As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use √3= 1.732]

Solution

Let ships are at distance x from each other,

$In ∆ APO \tan 45^{o} = \frac{100}{y} = 1 ∴ y = 100 m . . . . . (i) In ∆ POB \tan 30 ° = \frac{OP}{OB} = \frac{100}{x + y} = \frac{1}{\sqrt{3}} \sqrt{3} = \frac{x + y}{100} x + y = 100 \sqrt{3} x = 100 \sqrt{3} - y = 100 \sqrt{3} - 100 100 (\sqrt{3} - 1) ∴ x = 100 (1.732 - 1) = 100 x 0.732 = 73.2 m$

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Some Applications Of Trigonometry

As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use √3= 1.732]

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