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As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

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75 (\sqrt{3} - 1) m

75 (\sqrt{3} + 1) m

75 (2 \sqrt{3} - 1) m

75 (1 \sqrt{3} - 1) m

answer is A.

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Detailed Solution

Given,
PA and BD [Both the lines are parallel]
Transversal line = AB
Now, from alternate angles,

∠ ABD = ∠ PAB

Here in the given figure,

∠ ABD = 30^{\circ}

C:UserssoftberryDownloadsP0041 JPEGQ.3.jpg

In the same way we suppose
PA and BD [Both the lines are parallel]
Transversal line = AC
Now again by alternate angles

∠ ACD = ∠ PAC

In the given figure,

∠ ABD = 45^{\circ}

As, the lighthouse is perpendicular to the ground. Thus,

∠ ADB = 90^{\circ}

In the right angled

∆

ACD

\tan 45^{\circ} = \frac{AD}{CD}

1 = \frac{75}{CD}

[

ta n 45^{\circ} = 1]

CD = 75 m

In the same way
Now, in the right-angled

∆ ABD

, we have

\tan 30 ° = \frac{AB}{BD}

\frac{1}{\sqrt{3}} = \frac{75}{BD}

[

\tan 30 ° = \frac{1}{\sqrt{3}}]

BD = 75 \sqrt{3} m

Now, calculation for the two ships distance,

CD = BD - BC

BC = (75 \sqrt{3} - 75) m

BC = 75 (\sqrt{3} - 1) m

So, distance between two ships =

75 (\sqrt{3} - 1) m

.
Correct option is 1.

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