Question

$A$ man on the top of an observation tower finds an object at an angle of depression ${30}^0$. After the object was moved $30$ metres in a straight line towards the tower, he finds the angle of depression to be ${45}^0$. The distance of the object now from the foot of the tower in metres is

• A. $15\sqrt{3}$
• B. $15(\sqrt{3}+1)$
• C. $15(\sqrt{3}-1)$
• D. $15(2+\sqrt{3})$

Answer 1

The correct option is B $15(\sqrt{3}+1)$

Lets assume the height of the tower be $hm$

And $d$ be the initial distance of object from tower.

$\tan {45}^0=\frac{h}{d-30}$

$\tan {30}^0=\frac{h}{d}$

$\Rightarrow \sqrt{3}=\frac{d}{d-30}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{d-30}{d}$

$\Rightarrow \frac{30}{d}=1-\frac{1}{\sqrt{3}}$

$\Rightarrow d=\frac{30\times \sqrt{3}}{(\sqrt{3}-1)}$

Now, the distance $=d-30$
$=\frac{30\sqrt{3}}{(\sqrt{3}-1)}-30$
$=\frac{30}{(\sqrt{3}-1)}$
$=15(\sqrt{3}+1)m$