Question

The angle of elevation of the top of tower from the top and bottom of a building h meter high are $\alpha$ and $\beta$ then the height of tower is

• A. $\frac{h\sin \alpha \cos \beta }{\sin (\alpha +\beta )}$
• B. $\frac{h\cos \alpha \cos \beta }{\sin (\beta -\alpha )}$
• C. $\frac{h\cos \alpha \sin \beta }{\sin (\beta -\alpha )}$
• D. None of these

Answer 1

The correct option is C $\frac{h\cos \alpha \sin \beta }{\sin (\beta -\alpha )}$

Let the height of the tower be $H.$

In $△ABC,$

$\Rightarrow \tan \alpha =\frac{AC}{BC}=\frac{AE-CE}{BC}=\frac{H-h}{BC}$

$\Rightarrow BC=\frac{(H-h)}{\tan \alpha }⟶\left(1\right)$

In $△ADE,$

$\Rightarrow \tan \beta =\frac{AE}{DE}=\frac{H}{BC}$ $[\because BC=DE]$

$\Rightarrow BC=\frac{H}{\tan \beta }⟶\left(2\right)$

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

$\Rightarrow \frac{H}{\tan \beta }=\frac{(H-h)}{\tan \alpha }=\frac{H}{\tan \alpha }-\frac{h}{\tan \beta }$

$\Rightarrow \frac{H}{\tan \alpha }=\frac{H}{\tan \beta }=\frac{h}{\tan \beta }$

$\Rightarrow H\left[\frac{\tan \beta -\tan \alpha }{\tan \alpha .\tan \beta }\right]=\frac{h}{\tan \beta }$

$\Rightarrow H=\frac{h(\tan \alpha .\tan \beta )}{(\tan \beta -\tan \alpha )\tan \beta }=\frac{h\tan \alpha }{\tan \beta -\tan \alpha }$

$\Rightarrow H=\frac{h\sin \alpha }{\cos \alpha [\frac{\sin \beta }{\cos \beta }-\frac{\sin \alpha }{\cos \alpha }]}=\frac{h\sin \alpha }{\cos \alpha \left[\frac{\sin \beta \cos \alpha -\sin \alpha \cos \beta }{\cos \alpha \cos \beta }\right]}$

$\Rightarrow H=\frac{h\sin \alpha \cos \beta }{\sin (\beta -\alpha )}$

Hence, the answer is $\frac{h\sin \alpha \cos \beta }{\sin (\beta -\alpha )}.$