Question

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

Answer 1

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

${\bold{\therefore Height\:of\:remaining\:tree=6.9\:m}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a vertically straight tree ,15 m high is broken by the wind in such a way that its top just touched the ground and makes an angle of 60 degree with the ground .

• We have to height from the ground did the tree break?

$\underline \bold{Given : } \\ \implies Angle \: of \: elevation= 60 \degree \\ \\ \implies Height \: of \: tree = 15 \: m \\ \\ \underline \bold{To \: Find : } \\ \implies Height \: of \: remaining \: tree = ?$

• According to given question :

$\bold{ABC\: be \: the \: triangle \: in \: which \to} \\ \bold{ 90 \degree \: at \: B\: and \: \angle C= \theta=60\degree } \\ \\ \bold{Let \: remaining \: height \: be \: x \: m} \\ \implies sin \: \theta = \frac{p}{b} \\ \\ \implies sin \: 60 \degree = \frac{AB}{AC} \\ \\ \implies \frac{ \sqrt{3} }{2} = \frac{x}{15 - x} \\ \\ \implies 15 \sqrt{3} - \sqrt{3} x = 2x \\ \\ \implies 15 \sqrt{3} = 2x + \sqrt{3} x \\ \\ \implies x(2 + \sqrt{3} ) = 15 \sqrt{3} \\ \\ \implies x = \frac{15 \sqrt{3} }{ 2 + \sqrt{3} } \times \frac{ (2 - \sqrt{3} )}{ (2 - \sqrt{3} )} \\ \\ \implies x = \frac{30 \sqrt{3} - 15 \times 3}{4 - 3} \\ \\ \implies x = 30 \sqrt{3} - 45 \\ \\ \implies x = 30 \times 1.73 - 45 \\ \\ \implies x = 51.9 - 45 \\ \\ \bold{\implies x = 6.9 \: m}$

Answer 2

ANSWER:-

Given:

A Vertically straight tree, 15m high is broken by the wind in such a way that its top just touched the ground & makes an angle of 60° with the ground.

To find:

The height from the ground the tree break.

Solution:

Here, we have

•Total height of the tree is 15m.

⚫AB= 15m.

Let height at which trees is broken is h m

So,

⚫BC = h m

⚫CD= AB- BC

CD = 15 - h

Now,

In ∆DBC,

Formula used:

$sin \theta = \frac{Perpendicular}{Hypotenuse}$

Therefore,

$sin60 \degree = \frac{BC}{CD} \\ \\ = > \frac{ \sqrt{3} }{2} = \frac{h}{15 - h} \\ [Cross \: multiplication] \\ \\ = > \sqrt{3} (15 - h) = 2h \\ \\ = > h(2 + \sqrt{3} ) = 15 \sqrt{3} \\ \\ = > h = \frac{15 \sqrt{3} }{2 + \sqrt{3} } \\ [Rationalising] \\ \\ = > \frac{15 \sqrt{3} ( 2 - \sqrt{3}) }{(2 + \sqrt{3}) (2 - \sqrt{3} )} \\ \\ = > \frac{30 \sqrt{3} - 45 }{ {2}^{2} - {( \sqrt{3} )}^{2} } \: \: \: \: \: \: \: \: \: \: [ {a}^{2} - {b}^{2} = (a + b)(a - b)]\\ \\ = > \frac{30 \sqrt{3} - 45}{4 - 3} \\ \\ = > \frac{30 \sqrt{3} - 45 }{1} \\ \\ = > 30 \times 1.732 - 45 \\ \\ = > 51.96 - 45 \\ \\ = >h= 6.96m$

Thus,

The tree is broken at 6.96m from the ground.

Hope it helps ☺️