Question

The upper part of a tree broken by the wind in two parts, makes an angle of 60° with ground. if the top of the tree touches the ground at a distance of 2√3 m from the foot of the tree ,then find the height of the tree.​

Answer 1

Answer:

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Let the PQR be the whole tree & Form Right Angled ∆QRP,

$\tan60 = \frac{qr}{pr}$

$\sqrt{3 } = \frac{qr}{2 \sqrt{3} }$

QR = 6

$\cos60 = \frac{pr}{pq}$

$\frac{1}{2} = \frac{2 \sqrt{3} }{pq}$

PQ = 4√3

Thus,

________________________________

Height of tree = QR + PQ

$= (6 + 4 \sqrt{3} )m$

Hope it will be helpful :)

Answer 2

Answer:

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Let AC is the broken part of the tree and ABis standing part.

Given that:

BC=20 m   ∠ACB=30∘

Solution:

In △ABC

tan∠ACB=BCAB

or, tan30∘=20 mAB

or, AB=20 m×31=320m

Similarly,

cos30∘=ACBC

or, AC=cos30∘20 m

or, AC=320 m×2

or, AC=340m

Total length of the tree =AB+AC=320m+340m=360m=203m

Therefore, total length of the tree=203 m=34.64 m

hope help u