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A vertical pole with height more than $100 \mathrm{~m}$ consists of two parts, the lower being one-third of the whole. At a point on a horizontal plane through the foot and $40 \mathrm{~m}$ fromit, the upper
part subtends an angle uhose tangent is $\frac{1}{2}$. What is the height of the pole ?
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part subtends an angle uhose tangent is $\frac{1}{2}$. What is the height of the pole ?
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Verified Answer
The correct answer is:
120 m
Let h be the height of pole, upper portion CD subtend angle $\theta$ at $A$.
Then, $\tan \theta=\frac{1}{2}$
Let lower part $\mathrm{BC}$ subtend angle $\phi$ at A then $\operatorname{In} \Delta \mathrm{ABC}$
$\tan \phi=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{h} / 3}{40}=\frac{\mathrm{h}}{120}$
$\operatorname{In} \Delta \mathrm{ABD}$,
$\tan (\theta+\phi)=\frac{\mathrm{BD}}{\mathrm{AB}}$
$\Rightarrow \frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{\mathrm{h}}{40}$
$\Rightarrow \frac{\frac{1}{2}+\frac{\mathrm{h}}{120}}{1-\frac{\mathrm{h}}{240}}=\frac{\mathrm{h}}{40}$
$\Rightarrow \frac{2(60+\mathrm{h})}{(240-\mathrm{h})}=\frac{\mathrm{h}}{40}$
$\Rightarrow 80(60+\mathrm{h})=240 \mathrm{~h}-\mathrm{h}^{2} \Rightarrow 4800+80 \mathrm{~h}=240 \mathrm{~h}-\mathrm{h}^{2}$
$\Rightarrow \mathrm{h}^{2}-160 \mathrm{~h}+4800=0 \quad \Rightarrow(\mathrm{h}-120)(\mathrm{h}-40)=0$
$\Rightarrow \mathrm{h}=120$
$[\mathrm{h}=40$ is discarded, since $\mathrm{h}>100$ is given]
Then, $\tan \theta=\frac{1}{2}$
Let lower part $\mathrm{BC}$ subtend angle $\phi$ at A then $\operatorname{In} \Delta \mathrm{ABC}$
$\tan \phi=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{h} / 3}{40}=\frac{\mathrm{h}}{120}$
$\operatorname{In} \Delta \mathrm{ABD}$,
$\tan (\theta+\phi)=\frac{\mathrm{BD}}{\mathrm{AB}}$
$\Rightarrow \frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{\mathrm{h}}{40}$
$\Rightarrow \frac{\frac{1}{2}+\frac{\mathrm{h}}{120}}{1-\frac{\mathrm{h}}{240}}=\frac{\mathrm{h}}{40}$
$\Rightarrow \frac{2(60+\mathrm{h})}{(240-\mathrm{h})}=\frac{\mathrm{h}}{40}$
$\Rightarrow 80(60+\mathrm{h})=240 \mathrm{~h}-\mathrm{h}^{2} \Rightarrow 4800+80 \mathrm{~h}=240 \mathrm{~h}-\mathrm{h}^{2}$
$\Rightarrow \mathrm{h}^{2}-160 \mathrm{~h}+4800=0 \quad \Rightarrow(\mathrm{h}-120)(\mathrm{h}-40)=0$
$\Rightarrow \mathrm{h}=120$
$[\mathrm{h}=40$ is discarded, since $\mathrm{h}>100$ is given]
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