Question

A man standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60⁰. When he moves 40m away from the bank, he finds the angle of elevation to be 30⁰. Find the height of the tree.​

Answer 1

Answer:

REQUIRED ANSWER:

• Height of the tree=34.64 m
• width of the river=20m

☞Given:

• A man standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60⁰. When he moves 40m away from the bank, he finds the angle of elevation to be 30⁰.

☞To Find:

• height of the tree.
• width of the river

☞Solution:

Let CD=h be the height of the tree and BC=x be the breadth of the river.

From the figure ∠DAC=30 °and ∠DBC=60 °

$In \: right \: angled \: triangle \:  △BCD, \\ tan60∘= \frac{DC}{BC} </p><p>$

$= > \sqrt{3} = \frac{h}{x} \\ = > h = x \sqrt{3} ....(1)$

From the right-angled triangle △ACD

$tan {}^{o} = \frac{h}{40 + x} \\ = > \sqrt{3h} = 40 + x \: \: \: ....(2)$

From (1) and (2) we have

$\sqrt{3} (x \sqrt{3} )40 + x \\ = > 3x = 40 + x \\ 3x - x = 40 \\ = > 2x = 40 \\ = > x = \frac{40}{2} = 20$

$From  (1) we \: get \:  h= x \sqrt{3} = 20 \sqrt{3} = 20 \times 1.732 = 34.64m$

∴ Height of the tree=34.64 m and width of the river=20m

Answer 2

AnswEr :

Angle of elevation of top of a tree standing on opposite bank is 60⁰. When he moves 40m away from the bank, he finds the angle of elevation to be 30⁰.

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Let's consider that Height of the tree be h & CD = h, BC = x.

Now, In ∆BCD = $\sf\dfrac{DC}{BC}$

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From, Trigonometric ratios : $\sf tan 60^{\circ} = \sqrt{3}$

$:\implies\sf \sqrt{3} = \dfrac{DC}{BC} \\\\\\:\implies\sf \sqrt{3} = \dfrac{h}{x} \qquad \quad \quad \bigg\lgroup\bf DC = h \: \& \: BC = x\bigg\rgroup\\\\\\:\implies\sf \bf{h = x \sqrt{3}}$

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In ∆ACD :

$\star\: \boxed{\sf{\purple{tan 30^{\circ} = \dfrac{1}{\sqrt{3}}}}}$

$:\implies\sf tan \ 30^{\circ} = \dfrac{DC}{AC} \\\\\\:\implies\sf tan \ 30^{\circ} = \dfrac{h}{40 + x} \qquad \quad \quad \bigg\lgroup\bf DC = h \: \& \: AC = 40 + x \bigg\rgroup \\\\\\:\implies\sf \dfrac{1}{\sqrt{3}} = \dfrac{h}{40 + x} \\\\\\ \qquad \sf Cross \ Multiplying \ : \\\\\\:\implies\sf \sqrt{3}h = 40 + x$ ⠀⠀⠀⠀⠀⠀ ⠀-eq(1).

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$\qquad\quad \underline{\sf{\bf{Substituting \ Value \ of \ h \ in \ eq_{n} \ 1 \ : }}}$

$:\implies\sf\sqrt{3}(x \sqrt{3}) = 40 + x \qquad \quad \quad \bigg\lgroup\bf h = x \sqrt{3} \bigg\rgroup \\\\\\:\implies\sf 3x = 40 + x \\\\\\:\implies\sf 3x - x = 40 \\\\\\:\implies\sf 2x = 40 \\\\\\:\implies\sf x = \cancel\dfrac{40}{2} \\\\\\:\implies\sf\bf {x = 20}$

$\qquad\quad \underline{\sf{\bf{Substituting \ Value \ of \ x \ in \ value \ of \ h \ : }}}$

$:\implies\sf h = x \sqrt{3} \\\\\\:\implies\sf\bf {h = 20 \sqrt{3}}$

${\therefore} \ \underline{\sf{\bf{Height \ of \ the \ tree \ is \ 20\sqrt{3} \ m. }}}$