Question

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

Answer 1

The height of the tree is 5 (2 + √3) m

Step-by-step explanation:

Let PQ = x m

In ΔPQT, we have,

tan 45 = TQ/PQ = (h - 5)/x

⇒ 1 = (h - 5)/x

∴ x = h - 5

In ΔPOI, we have,

tan 60 = IQ/PQ = (h + 5)/8x

√3 = (h + 5)/x

Now, on substituting the value of x, we get,

√3 = (h + 5)/(h - 5)

√3h - 5√3 = h + 5

√3h - h = 5 + 5√3

h(√3 - 1) = 5(1 + √3)

h = 5(1 + √3)/(√3 - 1)

Now, on conjugating, we get,

h = 5(1 + √3)/(√3 - 1) × (√3 + 1)/(√3 + 1)

h = 5(1 + √3)²/(3 - 1)

h = 5(1 + √3)²/2

h = 5/2 (1 + 3 + 2√3)

h = 5/2 (4 + 2√3)

∴ h = 5 (2 + √3) m

Answer 2

$Let \:height \: of \:a \:coconut \:tree (AB) = h \:m$

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

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

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

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

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

$i) In \: \triangle FBD , \:we \: have , \\tan 69\degree = \frac{BD}{FB}$

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

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

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

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

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

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

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

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

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

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

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

$\implies h = \frac{5(4+2\sqrt{3})}{2}$

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

Therefore.,

$\red {Height \: of \:a \:coconut \:tree (AB)}\\\green {= 18.66 \:m}$

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