Question

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

A.173 m

B.200 m

C.273 m

D.300 m

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Answer 1

Let AB be the lighthouse and C and D be the position of the ship.

⇒GIVEN :-

• AB = 100 m
• ∠ACB = 30°
• ∠ADB = 45°

⇒SOLUTION :-

In △BCA

$\sf{ \frac{AB}{AC} = tan30° = \frac{1}{ \sqrt{3} }}$

$\sf{→AC = \sqrt{3}AB = 100 \sqrt{3}}$

In △BAD

$\sf{ \frac{AB}{AD} = tan45° = 1}$

$\sf{→AC = AB = 100m}$

$\therefore \sf{CD = AC + AD = 100 \sqrt{3} + 100}$

$\sf{→CD = 100(\sqrt{3} + 1)}$

$\sf{→CD = 100(1.73 + 1)}$

$\sf \fbox \green{→CD = 273m}$

Hence, the distance between the two ships is 273m.

OPTION C (273 m) is the right answer.

[NOTE :- Refer the attachment for figure.]