Question

as observed from the top of a 150 M tall height Lighthouse the angle of depression of two ships approaching it at 30 degree and 45 degree if one ship directly behind the other find the distance between the two ships.

plzz fast

Answer 1

Here is your solution



Height of tower = 150 m

Let

Bc = x and Cd = y

In Triangle ABC

$tan45 = \frac{ab}{bc} \\ 1 = \frac{150}{x} \\ x = 150m$

In Triangle ABD

$tan30 = \frac{ab}{bd} \\ \frac{1}{ \sqrt{3} } = \frac{150}{y} \\ y = 150 \sqrt{3}m$

So,

BD = BC + CD

Y = X + cD

Y - X = CD

$150 \sqrt{3} - 150 \\ 150( \sqrt{3} - 1)$
Distance between the two ships

$150 (\sqrt{3} - 1)m\:or 109.80m$

I hope it helps you

Answer 2

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solution : ♂
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