A river is $60 m$ wide. A tree of unknown height is on one bank. The angle of elevation of the top of the tree from the point exactly opposite to the foot of the tree on the other bank is $30^{o}$ . Find the height of the tree.

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Solution

Let us consider
$A B$ as the tree and $B C$ as the width of the river.

Let us take the height of the tree $A B = x m$

Width of the river $B C = 60 m$

In right $△ A B C,$

$tan θ = \frac{A B}{C B}$

$\Rightarrow tan 30^{o} = \frac{x}{60}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{x}{60}$

$\Rightarrow x = \frac{60}{\sqrt{3}}$

$\Rightarrow x = \frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow x = \frac{60 \sqrt{3}}{3}$

$= 20 \sqrt{3} m$

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A river is 60 m wide. A tree of unknown height is on one bank. The angle of elevation of the top of the tree from the point exactly opposite to the foot of the tree on the other bank is 30o. Find the height of the tree.

Let us considerAB as the tree and BC as the width of the river.Let us take the height of the tree AB=x mWidth of the river BC=60 mIn right △ABC,tanθ=ABCB⇒tan30o=x60⇒1√3=x60⇒x=60√3⇒x=60√3×√3√3⇒x=60√33 =20√3 m

A river is $60 m$ wide. A tree of unknown height is on one bank. The angle of elevation of the top of the tree from the point exactly opposite to the foot of the tree on the other bank is $30^{o}$ . Find the height of the tree.