Question

From the top of a lighthouse H m tall, a person observes the angle of depression of a boat to be ${60}^{\circ }$. Another person who is $\frac{H}{3}$ m from the top of a lighthouse observes the angle of depression of another boat directly behind the first boat to be ${45}^{\circ }$. Find the distance between the two boats.

(Take $\sqrt{3}$ = 1.7)

• A. $\frac{H}{10}$
• B. 0.6 H
• C. 1.7H
• D. $\frac{H}{3}(2-\sqrt{3})$

Answer 1

The correct option is D

$\frac{H}{3}(2-\sqrt{3})$

$\angle QAD=\angle ADC$ (alternate angles)
$\therefore tan{60}^{\circ }=\frac{AC}{CD}=\frac{H}{CD}\Rightarrow CD=\frac{H}{tan{60}^{\circ }}=\frac{H}{\sqrt{3}}\dots \dots \left(1\right)alsotan{45}^{\circ }=\frac{BC}{CE}=\frac{AC-AB}{CD+DE}=\frac{H-\frac{H}{3}}{CD+DE}=\frac{\frac{2H}{3}}{CD+DE}\Rightarrow CD+DE=\frac{2H}{3}\dots \dots \left(2\right)Subtracting\left(1\right)from\left(2\right)DE=\frac{2H}{3}-\frac{H}{\sqrt{3}}=\frac{2H-H\sqrt{3}}{3}=\frac{H}{3}(2-\sqrt{3})\therefore distancebetweenboatsis\frac{H}{3}(2-\sqrt{3})$