Question

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

Answer 1

Answer:

The tree broke at the height of 6.9 m from the ground.

Step-by-step explanation:

Given :  

Total height of tree = 15 m  

Let us assume that tree is broken at point A.  

Let the tree broke at  the height of 'x' m from the ground.

Then, AC = (15 - x) m

Angle made by broken part with the ground , (θ) ∠ACB = 60°

In right angle triangle, ∆ACB,

sin C (θ)  = P/H

sin 60° = AB/AC

√3/2  = x /(15 - x)  

√3(15 - x) = 2x

15√3 - √3x = 2x

15√3 = 2x + √3x

15√3 = x(2 + √3)

x = (15√3)/(2+√3)

h = 15√3 × (2 - √3) /[(2+ √3) (2 −√3)]

[By Rationalizing]  

h = 15√3 × (2 - √3) /(2² -√3²)  

[By using identity , (a + b) (a - b) = a² - b²]

h = 15√3 × (2 - √3) /(4 - 3)  

h = 15√3 × (2 - √3) /1

h = 15√3 × (2 - √3)  

h = 30√3 - 15×3

h = 30√3 - 45

h = 15(2√3 - 3)

h = 15(2 × 1.73 - 3)

h = 15 (3.46 - 3)

h = 15 × 0.46

h = 6.9 m

Hence, the tree broke at the height of 6.9 m from the ground.

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Answer 2

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Step-by-step explanation:

The height of the three = 15 m

suppose it broke at 'C' and its top 'A' touches the ground at 'D'

Now, AC = CD, and angle BDC = 60°

BC = ?

Let BC = 'x'

So, AC = 15 - x and CD = 15 - x

In right angle BCD,

BC/CD = sin 60°

x/(15-x) = √3/2

2x = (15 -x) (√3)

2x = 15√3 - √3x

2x + √3x = 15√3

x(2 + √3) = 15√3

x = (15√3)/(2 + √3)

= {(15√3)/(2 + √3)} × {(2 - √3)/(2 - √3)}

= {(30√3) - (15 ×3)}/(4 - 3)

= {(30 × 1.73) - 45}/1

x = 51.9 - 45

x = 6.9 m

So, the height above the ground from the tree broke is 6.9 meter.

Answer.