Question

A tree breaks due to storm and the broken part bends so that the top of the tree

touches the ground by making 30º angle with the ground. The distance between

the foot of the tree and the top of the tree on the ground is 6m. Find the height of

the tree before falling down.​

Answer 1

Answer:

tan 30=

$\frac{opp}{adjacent}$

$\frac{1}{ \sqrt{3} } = \frac{opp}{6}$

$2 \sqrt{3} = opposite$

cos 30=

$\frac{adacent}{hypotenuse}$

$\frac{1}{2} = \frac{6}{hypo}$

hypotenuse =12

the height of the tree is

$14 \sqrt{3}$

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Answer 2

Answer:

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, ADC = 30° and CD = CB.

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

From right ∆DAC, we have

Refer the attachment no 1 after this

Also, from right ∆DAC, we have

Refer the attachment no 2 after this

Total height of tree = AC + BC.

Refer the attachment no 3 after this

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