Given (90^ - theta) = 0.8, find the value of theta.

Given that $\sin(90^\circ - \theta) = 0.8$.
We know the trigonometric identity $\sin(90^\circ - \theta) = \cos \theta$.
Therefore, we have:

$$\cos \theta = 0.8$$

$$\cos \theta = \frac{8}{10} = \frac{4}{5}$$

Recall that in a right-angled triangle, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$. So, the adjacent side is 4 and the hypotenuse is 5.

Let the opposite side be $y$.

$$(\text{opposite})^2 + (\text{adjacent})^2 = (\text{hypotenuse})^2$$ $$y^2 + 4^2 = 5^2$$ $$y^2 + 16 = 25$$ $$y^2 = 25 - 16$$ $$y^2 = 9$$ $$y = \sqrt{9}$$

Since $y$ represents a length, it must be positive:

$$y = 3$$

So, the opposite side is 3.

We know that $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$.

$$\tan \theta = \frac{3}{4}$$

The value of $\tan \theta$ is $\boxed{\frac{3}{4}}$.