How to find triangle height? I need to know the height ($h$) of a triangle with two unknown angles ($\alpha$ and $\beta$) and the known length of two sides $AB$ and $BC$.

Is it possible to have that value of $h$ (height)?
 A: Your post says that the angles $\alpha$ and $\beta$ are unknown.  I assume this means that only $AB$ and $BC$ are known.  Remove the line $AC$ from your diagram. Then you can turn the line $BC$ around the "hinge" at $B$, changing the height.  So the answer is that the height cannot be determined.
The only thing you can say is that the height is greater than $0$ and less than or equal to $BC$.
A: If you let D denote the point of intersection of the perpendicular point C to segment AB (so that the length of CD = h), then we have a right triangle BCD ($\angle BDC = 90^\circ = \pi/2$ radians), 
then you have that $\sin(\beta)$ = opposite side / hypotenuse = $\displaystyle \frac{h}{|BC|}$
Then solve for h: $h = \sin(\beta)\times |BC|$.
Since you have the values for $\beta$ and for $|BC|$ (the length of segment BC), simply compute h at those values.
Note: this answer was posted when the question stated $\alpha, \beta$ were known.
With only two sides of the triangle known (and that the line perpendicular to AB, and intersecting C, forms divides ABC into two right triangles), we still don't have enough information to determine height (h).
See Americo's comment above for a list of values (side lengths and/or angles) necessary to define a triangle. (And his subsequent comment re: "case 4")!
A: From the definition of $\sin$ function, $h = |BC|\sin\beta$.
Even if you know only $|AB| = a$ (which seems to be the case from your picture) then:
$h = ka\cos\alpha$ and $h = (1-k)a\cos\beta$ for some $k<1$. This $k$ is a proportion in which $h$ divides $AB$. So, we have to equation on two variables: Jejeje
$$
k = \frac{\cos\beta}{\cos\alpha+\cos\beta}
$$
and 
$$
h = \frac{\cos\alpha\cos\beta}{\cos\alpha+\cos\beta}a.
$$
