# 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°.

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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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Let, height of light house from sea level (AB)=100m

Let, two ships be at the positions be C and D

In △ABC,

tan 45° = $\frac{AB}{BC}$

⇒ 1 = $\frac{100}{BC}$

⇒ BC = 100m

In △ABD,

tan 30° = $\frac{AB}{BD}$

⇒ $\frac1{\sqrt3}$ = $\frac{AB}{BD}$

BD = AB × 3​

=100 × √3​

=173.2       [√3​=1.732]

∴ CD = BD − BC

= 173.2 - 100

= 73.2m

∴ The distance between two ships 73.2 m