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From a 100 m high lighthouse, it is observed that two ships are approaching it from west and south. If angles of depression of the two ships are $30^{\circ}$ and $45^{\circ}$ respectively, then the distance between the ships in metres is

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100 (\sqrt{3} + 1)

100 (\sqrt{3} - 1)

200

400

answer is C.

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Detailed Solution

Given that,
Height of the light house =

100 m

Angles of elevation of the ships =

30^{\circ}

and

45^{\circ}

We have the formula,

tanθ = \frac{Height}{Distance}

We have to find the distance between the two ships.
When the angle of elevation is

30^{\circ}

, we take the distance as

d_{a}

.
We have,

\tan 30^{\circ} = \frac{100}{d_{a}}

\Rightarrow

d_{a} = 100 \sqrt{3} m

When the angle of elevation of the ships is

45^{\circ}

, we take the distance as

d_{b}

.
We have,

\tan 45^{\circ} = \frac{100}{d_{b}}

\Rightarrow d_{b} = 100 m

Distance (d) between the ships will be,

\Rightarrow d

\sqrt{d_{a}^{2} + d_{b}^{2}}

\Rightarrow d

\sqrt{(100 \sqrt{3})^{2} + 100^{2}} m

\Rightarrow d

100 (\sqrt{4}) m

\Rightarrow d

200 m

Thus, the distance between the ships is 200m.
Therefore, the correct option is 3.

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