Question

The angle of elevation of the top of a vertical tower from a point P on the horizontal ground was observed to be $\alpha$. After moving a distance 2 metres from P towards the foot of the tower, the angle of elevation changes to $\beta$. Then the height (in metres) of the tower is :

• A. $\frac{\cos (\beta -\alpha )}{\sin \alpha \sin \beta }$
• B. $\frac{2\sin (\beta -\alpha )}{\sin \alpha \sin \beta }$
• C. $\frac{2\sin \alpha \sin \beta }{\sin (\beta -\alpha )}$
• D. $\frac{\sin \alpha \sin \beta }{\cos (\beta -\alpha )}$

Answer 1

Answer: (C)

$\frac{2\sin \alpha \sin \beta }{\sin (\beta -\alpha )}$

$L\tan \alpha =h$

$(L-2)\tan \beta =h$

$\therefore \frac{h}{\tan \alpha }=\frac{h}{\tan \beta }+2$

$\therefore h=\frac{2\tan \alpha \tan \beta }{\tan \beta -\tan \alpha }=\frac{2\frac{\sin \alpha }{\cos \alpha }\times \frac{\sin \beta }{\cos \beta }}{\frac{\sin \alpha \cos \beta -\cos \alpha \sin \beta }{\cos \alpha \cos \beta }}=\frac{2\sin \alpha \sin \beta }{\sin (\beta -\alpha )}$