Question

As observed from the top of a light-house, 60m high above the sea level, the angles of depression of a ship sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship during this period of observation.Two triangles (type 4)60 m60 x (root 3 -1) m0

Answer 1

This is the answer to your question

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Answer 2

Hey there !!

→ Let AC be the lighthouse and C and D be the two positions of the ship.

→ And, BD (y) be the distance travelled by the ship during this period of observation.

▶ From right ∆ACD, we have

=> tan 45° = $\frac{AC}{DA} .$

[ => tan 45° = 1 ]

=> 1 = $\frac{60 m}{DA} .$

=> DA = 60 m.

➡ Now,

▶ From right ∆ABC, we have

=> tan 30° = $\frac{AC}{AB} .$

[=> tan 30° = $\frac{1}{ \sqrt{3} }$

=> $\frac{1}{ \sqrt{3} } = \frac{60}{60 + y} .$

=> y + 60 = 60√3.

=> y = 60√3 - 60.

$\huge \boxed{ \boxed{ \bf => y = 60( \sqrt{3} - 1 ). }}$

✔✔ Hence, the distance travelled by the ships during the period of observation is 60( √3 - 1 ) m. ✅✅

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