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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 =  y
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}=\frac{1}{\sqrt{3}}
So,  the value of x = 8/\sqrt{3}

Similarily, 
\cos 30^o = \frac{BC}{AC} = \frac{8}{y}
the value of y is 16/\sqrt{3}

So, the total height of the tree is-

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

= 8 (1.732) = 13.856 m (approx)
                                                               

Posted by

manish

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