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 8 m. Find the height of the tree.

Answer 1

Hey there !!

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 = 8m, $\angle$ ADC = 30° and CD = CB.

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

From right ∆DAC, we have

=> $\frac{AC}{AD} = tan 30 \degree = \frac{1}{ \sqrt{3} } .$

=> $\frac{x}{8} = \frac{1}{ \sqrt{3} } .$

=> x = $\frac{8}{ \sqrt{3} } .$

=> x = $\frac{8}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } .$

=> x = $\frac{ 8 \sqrt{3} }{ 3 } .$

▶ Also, from right ∆DAC, we have

=> $\frac{CD}{AD} = sec 30 \degree = \frac{2}{ \sqrt{3} } .$

=> $\frac{y}{8} = \frac{2}{ \sqrt{3} } .$

=> y = $\frac{16}{ \sqrt{3} } .$

=> y = $\frac{16}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } .$

=> y = $\frac{ 16 \sqrt{3} }{ 3 } .$

$\therefore AC = \frac{ 8 \sqrt{3} }{ 3 } m and CB = \frac{ 16 \sqrt{3} }{ 3 } m.$

▶ Total height of tree = AC + BC.

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

$= \frac{ 8 \sqrt{3} + 16 \sqrt{3} }{3} .$

$= \frac{ 24 \sqrt{3} }{3} .$

$\huge \boxed{ \boxed{ \bf = 8 \sqrt{3} m. }}$

✔✔ Hence, it is solved ✅✅.

____________________________________


$\huge \boxed{ \boxed{ \boxed{ \mathbb{THANKS}}}}$


$\huge \bf{ \#BeBrainly.}$

Answer 2

$\bold{\large{\underline{Answer\: \colon}}}$

$\bold{\large{\underline{Given\: \colon}}}$

The broken part bends so that the top of the tree touches the ground making an angle 30 °.

$\bold{\angle{BAC}\: = \: 30^{\circ}}$

The distance between the foot of the tree to the point where the top touches the ground is 8 m.

$\bold{\large{\underline{To\: find\: \colon}}}$

Original height of tree that is A`BC

$\bold{\large{\underline{Solution\: \colon}}}$

Let we consider that the tree bend by strom from the point B and angle formed by the top of the tree is 30°.

According to the diagram

Height of tree(A`BC) = AB + BC

Imagine A`BC = AC

A` is a top of tree which will be on the land form an angle but after bending height of tree remain same.

$\bold{\underline{\underline{In \: \triangle{ABC} \: \colon}}}$

$\bold{ tan\: 30^{\circ} \: = \: \dfrac{BC}{AC}}$

$\bold{ \dfrac{1}{\sqrt{3}} \: = \: \dfrac{BC}{8} }$

$\bold{ BC \: = \: \dfrac{8}{\sqrt{3}}}$

Rationalising the denominator and numerator

$\bold{ BC \: = \: \dfrac{8}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}}$

$\bold{ BC \: = \: \dfrac{8\sqrt{3}}{(\sqrt{3})^{2}}}$

$\bold{ BC \: = \: \dfrac{8\sqrt{3}}{3}}$

So,

$\bold{ BC \: = \: \dfrac{8\sqrt{3}}{3}\: m }$

Now in same ∆ ABC

$\bold{cos\: 30^{\circ} \: = \: \dfrac{AC}{AB}}$

$\bold{ \dfrac{\sqrt{3}}{2} \: = \: \dfrac{8}{AB} }$

$\bold{ AB \: = \: \dfrac{8\times 2}{\sqrt{3}} }$

$\bold{ AB \: = \: \dfrac{16}{\sqrt{3}} }$

After rationalising

$\bold{ AB \: = \: \dfrac{16\sqrt{3}}{3} }$

$\bold{AB \: = \: \dfrac{16\sqrt{3}}{3}}\: m$

Now we have ,

A`BC = AB + BC

$\bold{A`BC \: = \: \dfrac{16\sqrt{3}}{3} \: + \: \dfrac{8\sqrt{3}}{3}}$

Take LCM

$\bold{A`BC \: = \: \dfrac{16\sqrt{3} \: + \: 8\sqrt{3}}{3}}$

$\bold{A`BC \: = \: \dfrac{24\sqrt{3}}{3}}$

$\bold{A`BC \: = \: \dfrac{\cancel{24}\sqrt{3}}{3}}$

$\bold{A`BC \: = \:8\sqrt{3}\: m}$

$\bold{\large{\underline{Answer}}}$

So , original height of tree is $\bold{\boxed{\boxed{8\sqrt{3}\: m}}}$