Question

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 9 m. Find the height of the tree.
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Answer 1

➭ Given :-

⟶ angle of elevation from the ground to broken part of the tree and,

⟶ The distance between the foot of the tree to the point where the top touches the ground is 9 m.

➭ To find :-

⟶ height of the tree

➭ Solution :-

Let AB be the original height of the tree.

Suppose it got bent at a point C and let the part CB take the position CD, meeting the ground at D. Then,

→ AD = 9 m

→ ∠ADC = 30°

→ CD =CB.

Let AC = x metres and CD = CB = y metres.

From right ∆DAC, we have

$⟹ \frac{AC}{AD} = \tan \: 30° = \frac{1}{ \sqrt{3} }$

$⟹ \frac{x}{6} = \frac{1}{ \sqrt{3} }$

$⟹ x = \frac{9}{ \sqrt{3} }$

$⟹ x = \frac{9}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = 3 \sqrt{3}$

Also, from right ∆DAC, we have

$⟹ \frac{CD}{AD} = \sec30 = \frac{2}{ \sqrt{3} }$

$⟹ \frac{y}{9} = \frac{2}{ \sqrt{3} }$

$⟹ y = <strong>\frac{18}{ \sqrt{3} } </strong>$

$⟹ y = <strong>\frac{18}{ \sqrt{3} }</strong> \times <strong>\frac{ \sqrt{3} }{ \sqrt{3} } = 6 \sqrt{3}</strong>$

∴ Ac = $3\sqrt{3}m$

CB = $6\sqrt{3}m$

total height of the tree = AC + CB

= $3\sqrt{3}m$ + $6\sqrt{3}m$

= $9\sqrt{3}m$

Hence height of the tree is $9\sqrt{3}m$

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Note :-

refer the above attachment

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\colorbox{navy}{\color{silver}{\tt{❝Answer❞}}}$

Let, AC be the broken part of the tree.

∴ Total height of the tree =AB + AC

$\: \: \: \: \: \: \: \: \: \sf{In \: right \: ∆ ABC,}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \sf{cos 30°= \frac{BC}{AC} }$

$\: \: \: \: \: \: \: \: \: \: \: \sf\rightarrow{ \frac{1}{ \sqrt{3} } = \frac{8}{AC} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\rightarrow\sf{AC = \frac{16}{ \sqrt{3} } }}}$

$\sf{Also, tan30° = \frac{AB}{BC} }$

$\: \: \: \: \: \: \: \: \: \: \: \sf\rightarrow{ \frac{1}{ \sqrt{3} } = \frac{AB}{8} }$

$\: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\rightarrow\sf{AB = \frac{8}{ \sqrt{ 3} } }}}$

$\sf{∴ Total \: height \: of \: tree = AB + AC}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \frac{8}{ \sqrt{3} } + \frac{16}{ \sqrt{3} } }$

$\sf\underline{\underline{\bold{∴ Total \: height \: of \: tree = \frac{24}{ \sqrt{3} } m}}}$

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