The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance ‘d’ towards the foot of the tower the angle of elevation is found to be β. The height of the tower is
A. $\frac{d}{\cot \alpha+\cot \beta}$ B. $\frac{d}{\cot \alpha-\cot \beta}$ C. $\frac{d}{\tan \beta-\tan t \alpha}$ D. $\frac{d}{\tan \beta+\tan t \alpha}$ Answer: Option BShow Answer
Solution(By Apex Team)
Let AB be the tower and C is a point such that the angle of elevation of A is α.
After walking towards the foot B of the tower, at D the angle of elevation is β.
Let h be the height of the tower and DB = x Now in ΔACB,
$\begin{array}{l}\tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{AB}{CB}\\
\tan\alpha=\frac{h}{d+x}\\
\Rightarrow d+x=\frac{h}{\tan\alpha}\\
\Rightarrow d+x=h\cot\alpha\\
\Rightarrow x=h\cot\alpha-d\ldots\ldots\text{ (i) }\\
\text{Similarily in right }\triangle ADB,\\
\tan\beta=\frac{h}{x}\\
\Rightarrow x=\frac{h}{\tan\beta}\\
\Rightarrow x=h\cot\beta\ldots\ldots..\text{ (ii) }\\
\text{ From (i) and (ii) }\\
h\cot\alpha-d=h\cot\beta\\
\Rightarrow h\cot\alpha-h\cot\beta=d\\
\Rightarrow h(\cot\alpha-\cot\beta)=d\\
\Rightarrow h=\frac{d}{\cot\alpha-\cot\beta}\\
\therefore\text{ Height of the tower },\\
=\frac{d}{\cot\alpha-\cot\beta}\end{array}$
$\begin{array}{l}\tan\theta=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{AB}{CB}\\
\tan\alpha=\frac{h}{d+x}\\
\Rightarrow d+x=\frac{h}{\tan\alpha}\\
\Rightarrow d+x=h\cot\alpha\\
\Rightarrow x=h\cot\alpha-d\ldots\ldots\text{ (i) }\\
\text{Similarily in right }\triangle ADB,\\
\tan\beta=\frac{h}{x}\\
\Rightarrow x=\frac{h}{\tan\beta}\\
\Rightarrow x=h\cot\beta\ldots\ldots..\text{ (ii) }\\
\text{ From (i) and (ii) }\\
h\cot\alpha-d=h\cot\beta\\
\Rightarrow h\cot\alpha-h\cot\beta=d\\
\Rightarrow h(\cot\alpha-\cot\beta)=d\\
\Rightarrow h=\frac{d}{\cot\alpha-\cot\beta}\\
\therefore\text{ Height of the tower },\\
=\frac{d}{\cot\alpha-\cot\beta}\end{array}$
