1}{4}\left(4\cos^3\alpha - 3\cos\alpha\right) \\
= \frac{\cos(3\alpha)}{4}
$$
With this in mind, we can collapse our identity to:
$$
\sin\left(\frac{2\pi}{15}\right)\cos\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right)
$$
This is beginning to look a lot more convenient. Using the double angle formula on $\sin(2\cdot\pi/15)$ and evaluating $\cos(\pi/3)$ we obtain:
$$
2\sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right)\frac{1}{2} = \sin\left(\frac{\pi}{15}\right)\cos\left(\frac{\pi}{15}\right)
$$
Which is obviously true.
Since we know that the solution is unique, this proves that the solution is $34^{\circ}$. As to why, the only answer I can offer is "because it makes the beautiful cancellations above possible."
