Question

Angles of elevation of top of a tower when observed from ground floor and roof of a building of height $h$ are $\alpha $ and $\beta $ respectively, then height of tower will be

• A. $\frac{h cos \beta }{cot \beta - cot \alpha }$
• B. $\frac{h cot \beta }{cot \beta - cot \alpha }$
• C. $\frac{h cot \alpha }{cot \beta - cot \alpha }$
• D. $\frac{h cos \alpha }{cot \beta - cot \alpha }$

Answer 1

Answer: (B)

Let $AB$ be the building whose height is $h$ and $CD$ be the tower whose height be $H$ (let say).
Since, $AB=EC=h$ .
So, $DE=DC-EC=H-h$ .
Let $BC=AE=x$ .
Now, in $\triangle BCD,cot\alpha =\frac{B C}{D C}=\frac{x}{H}\Rightarrow x=Hcot\alpha ............\left(i\right)$
Also, in $\triangle DAE,cot\beta =\frac{A E}{D E}=\frac{x}{H - h}\Rightarrow x=\left(H - h\right)cot\beta ...............\left(i i\right)$
From equations $\left(i\right)\&\left(i i\right)$ , we get
$Hcot\alpha =\left(H - h\right)cot\beta $
$\Rightarrow Hcot\alpha =Hcot\beta -hcot\beta $
$\Rightarrow hcot\beta =H\left(c o t \beta - c o t \alpha \right)$
$\Rightarrow H=\frac{h c o t \beta }{c o t \beta - c o t \alpha }$