Question

There is a coconut tree on the bank of a river from a boat 5 m above the water, the angle of
elevation of the top of the tree is 45° and the angle of depression of reflection of tree top is
60'. Find the height of the tree.​

Answer 1

$Let\:Height \:of \: tree (AB) = h \:m$

$Height \:of \: boat (EF) = 5\:m$

$Height \:of \: reflection \:of \: the \: tree\: (BD)\\ =( h+5)\:m$

$Let \: distance \: between \: boat \:to \\foot \:of \:the \:tree (FB) = x \:m$

$i ) In \:\triangle ABF , \:we \:have ,\\tan \:45\degree = \frac{AB}{FB}$

$\implies 1 = \frac{h-5}{x}$

$\implies x = h - 5 \: --(1)$

$ii ) In \:\triangle FBD , \:we \:have ,\\tan \:60\degree = \frac{BD}{FB}$

$\implies \sqrt{3} = \frac{h+5}{x}$

$\implies x = \frac{h + 5}{\sqrt{3}} \: --(2)$

/* From (1) and (2) , we get */

$h - 5 = \frac{h + 5}{\sqrt{3}}$

$\implies \sqrt{3}(h-5) = h + 5$

$\implies \sqrt{3}h-5\sqrt{3} = h + 5$

$\implies \sqrt{3}h - h = 5\sqrt{3} + 5$

$\implies (\sqrt{3} - 1) h = 5(\sqrt{3}+1)$

$\implies h = \frac{5(\sqrt{3}+1)}{(\sqrt{3}-1)}$

$\implies h = \frac{5(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

$\implies h = \frac{5(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-1^{2}}$

$\implies h = \frac{5( 3+1+2\sqrt{3})}{3-1}\\= \frac{5(4+2\sqrt{3})}{2}\\= \frac{5\times 2(2+\sqrt{3})}{2}\\= \frac{5(2+\sqrt{3})}{1}\\= 5(2+1.732)\\= 5\times 3.732 \\= 18.66 \: m$

Therefore.,

$\red { Height \:of \:the \:tree} \green {= 18.66 \: m }$

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