Question

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is ${60}^o$. When he moves 40 m away from the bank, he finds the angle of elevation to be ${30}^o$. Find the height of the tree and the width of the river. ($\sqrt{3}=1.73$)

Answer 1

Let the height of the tree be ${}^{\prime }{h}^{\prime }$ meters and width of the river be ${}^{\prime }{x}^{\prime }$ meters.

In $△ABC,$

$\Rightarrow \tan 60°=\frac{AB}{AC}=\frac{h}{x}$

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

$\Rightarrow h=\sqrt{3}x\to \left(1\right)$

In $△ABD,$

$\Rightarrow \tan 30°=\frac{AB}{AD}=\frac{h}{x+40}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+40}$

$\Rightarrow \sqrt{3}h=(x+40)\to \left(2\right)$

From $\left(1\right)\&\left(2\right),$ we get

$\Rightarrow \sqrt{3}\left(\sqrt{3}x\right)=x+40$

$\Rightarrow 3x-x=40$

$\Rightarrow 2x=40$

$\Rightarrow x=20m$

$\therefore h=\sqrt{3}x$

$=20\times 1.73$

$=34.60m$

Hence, the answer is $34.60m,20m$