Question

From the top of a light house, the angles of
depression of two ships on the same side of
the light house are 60° and 45°. Find the height
of the light house, if the distance between the
ships is 36 m. Answer to the nearest metre.​

Answer 1

Given -

From the top of a light house, the angles of depression of two ships on the same side of the light house are 60° and 45°. The distance between the ships is 36 m.

To find -

• Height of the light house

Solution -

Let

• AB = Height of the light house
• DC = distance between light house = 36m

In ∆ACB

$\implies\sf tan60\degree=\dfrac{AB}{CB}$

$\implies\sf \sqrt{3}=\dfrac{AB}{CB}$

$\implies\sf AB=\sqrt{3}CB$

Now,In ∆ADB

$\implies\sf tan45\degree=\dfrac{AB}{DB}$

$\implies\sf 1=\dfrac{AB}{DC + CB}$

• Put the value of AB

$\\\implies\sf 1=\dfrac{\sqrt{3}CB}{36 + CB}$

$\implies\sf 36 + CB = \sqrt{3}CB$

$\implies\sf 36= \sqrt{3} CB-CB$

$\implies\sf 36= CB(\sqrt{3}-1)$

$\implies\sf CB=\dfrac{36}{(\sqrt{3}-1)}$

Height of a light house

$\implies\sf AB=\sqrt{3}CB$

• Put the value of CB

$\\\implies\sf AB=\sqrt{3} \times\dfrac{36}{(\sqrt{3}-1)}$

$\implies\sf AB=1.73\times\dfrac{36}{(1.73-1)}$

$\implies\sf AB=1.73\times\dfrac{36}{0.73}$

$\implies\sf AB=\cancel\dfrac{62.28}{0.73}$

$\implies\sf AB=85.31m$

•°• Height of a light house is 85.31m

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Answer 2

${\rm{\underline{\underline{Question:-}}}}$

→ From the top of a light house, the angles of  depression of two ships on the same side of  the light house are 60° and 45°. Find the height  of the light house, if the distance between the  ships is 36 m. Answer to the nearest metre.​

${\rm{\underline{\underline{Given:-}}}}$

→ The angles of  depression of two ships on the same side of  the light house =  60° and 45°

→ The distance between the  ships = 36 m.

${\rm{\underline{\underline{To\;Find:-}}}}$

→ Height  of the light house

${\rm{\underline{\underline{Solution:-}}}}$

→ Let ,

JK = Height of the light house

LM = distance between light house

→ In , Δ JKM we have

$\sf tan\;60 = \dfrac{JK}{MK} \\\\ (cross\;multiply\;)\\\\\\\rightarrow \sqrt{3} = \dfrac{JK}{MK} \\\\\\\rightarrow JK = \sqrt{3}MK$

→ In Δ JLK we have ,

$\sf tan\;45 = \dfrac{JK}{LK} \\\\\\\rightarrow 1 = \dfrac{JK}{LM+MK} \\\\\\\rightarrow 1 = \dfrac{\sqrt{3}KM }{36+MK} \\\\\\(cross\;multiply\;)\\\\\rightarrow 36 + MK = \sqrt{3}KM\\\\\rightarrow 36 = \sqrt{3}KM - MK \\\\\rightarrow 36 = KM(\sqrt{3} - 1 ) \\\\\\\rightarrow KM = \dfrac{36}{\sqrt{3}-1 }$

→ Now , finding the height of the lighthouse !!

$\sf JK = \sqrt{3}MK\\\\\\\rightarrow JK = \sqrt{3} \times \dfrac{36}{\sqrt{3}-1 } \\\\\\JK = 1.73 \times \dfrac{36}{0.73} \\\\\\JK = \dfrac{62.28}{0.73}$

∴ JK = 85.31 m = Height of the lighthouse

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All Done !! :D