Question

The angle of depression of two ships from an aeroplane flying at the height of 7500m are 300

and 450. If both the ships are in the same line that one ship is exactly behind the other, find

the distance between the ships.​

Answer 1

$\large \sf\underline{ \underline{ \red{given : }}}$

• Plane is at the height of 7500m.

• Angle of depression of two ships are 30° and 45°.

$\\ \\ \large \sf\underline{ \underline{ \red{to \: find : }}}$

• Distance between two ships.

$\\ \\ \large \sf\underline{ \underline{ \red{solution : }}}$

The first main approach is to make a rough diagram of the given problem.

Kindly refer the attachment.

According to the diagram , we have to find BC , which is distance between two ships.

▬▬▬▬▬▬▬

In ∆ADB ,

tan 45 = BD / AD

1 = BD / 7500

∴ BD = 7500m

▬▬▬▬▬▬▬

In ∆ADC ,

tan 60 = DC / AD

√3 = DC / 7500

∴ DC = 7500√3 m

▬▬▬▬▬▬

FROM DIAGRAM ,

BC = BD + DC

BC = 7500 + 7500√3

BC = 7500 ( 1 + √3 )m

$\\ \therefore \sf{distance \: between \: two \: ships} \\ \: \: \: \: \: \sf is \: \pink{7500(1 + \sqrt{3})m}.$

Answer 2

EXPLANATION.

The angle of depression of two ships from an aeroplane flying at the height of 7500 m are 30° and 45°.

Both the ships are in the same line and one ship is exactly behind the other.

As we know that,

tanθ = Perpendicular/Base.

Using this formula in this question, we get.

In Δ ADC.

tan(45°) = AC/DC.

⇒ 1 = 7500/y.

⇒ y = 7500. - - - - - (1).

In Δ ABC.

⇒ tan(30°) = AC/BC.

⇒ 1/√3 = 7500/(x + y).

⇒ x + y = 7500√3.

Put the value of equation (1) in expression, we get.

⇒ x + 7500 = 7500√3.

⇒ x = 7500√3 - 7500.

⇒ x = 7500(√3 - 1).

⇒ x = 7500 x (1.73 - 1).

⇒ x = 7500 x 0.73.

⇒ x = 5475 m.

∴ The distance between the ships is 5475 m.