Question

A person is standing on the bank of river observer that angle of elevation of the top of the tower standing on the opposite of bank is ${60}^{\circ }$. When he moves $40m$ away from the bank he find the angle of elevation to be ${30}^{\circ }$. Find the height of the tower.

• A. $40\sqrt{3}m$
• B. $10\sqrt{3}m$
• C. $\sqrt{3}m$
• D. $20\sqrt{3}m$

Answer 1

Answer: (D)

$20\sqrt{3}m$

Let $BC$ be the height of the tree,

$AB$ be the breadth of the river

$A$ be the initial position of the man

$D$ be the final potion of the man $\angle CAB={60}^{\circ }$

and $\angle CDB={30}^{\circ }$

$DA=40$m

Let $AB=x$

Let $BC=h$

Consider the $△DBC$

$\tan {30}^{\circ }=\frac{BC}{DB}=\frac{BC}{DA+AB}=\frac{h}{40+x}$

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

$\Rightarrow h=\frac{40+x}{\sqrt{3}}$ .........$\left(1\right)$

Consider the $△ABC$

$\tan {60}^{\circ }=\frac{BC}{AB}=\frac{h}{x}$

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

$\Rightarrow h=x\sqrt{3}$ .........$\left(2\right)$

Using $\left(2\right)$ in $\left(1\right)$ we have

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

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

$\Rightarrow 3x=40+x$

$\Rightarrow 2x=40$

$\therefore x=20$

From $\left(1\right)$

height of the tower $h=\frac{40+x}{\sqrt{3}}$

$=\frac{40+20}{\sqrt{3}}$

$=\frac{60}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$ by rationalising the denominator.

$=20\sqrt{3}m$

Hence, the height of the tower is $20\sqrt{3}m$.