Question

Two ships are sailing on the opposites of a light
house 150 metre high and in the same line with
the foot of the light house. Find the distance
between the ships when the angles of depression
of the ships observed from the top of light house
are 60° and 45°
6​

Answer 1

Given:-

• Two ships are sailing on the opposite sides of a light house which is 150 m high.
• The angle of depression of both the ships from the top of the light house is 60° and 45°

To Find:-

• The distance between the two ships.

Note:-

• Refer to the attachment for the diagram.

Solution:-

From the attachment we can clearly see,

• ∠CAD = ∠ACE = 60° .... [Alternate angles]
• ∠CBD = ∠BCF = 45° .... [Alternate angles]
• CD = 150 m [Height of the light house]

Now,

In ∆CAD,

$\sf{Tan60^\circ = \dfrac{CD}{AD}}$

$= \sf{\sqrt{3} = \dfrac{150}{AD}}$

$\implies \sf{AD = \dfrac{150}{\sqrt{3}}}$

By rationalising the denominator:-

$= \sf{AD = \dfrac{150\times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}}$

$= \sf{AD = \dfrac{150\sqrt{3}}{3}}$

$= \sf{AD = 50\sqrt{3}}$

∴ AD = 50√3 m

In ∆CBD,

$\sf{Tan45^\circ = \dfrac{CD}{BD}}$

$= \sf{1 = \dfrac{150}{BD}}$

$\implies \sf{BD = 150}$

∴ BD = 150 m

Now,

Distance between both the ships = AD + BD

Hence,

AD + BD = 50√3 + 150

=> AB = 50(√3 + 3) m

∴ The distance between both the ships is 50(√3 + 3) m.

______________________________________

Answer 2

⠀⠀⠀⠀⠀★ Required figure ★

⠀

Given -

• Two ships are sailing on the opposites of a light house 150 metre high and in the same line with the foot of the light house.
• The angles of depression of the ships observed from the top of light house are 60° and 45°.

⠀

To find -

• Distance between the two ships.

⠀

Solution -

As we can see that height of the lighthouse (CD) is 150 metres. Now, we know that

$\: \: \: \: \: \: \: \: \bullet\bf\: \: \: {tan \theta = \dfrac{Perpendicular}{Base}}$

⠀

☆ In right triangle CDA

$\tt\dashrightarrow{tan60^{\circ} = \dfrac{CD}{AD}}$

⠀

$\tt\dashrightarrow{\sqrt{3} = \dfrac{150}{AD}}$

⠀

$\tt\dashrightarrow{AD = \dfrac{150}{\sqrt{3}}}$

⠀

Rationalising the denominator

$\tt\dashrightarrow{AD = \dfrac{150}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}}$

⠀

$\tt\dashrightarrow{AD = \dfrac{150 \sqrt{3}}{3}}$

⠀

$\bf\dashrightarrow{AD = 50 \sqrt{3}}$

⠀

☆ Similarly, in right triangle CDB

$\tt\dashrightarrow{tan45^{\circ} = \dfrac{CD}{DB}}$

⠀

$\tt\dashrightarrow{1 = \dfrac{150}{DB}}$

⠀

$\bf\dashrightarrow{DB = 150}$

⠀

Since, distance between the ships is the sum of distance AD and DB.

•°• Required distance = AD + DB

⠀

$\tt\longmapsto{Required\: distance = 50 \sqrt{3} + 150}$

$\bf\longmapsto{Required\: distance = 50( \sqrt{3} + 3)}$

⠀

Hence,

• Distance between the two ships A and B is 50(√3 + 3) metres.

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