From the top of a tower,the angles of depression of two vehicles on the same side of the tower are found to be $\alpha$ and $\beta$ $(\alpha > \beta)$. If the distance between the vehicles is $b$,show that the height of the tower is $\frac{b \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}$.

(N/A) Let $AB$ be the tower of height $h$ and $C$ and $D$ be the two vehicles on the same side of the tower,where $C$ is closer to the tower.
Given $CD = b$. Let $BC = x$.
Then $BD = BC + CD = x + b$.
The angles of depression from the top $A$ to $C$ and $D$ are $\alpha$ and $\beta$ respectively.
Thus,$\angle ACB = \alpha$ and $\angle ADB = \beta$ (alternate interior angles).
In right-angled $\Delta ABC$,$\tan \alpha = \frac{AB}{BC} = \frac{h}{x} \implies x = \frac{h}{\tan \alpha} \quad (1)$.
In right-angled $\Delta ABD$,$\tan \beta = \frac{AB}{BD} = \frac{h}{x + b} \implies x + b = \frac{h}{\tan \beta} \implies x = \frac{h}{\tan \beta} - b \quad (2)$.
Equating $(1)$ and $(2)$,we get $\frac{h}{\tan \alpha} = \frac{h}{\tan \beta} - b$.
Rearranging the terms: $b = \frac{h}{\tan \beta} - \frac{h}{\tan \alpha} = h \left( \frac{\tan \alpha - \tan \beta}{\tan \alpha \tan \beta} \right)$.
Therefore,$h = \frac{b \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}$.