Question

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are are 30° and 45°.if one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.​

Answer 1

Answer:

The distance between the two ships are

75(1-√3) m.

Step-by-step explanation:

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Answer 2

$\huge\underline\mathrm{SOLUTION:-}$

AnswEr:

• The distance between two ships be = $\mathsf {75 \: ( \sqrt{3} - 1) \: metre}$.

ExPlanation:

Let CD be the lighthouse whose height is 75 metre.

Let the two ships be at A and B such that their angles of depression from D are 30° and 45° respectively.

Let AB = x metre and BC = y metre

In right triangle BCD, We Have:

$\mathsf { \tan \: 45\degree = \frac{CD}{BC} }$

➠ $\mathsf {1 = \frac{75}{y} }$

➠ $\mathsf {y = 75 \: metre }$...... (i)

In right triangle ACD, We Have:

$\mathsf { \tan \: 30\degree = \frac{CD}{AC} }$

➠ $\mathsf { \tan \: 30\degree = \frac{CD}{AB + BC} }$

➠ $\mathsf { \frac{1}{ \sqrt{3} } = \frac{75}{x + y} }$

➠ $\mathsf {x + y = 75 \sqrt{3} }$

➠ $\mathsf {y = (75 \sqrt{3} - x) \: metre }$...... (ii)

Comparing (i) and (ii), We Get:

$\mathsf {75 = 75 \sqrt{3} - x}$

➠ $\mathsf {x = 75 \sqrt{3} - 75 }$

➠ $\mathsf {75 \: ( \sqrt{3} - 1) \: metre}$

Hence:

• The distance between two ships be = $\mathsf {75 \: ( \sqrt{3} - 1) \: metre}$.

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