Question

A tower subtends an angle $\alpha$ at a point $A$ in the plane of its base and the angle of depression of the foot of the tower at a height '$b$' just above $A$ is $\beta$, then the height of the tower is

• A. $b\tan \alpha \cot \beta$
• B. $b\cot \alpha \tan \beta$
• C. $b\tan \alpha \tan \beta$
• D. $b\cot \alpha \cot \beta$

Answer 1

Answer: (A)

$b\tan \alpha \cot \beta$

Let, $PQ$ be the height of the tower

$AP=BR=x\left(say\right)$

$AB=PR=b$

From $△APQ$

$\tan \alpha =\frac{PQ}{AP}$

$\Rightarrow \tan \alpha =\frac{PQ}{x}$

$\Rightarrow PQ=x\tan \alpha ....\left(1\right)$

Now. frpm $△BPR$

$\tan \beta =\frac{PR}{BR}$

$\Rightarrow \tan \beta =\frac{b}{x}$

$\Rightarrow x=\frac{b}{\tan \beta }$

$\Rightarrow x=b\cot \beta ....\left(2\right)$

Putting in $\left(1\right)$,

$PQ=b\cot \beta \tan \alpha$

Hence, $Op-A$ is correct.