The elevation of an object on a hill is observed from a certain point in the horizontal plane through its base, to be $30^{\circ}$. After walking 120 metres towards it on level ground the elevation is found to be $60^{\circ}$. Then the height of the object (in metres) is :

• A

  120
• Correct answer
  B

  $60 \sqrt{3}$
• C

  $120 \sqrt{3}$
• D

  60

Solution

Let $h$ be the height of the object. In $\triangle A C D$
$\tan 30^{\circ}=\frac{C D}{A C}$ $\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{120+x}$ $\Rightarrow \quad \sqrt{3} h=120+x$ ...(i) and in $\triangle B C D$ $\tan 60^{\circ}=\frac{C D}{B C}$ $\Rightarrow \quad \sqrt{3}=\frac{h}{x}$ $\Rightarrow \quad h=\sqrt{3} x$ ...(ii) From (i) and (ii), we get $3 x=120+x \Rightarrow x=60$ From Eq. (ii) Height of the object $=60 \sqrt{3} \mathrm{~m}$.

Hence, the correct answer is option: 'B'.