Explanation:

Let $\mathrm{BD}$ be horizontal position of the lake and $\mathrm{A}$ be the given point at height $\mathrm{h}$ above it. If position of the cloud and its reflection be $\mathrm{P}$ and $\mathrm{Q}$ respectively, then from diagram $\angle \mathrm{PAC}=\mathrm{a}$ and $\angle \mathrm{QAC}=\mathrm{b}$
Let $\mathrm{h}$ be the height of the cloud, then from triangles PCA and QCA $\displaystyle \frac{{h-a}}{{AC}}$ ​= tan α
and $\displaystyle \begin{array}{l}\dfrac{{h-a}}{{AC}}=\tan \beta \\\left( 2 \right)\div \left( 1 \right)=\dfrac{{h+a}}{{h-a}}=\dfrac{{\tan \alpha }}{{\tan \beta }}\\=\dfrac{{2h}}{{2a}}=\dfrac{{\tan \beta +\tan \alpha }}{{\tan \beta -\tan \alpha }}=\dfrac{{\sin \beta \cos \alpha +\sin \alpha \cos \beta }}{{\sin \beta \cos \alpha -\sin \alpha \cos \beta }}\\h=a\dfrac{{sin\left( {\alpha +\beta } \right)}}{{sin\left( {\beta -\alpha } \right)}}\end{array}$