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A man standing on the bank of a river observes that the angle of elevation of the top of a tree just on the opposite bank is $60^{\circ}$. The angle of elevation is $30^{\circ}$ from a point at a distance $y m$ from the bank of the river. What is the height of the tree?
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Verified Answer
The correct answer is:
$\frac{\sqrt{3} y}{2} m$
Let DC be the tree of height $h$ metre. Let a man is standing on the point $B$ (bank of a river) Let $\mathrm{BC}=x$ and angle of elevation i.e. $\angle \mathrm{DBC}=60^{\circ} .$
$$
\text { Also, let } \mathrm{AB}=y \text { and } \angle \mathrm{DAC}=30^{\circ}
$$
$\operatorname{In} \Delta \mathrm{ACD}$
$\tan 30^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}$
$\Rightarrow x+y=h \sqrt{3}$
and in $\Delta \mathrm{BCD}$,
$$
\tan 60^{\circ}=\frac{\mathrm{CD}}{\mathrm{BC}} \Rightarrow \sqrt{3}=\frac{h}{x}
$$
$\Rightarrow x=\frac{h}{\sqrt{3}}$
From eqns. (i) and(ii),
$$
\frac{h}{\sqrt{3}}+y=h \sqrt{3}
$$
$\Rightarrow y=h\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=\frac{2 h}{\sqrt{3}} \Rightarrow h=\frac{\sqrt{3} y}{2} \mathrm{~m}$
$$
\text { Also, let } \mathrm{AB}=y \text { and } \angle \mathrm{DAC}=30^{\circ}
$$
$\operatorname{In} \Delta \mathrm{ACD}$
$\tan 30^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}$
$\Rightarrow x+y=h \sqrt{3}$
and in $\Delta \mathrm{BCD}$,
$$
\tan 60^{\circ}=\frac{\mathrm{CD}}{\mathrm{BC}} \Rightarrow \sqrt{3}=\frac{h}{x}
$$
$\Rightarrow x=\frac{h}{\sqrt{3}}$
From eqns. (i) and(ii),
$$
\frac{h}{\sqrt{3}}+y=h \sqrt{3}
$$
$\Rightarrow y=h\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=\frac{2 h}{\sqrt{3}} \Rightarrow h=\frac{\sqrt{3} y}{2} \mathrm{~m}$
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