## University Calculus: Early Transcendentals (3rd Edition)

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 $\alpha=\theta_{1}-\theta_{2}$ $\displaystyle \alpha=\cot^{-1}\frac{15}{x}-\cot^{-1}\frac{3}{x}$