Can someone help me solve this question which is from the chapter Some Applications of Trigonometry.

Dear Student

Let $AB$ be the surface of the lake and $P$ be the point vertically above $A$ such that $AP = h$ m.
Let $C$ be the position of the cloud and $D$ be its reflection on the lake.
Let the height of the cloud be $H$ metres. Thus, we have:
$BC = BD = H$ m
Draw $PQ$⊥$CD$.
Thus, we have:
∠$CPQ = \alpha$ and ∠$QPD = \beta$
$AP = BQ = h$ m
$CQ = (BC - BQ) = (H - h)$ m and $DQ = (H + h)$ m


From the right ∆$CQP$, we have:
$\frac{CQ}{PQ} = \tan \alpha$
⇒ $\frac{(H - h)}{PQ} = \tan \alpha$
⇒ $PQ = \frac{(H - h)}{\tan \alpha }$                  ...(i)
From the right ∆$QPD$, we have:
$\frac{DQ}{PQ} =\tan \beta$
⇒ $\frac{(H + h)}{PQ} = \tan \beta$
⇒ $PQ = \frac{(H + h)}{\tan \beta }$                 ...(ii)

From (i) and (ii), we get:
$\frac{(H - h)}{\tan \alpha } = \frac{(H + h)}{\tan \beta }$
 ⇒ $H \tan \beta - h \tan \alpha = H \tan \alpha + h \tan \beta$
 ⇒ $H (\tan \beta - \tan \alpha ) = h (\tan \alpha + \tan \beta )$
 ⇒ $H = \frac{h (\tan \alpha + \tan \beta )}{(\tan \beta - \tan \alpha )}$   ...(iii)
Using the value of $H$ in (i), we get:
$PQ = \frac{(H - h)}{\tan \alpha } =\frac{H}{\tan \alpha } -\frac{h}{\tan \alpha } = \frac{h(\tan \alpha + \tan \beta )}{\tan \alpha (\tan \beta - \tan \alpha )} -\frac{h}{\tan \alpha }$

                                                  $$\begin{gathered} = \frac{h \tan \alpha +h \tan \beta -h \tan \beta +h \tan \alpha }{\tan \alpha \left(\tan \beta -\tan \alpha \right)} \\ = \frac{2h \tan \alpha }{\tan \alpha \left(\tan \beta -\tan \alpha \right)} \\ = \frac{2h}{(\tan \beta - \tan \alpha )} \end{gathered}$$    ...(iv)

Now, to find $PC$, we have:
$\cos \alpha = \frac{PQ}{PC}$
⇒ $PC= \frac{PQ}{\cos \alpha }= PQ sec \alpha$
Putting the value of $PQ$ from (iv), we get:
$PC = \frac{2h sec \alpha }{(\tan \beta - \tan \alpha )}$

The distance of the cloud =
                                          $PC =\frac{2h sec\alpha }{(\tan \beta -\tan \alpha )}$







​Hence proved


Regards