Question

A person, standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is ${60}^{\circ }$ when he retreats $20$ m from the bank, he finds the angle to be ${30}^{\circ }$. The height of the tree and the breadth of the river.

• A. $10$ $\sqrt{3}$ m, $10$ m
• B. $10$; $10$ $\sqrt{3}$ m
• C. $20$ m, $30$ m
• D. None of these

Answer 1

The correct option is A $10$ $\sqrt{3}$ m, $10$ m

Let AB be the breadth of the river and BC be the height of the tree which makes a $\angle$ of 60${}^{\circ }$ at a point A on the opposite bank.

Let D be the position of the person after retreating 20 m from the bank.

Let AB $=x$ metres and BC $=h$ metres.

We know, $\tan \left(\theta \right)$ = Opposite / Adjacent

From right $\angle ed△$ ABC and DBC,

we have $\tan {60}^{\circ }=\frac{BC}{AB}$ and $\tan {30}^{\circ }=\frac{h}{20+x}$

$\Rightarrow \sqrt{3}=\frac{h}{x}$ and $\frac{1}{\sqrt{3}}=\frac{h}{x+20}$

$\Rightarrow h=x\sqrt{3}$ and $h=\frac{x+20}{\sqrt{3}}$

$\Rightarrow x\sqrt{3}=\frac{x+20}{\sqrt{3}}\Rightarrow 3x=x+20\Rightarrow x=10m$

Putting $x=10$ in $h=\sqrt{3}x$, we get

$h=10\sqrt{3}=17.32m$

Hence, the height of the tree $=17.32$ m and the breadth of the river $=10$ m.