Given (5x-28)^ = (3x-50)^, find the value of x where 0 < x < 90^.

Given $\sin(5x-28)^\circ = \cos(3x-50)^\circ$.
We can rewrite $\sin(5x-28)^\circ$ as $\cos(90^\circ - (5x-28)^\circ)$.

$$\cos(90^\circ - (5x-28)^\circ) = \cos(3x-50)^\circ$$

$$90 - (5x - 28) = 3x - 50$$ $$90 - 5x + 28 = 3x - 50$$

$$118 - 5x = 3x - 50$$

Add $5x$ to both sides:

$$118 = 8x - 50$$

Add 50 to both sides:

$$118 + 50 = 8x$$ $$168 = 8x$$

Divide by 8:

$$x = \frac{168}{8}$$ $$x = 21$$

Since $x = 21$, and $0 < 21 < 90$, the value is valid.

The value of $x$ is $\boxed{\text{21}}$.