The angle of elevation of a tower at a level ground is $30^{\circ}$. The angle of elevation becomes $\theta$ when moved 10 m towards the tower. If the height of tower is $5 \sqrt{3} \mathrm{~m}$, then what is $\theta$ equal to?

• A

  $45^{\circ}$
• Correct answer
  B

  $60^{\circ}$
• C

  $75^{\circ}$
• D

  None of the above

Solution

In $\triangle \mathrm{ABC}$, $\frac{\mathrm{BC}}{\mathrm{AB}}=\tan 30^{\circ}$ $\begin{aligned} \mathrm{AB} & =\frac{\mathrm{BC}}{\tan 30^{\circ}} \\ & =\frac{5 \sqrt{3}}{1 / \sqrt{3}} \\ & =15 \\ \mathrm{AB} & =15 \end{aligned}$ Given In $\triangle \mathrm{BCD}$, $\begin{aligned} \frac{\mathrm{BC}}{\mathrm{BD}} & =\tan \theta \\ \frac{5 \sqrt{3}}{5} & =\tan \theta \\ \theta & =60^{\circ} . \end{aligned}$

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