Question

The angle of depression of two ships from the top of a light house and on the same side of it are found to be 45° and 30° respectively. If the ships are 200m apart, find the height of the light house. ​

Answer 1

Imagine the given situation in the attached diagram which I have expressed very easily.

In ∆ABC,

⇒ tan θ = Perpendicular / Base

⇒ tan θ = AC / BC

Let, DC = x and height of lighthouse be h

so,

⇒ tan 30° = h / (BD + x)

⇒ 1 / √3 = h / (200 + x)

⇒ √3 h = 200 + x

⇒ x = √3 h - 200 ...(1)

Similarly, In ∆ADC,

⇒ tan θ = Perpendicular / Base

⇒ tan θ = AC / DC

⇒ tan 45° = h / x

⇒ 1 = h / x

⇒ x = h ...(2)

From (1) & (2),

⇒ h = √3h - 200

⇒ √3h - h = 200

⇒ h(√3 - 1) = 200

⇒ h = 200 / (√3 - 1) m

Hence,

The height of the lighthouse is 200 / (√3 - 1 ) m

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• The angle of depression of two ships from the top of a light house and on the same side of it are found to be 45° and 30° respectively .

• The ships are 200 m apart .

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

• The height of the light house .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

☞︎︎︎ See the attachment diagram .

Let,

• B & C be given ship's .

• And AD be height of the light house.

Let,

• $\bf\blue{Height\:of\:the\:light\:house\:(AD)\:is\:\:h\:.}$

✈︎ Now in ∆DAC , we have

$\\ \bf{\implies\:\tan{45^{\degree}}\:=\:\dfrac{AD}{CD}\:}$

$\\ \bf{\implies\:1\:=\:\dfrac{AD}{CD}\:}$

$\\ \bf{\implies\:CD\:=\:AD\:}$

$\\ \bf{\implies\:CD\:=\:h\:}----(1)$

✈︎ Again in ∆ABD , we have

$\\ \bf{\implies\:\tan{30^{\degree}}\:=\:\dfrac{AD}{BD}\:}$

$\\ \bf{\implies\:\dfrac{1}{\sqrt{3}}\:=\:\dfrac{AD}{BC\:+\:CD}\:}$

Where,

• $\bf\red{BC}$ = 200 m

$\\ \bf{\implies\:\dfrac{1}{\sqrt{3}}\:=\:\dfrac{h}{200\:+\:CD}\:}$

☞︎︎︎ Putting the value of 'CD' from equation (1), we get

$\\ \bf{\implies\:\dfrac{1}{\sqrt{3}}\:=\:\dfrac{h}{200\:+\:h}\:}$

$\\ \bf{\implies\:200\:+\:h\:=\:h\times{\sqrt{3}}\:}$

$\\ \bf{\implies\:\sqrt{3}\:h\:-\:h\:=\:200\:}$

$\\ \bf{\implies\:h\:(\sqrt{3}\:-\:1)\:=\:200\:}$

$\\ \bf{\implies\:h\:=\:\dfrac{200}{\sqrt{3}\:-\:1}\:}$

➪ By rationalization and simplification, we get

$\\ \bf\purple{\implies\:h\:=\:100\:(\sqrt{3}\:+\:1)\:m\:}$

$\huge\green\therefore$ The height of the light house is $\bf\gray{100\:(\sqrt{3}\:+\:1)\:m\:}$ .