Question

As observed from the top of a 100 m high lighthouse (from 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 lighthouse, find the distance between the ships. [Use √3=1.73.]

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Answer 1

More to know:

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Solution:

Let the observer on the lighthouse CD be at D.

★ Height of the lighthouse CD = 100m

=> From the diagram,

∠XDA = 30° = ∠DAC &

∠XDB = 45° = ∠DBC

★ In the right angled ∆DCB,

$tan45 \degree = \frac{DC}{BC}$

$\rightarrow \: 1 = \frac{100}{BC} \\ \\ \color{red}BC = 100m$

In the right angled ∆DCA,

BD= AB × √3

= 100 × √3

= 173.2

Distance:

AB = AC - BC

= 73.2 m