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. $\frac{hcos\beta }{cot\beta -cot\alpha }$
• B. $\frac{hcot\beta }{cot\beta -cot\alpha }$
• C. $\frac{hcot\alpha }{cot\beta -cot\alpha }$
• D. $\frac{hcos\alpha }{cot\beta -cot\alpha }$

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 $△BCD,cot\alpha =\frac{BC}{DC}=\frac{x}{H}\Rightarrow x=Hcot\alpha ............\left(i\right)$
Also, in $△DAE,cot\beta =\frac{AE}{DE}=\frac{x}{H-h}\Rightarrow x=\left(H-h\right)cot\beta ...............\left(ii\right)$
From equations $\left(i\right)\&\left(ii\right)$ , we get
$Hcot\alpha =\left(H-h\right)cot\beta$
$\Rightarrow Hcot\alpha =Hcot\beta -hcot\beta$
$\Rightarrow hcot\beta =H\left(cot\beta -cot\alpha \right)$
$\Rightarrow H=\frac{hcot\beta }{cot\beta -cot\alpha }$