Question

The height of a tower is 'h' and $\alpha$ is the angle of elevation of the top of the tower. On moving a distance $\frac{h}{2}$ towards the tower, the angle of elevation becomes $\beta$. What is the value of $\cot \alpha -\cot \beta$?

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. $2$

Answer 1

Answer: (A)

$\frac{1}{2}$

In, $△ACB,$

$\cot \beta =\frac{BC}{AB}=\frac{x}{h}$

$\Rightarrow \cot \beta =\frac{x}{h}$

$x=h\cot \beta ⟶\left(1\right)$

In $△ADB,$

$\cot \alpha =\frac{BD}{AB}=\frac{x+\frac{h}{2}}{h}$

$h\cot \alpha =x+\frac{h}{2}$

$x=h\cot \alpha -\frac{h}{2}⟶\left(2\right)$

Subtracting $\left(2\right)\&\left(1\right),$ we get

$h\cot \alpha -\frac{h}{2}-h\cot \beta =0$

$(\cot \alpha -\cot \beta )=\frac{h}{2}$

$\cot \alpha -\cot \beta =\frac{1}{2}$

Hence, the answer is $\frac{1}{2}.$