Step-by-step Solution:

\[ \begin{aligned} \text{Let AD} &= h \text{ be the height of the building and EB be the height of the hill.} \\ \text{We have } \tan q &= \frac{h+x}{y} \\ \text{and } \tan p &= \frac{x}{y} \\ \Rightarrow y &= x \cot p \\ \text{Now, } \tan q &= \frac{h+x}{x \cot p} \\ \Rightarrow x \cot p &= (h+x) \cot q \\ \Rightarrow x &= \frac{h \cot q}{\cot p - \cot q} \\ \therefore h+x &= h + \frac{h \cot q}{\cot p - \cot q} \\ &= \frac{h \cot p}{\cot p - \cot q} \end{aligned} \]