A tree 10(2+ $\sqrt{3}$ ) metres high is broken by the wind at a height 10 $\sqrt{3}$ metres from its root in such a way that top struck the ground at certain angle and horizontal distance from the root of the tree to the point where the top meets the ground is 10 m. Find the angle of elevation made by the top of the tree with the ground.

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Solution

Given that BC = 10 $\sqrt{3}$

$⟹$ AC = 10(2+ $\sqrt{3}$ ) - 10 $\sqrt{3}$ = 20m

AB = 10m

Hence, cosA = $\frac{A B}{A C}$ = $\frac{10}{20}$ = $\frac{1}{2}$

$⟹$ A = $60^{\circ}$ .

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Similar questions

A tree is broken by the wind. The top struck the ground at an angle of $30^{\circ}$ and at a distance of $30$ metres from the foot of the tree. The height of the tree in metres is

Q. A tree is broken by the wind. The top of that tree struck the ground at an angle of

30^{o}

and at a distance of

30

m from the root. Find the height of the whole tree.

Q. A tree

12 m

high, is broken by the wind in such a way that its top touches the ground and makes an angle

60^{\circ}

with the ground. At what height from the bottom the tree is broken by the wind?

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Given that BC = 10√3 ⟹ AC = 10(2+√3) - 10√3 = 20m AB = 10m Hence, cosA = ABAC = 1020 = 12 ⟹ A = 60∘.

Given that BC = 10 $\sqrt{3}$

$⟹$ AC = 10(2+ $\sqrt{3}$ ) - 10 $\sqrt{3}$ = 20m

AB = 10m

Hence, cosA = $\frac{A B}{A C}$ = $\frac{10}{20}$ = $\frac{1}{2}$

$⟹$ A = $60^{\circ}$ .