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  • A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

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Suppose DB is a tree and the AD is the broken height of the tree which touches the ground at C.
GIven that,
\angle ACB = 30^o, BC = 8 m 
let AB = x m and AD =  ym 
So, AD+AB = DB = x+y

In right angle triangle \Delta ABC,
\tan \theta = \frac{P}{B}=\frac{x}{8}
\tan 30^o =\frac{x}{8}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{8}

x = \frac{8}{\sqrt{3}}

Similarily, 
\cos 30^o = \frac{BC}{AC}

\Rightarrow\frac{1}{2} = \frac{8/\sqrt{3}}{y}
the value of y is \frac{16}{\sqrt{3}}

So, the total height of the tree is-

  x+y=\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}

x + y =\frac{24 \times \sqrt{3}}{\sqrt{3}\times \sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3}          

= 8 (1.732) = 13.856 m (approx)
                                                               

Posted by

manish

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