A boat is sailing towards a lighthouse of height 20√3 m at a certain speed. The angle of elevation of the top of the lighthouse changes from 30° to 60° in 10 seconds. What is the time taken (in seconds) by the boat to reach the lighthouse from its initial position?
15
Let C be the base of the lighthouse, A the initial boat position and B the position after 10 s. Height \(CD = 20\sqrt{3}\) m.
Step 1 — Find AC and BC:
\[AC = \frac{20\sqrt{3}}{\tan 30^\circ} = 20\sqrt{3}\cdot\sqrt{3} = 60 \text{ m}\]
\[BC = \frac{20\sqrt{3}}{\tan 60^\circ} = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \text{ m}\]
Step 2 — Find speed: Distance AB = 60 - 20 = 40 m covered in 10 s, so speed = 4 m/s.
Step 3 — Total time from A to C:
\[t = \frac{AC}{\text{speed}} = \frac{60}{4} = 15 \text{ s}\]
Therefore 15 seconds.
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