A vertically straight tree, 15 m high,


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?


Let  be the tree of desired height m and tree is broken by wind then tree makes an angle . Let AC=15 − x

Here we have to find height x

So we use trigonometric ratios.

In a triangle,

$\Rightarrow \quad \sin C=\frac{A B}{A C}$

$\Rightarrow \quad \sin 60^{\circ}=\frac{x}{15-x}$

$\Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{x}{15-x}$

$\Rightarrow 15 \sqrt{3}-\sqrt{3} x=2 x$

$\Rightarrow \quad 15 \sqrt{3}=2 x+\sqrt{3} x$

$\Rightarrow \quad 15 \sqrt{3}=x(2+\sqrt{3})$

$\Rightarrow \quad x=\frac{15 \sqrt{3}}{2+\sqrt{3}}$

$\Rightarrow \quad x=6.9$

Hence the height of tree is $6.9 \mathrm{~m}$.


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