#### Answer

See below.

#### Work Step by Step

The angle $\theta_{1}$, under which you see the whole height,
from the floor to the upper edge of the blackboard, is such that
$\displaystyle \cot\theta_{1}=\frac{15}{x}$
The angle $\theta_{2}$, under which you see the height
from the floor to the lower edge of the blackboard, is such that
$\displaystyle \cot\theta_{2}=\frac{3}{x}$
The angle $\alpha$ is such that
$\alpha=\theta_{1}-\theta_{2}$
$\displaystyle \alpha=\cot^{-1}\frac{15}{x}-\cot^{-1}\frac{3}{x}$