Question

The angle of depression of a ship from the top of a tower $30$ m height has measure ${60}^o$. Then, the distance of the ship from the base of the tower is ........

• A. $10$
• B. $30$
• C. $10\sqrt{3}$
• D. $30\sqrt{3}$

Answer 1

The correct option is C $10\sqrt{3}$

As shwon in the fig. let $AB$ be the height of the column.

$\angle BAC=90-{60}^o$

$={30}^o$

Thus, $\angle A={30}^o$

$\tan {30}^o=\frac{BC}{AB}$

$\frac{1}{\sqrt{3}}=\frac{x}{30}$

$x=\frac{30}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$x=10\sqrt{3}$m.