Question

A vertically straight tree 15 m high is broken by the wind in such a way that its top just touch the ground and makes an angle of degree 60 with the ground. At what height from the ground did the tree break

Answer 1

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Answer 2

Solution:

Given:

➜ A vertically straight tree 15 m high is broken by the wind in such a way that its top just touch the ground and makes an angle of degree 60 with the ground.

Find:

➜ At what height from the ground did the tree break.

According to the given question:

➜ 15 m = height of tree.

➜ Let "x" be the point at broken.

➜ (θ) = 60° = angle by broken part with the ground.

➜ AB = height from ground from the broken points.

Calculations:

$\sf⇝H = AC + h \\ \sf⇝AC = (H - h) m \\ \sf⇝ sin \: θ = \frac{Opposite \: side}{hypotenuse} \\ \sf⇝60° = \frac{AB }{BC} \\ \sf⇝ \sqrt{ \frac{3}{2} } = \frac{h}{H - h} \\ \sf⇝ \sqrt{3} (15 - h) = 2h \\ \: \sf⇝(2 + \sqrt{3})h = 15 \sqrt{3 } \\ \sf ⇝ h = \frac{15 \sqrt{3} }{2 + \sqrt{ 3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

Rationalizing denominator:

$\sf ⇝ h= \frac{(15 \sqrt{3)}(2 - \sqrt{3)} }{2 {}^{2} - ( \sqrt{3) {}^{2} } } \\ \sf ⇝ h = \frac{15(2 \sqrt{3 - 3)} }{1}$

Hence, 15(2 √3 - 3) m is the height of broken point from the ground.

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