The angle of elevation of the top of a vertical tower from a point $\mathrm{P}$ on the horizontal ground was…

1. $\frac{2 \sin \alpha \sin \beta}{\sin (\beta-\alpha)}$
2. $\frac{\sin \alpha \sin \beta}{\cos (\beta-\alpha)}$
3. $\frac{2 \sin (\beta-\alpha)}{\sin \alpha \sin \beta}$
4. $\frac{\cos (\beta-\alpha)}{\sin \alpha \sin \beta}$

Solution

$ \text { Let } \mathrm{AB} \text { be the tower of height ' } \mathrm{h} \text { '. } $
Given : In $\triangle \mathrm{ABP}$ $\tan \alpha=\frac{\mathrm{AB}}{\mathrm{PB}}$ or $\frac{\sin \alpha}{\cos \alpha}=\frac{h}{x+2}$ $\Rightarrow(x+2) \sin \alpha=\mathrm{h} \cos \alpha$ $\Rightarrow h=\frac{x \sin \alpha+2 \sin \alpha}{\cos \alpha}$ Now, In $\Delta \mathrm{ABC}, \tan \beta=\frac{\mathrm{AB}}{\mathrm{BC}}$ $ \Rightarrow \frac{\sin \beta}{\cos \beta}=\frac{h}{x} \Rightarrow x=\frac{h \cos \beta}{\sin \beta} $ Putting the value of $x$ in eq. (2) to eq. (1), we get $ \begin{aligned} &h=\frac{\frac{h \cos \beta \sin \alpha}{\sin \beta}+\frac{2 \sin \alpha}{1}}{\cos \alpha} \\ &\Rightarrow h=\frac{h \cos \beta \cdot \sin \alpha+2 \sin \alpha \sin \beta}{\sin \beta \cdot \cos \alpha} \\ & \end{aligned} $ $\Rightarrow h(\sin \beta \cdot \cos \alpha-\cos \beta \cdot \sin \alpha)=2 \sin \alpha \cdot \sin \beta$ $\Rightarrow h[\sin (\beta-\alpha)]=2 \sin \alpha \cdot \sin \beta$ $\Rightarrow h=\frac{2 \sin \alpha \cdot \sin \beta}{\sin (\beta-\alpha)}$

Asked in: JEE Main 2014 (11 Apr Online)

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