Question : As observed from the top of a lighthouse, 45 m high above the sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. The distance travelled by ship during the period of observation is: (Your answer should be correct to one decimal place.)
Option 1: 32.9 m
Option 2: 33.4 m
Option 3: 36.9 m
Option 4: 24.8 m
Correct Answer: 32.9 m
Solution :
Given, AB = 45 m
In $\triangle$ABC
$\tan$ 45° = $\frac{\text{AB}}{\text{BC}}$
⇒ 1 = $\frac{45}{\text{BC}}$
⇒ BC = 45 m
In $\triangle$ABD
$\tan 30° = \frac{\text{AB}}{\text{BD}}$
⇒ $\frac{1}{\sqrt{3}} = \frac{45}{\text{BD}}$
⇒ BD = $45\sqrt{3}$
Also, BD = CD + BC
⇒ $45\sqrt{3}$ = CD + 45
⇒ CD = $45 (\sqrt{3} - 1)$
⇒ CD = $45 × (1.732 - 1)$
⇒ CD = 45 × 0.732 = 32.9
Hence, the correct answer is 32.9 m.
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