Correct Answer: $\frac{h\tan\alpha}{(\tan\beta–\tan\alpha)}$

Solution : Given: A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. The angle of elevation at the bottom of the flagstaff is $\alpha$ and that of the top of the flagstaff is $\beta$.
We know that $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$
Let AB be $x$.

From the figure, it is clear that in $\triangle ADB$, $\tan\alpha =\frac{AD}{AB}$
$\tan\alpha =\frac{H}{x}$
⇒ $x =\frac{H}{\tan\alpha}$
Similarly, in $\triangle CAB$, $\tan\beta =\frac{CA}{AB}$
$\tan\beta =\frac{h+H}{x}$
⇒ $x =\frac{h+H}{\tan\beta}$
So, $\frac{h+H}{\tan\beta}=\frac{H}{\tan\alpha}$
⇒ $H\tan\beta=h\tan\alpha+H\tan\alpha$
⇒ $H (\tan\beta–\tan\alpha)=h\tan\alpha$
⇒ $H =\frac{h\tan\alpha}{(\tan\beta–\tan\alpha)}$
Hence, the correct answer is $\frac{h\tan\alpha}{(\tan\beta–\tan\alpha)}$.