Question

A tower subtends an angle $\alpha$ at a point in the plane of its base and the angle of depression of the foot of the tower at a point $b ft$ just above $A$ is $\beta$. Then 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)

Let the height of the tower be $h$.
In $\triangle A B C$,
$\tan \alpha=\frac{h}{A B}$

$\Rightarrow A B =h \cot \alpha$ ...(i)
In $\Delta A B D,$
$\tan \beta =\frac{b}{A B}$
$\Rightarrow A B =b \cot \beta$ ...(ii)
From Eqs. (i) and (ii)
$h \cot \alpha=b \cot \beta$
$\Rightarrow h=b \tan \alpha \cot \beta$