Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag of height h. From a point, lying on the top of another tower on the same plane of height H, the angles of elevation of the top and bottom of the flag are $\alpha$ and $\beta$ respectively. The height of the first tower is

• A. $H+h\left(\frac{\tan \beta }{\tan \alpha -\tan \beta }\right)$
• B. $\frac{H}{2}+h\left(\frac{\tan \alpha -\tan \beta }{\tan \alpha }\right)$
• C. $\frac{1}{2}H+h\left(\frac{\tan \alpha }{\tan \alpha -\tan \beta }\right)$
• D. $H+h\left(\frac{2}{\tan \alpha -\tan \beta }\right)$

Answer 1

The correct option is A $H+h\left(\frac{\tan \beta }{\tan \alpha -\tan \beta }\right)$

Let CB represent the flag, BF represent the first tower and OE represent the another tower.
Given that the angle of elevation of the top and the bottom of the flag are α and β respectively, from the height H.
Let AB = y and OA = x
∴ Height of the first tower, BF = y + H
In ΔOAC,
$tan\alpha =\frac{CA}{OA}$
$\Rightarrow tan\alpha =\frac{y+h}{x}$
$\Rightarrow x=\frac{y+h}{tan\alpha }$ ....(i)
Now, in ΔOAB,
$tan\beta =\frac{AB}{OA}$
$tan\beta =\frac{y}{x}$
$\Rightarrow x=\frac{y}{tan\beta }$ .....(ii)
From (i) and (ii), we get
$\frac{y}{\tan \beta }=\frac{y+h}{\tan \alpha }$
$\Rightarrow ytan\alpha =(y+h)tan\beta$
$\Rightarrow y(tan\alpha -tan\beta )=htan\beta$
$\Rightarrow \frac{h\tan \beta }{(\tan \alpha -\tan \beta )}$
Thus, the height of the first tower = $H+y$ = $H+h\left(\frac{\tan \beta }{\tan \alpha -\tan \beta }\right)$
Hence, the correct answer is option (a).