Question

A pilot in an aeroplane observe that a light
house is on one side of aeroplane and a
watchtower is just on the opposite side. The angle
of depressions made by pilot with light house
and watch tower are of 60° and 30 respectively.
If the areoplane is at the height of 550√3
metre at that time, then what is the distance
between light house and watch tower?​

Answer 1

Step-by-step explanation:

Let V=Vashi Bridge

W=Worli sea-link

A=aero plane

From triangle ABV,

From triangle ABW,

Distance VW= x+y=5500+16500=22000 m

solution

Answer 2

Given :

• Height of the plane above the ground = 550√3 m.

• Angle of Depression = d = 60°

• Angle of Depression = d' = 30°

To find :

Distance between the lighthouse and the tower.

Solution :

To find the distance between the lighthouse and the tower , first we have to find the individual distance between the A & D and D & C.

Here , the sum of AD and DC will be the distance between the lighthouse and the the tower.

So , let's first find the height of the tower !

Distance between B and D :

Let the distance between B and D be x m.

According to the figure and given information , ABD is a Right-angled triangle .

Here :

• AD is the Height (P)
• BD is the Base (B)
• AB is the Hypotenuse (H)

We are provided with the height and to find the base , hence we will use tan θ.

Since,

$\boxed{\bf{tan\:\theta = \dfrac{P}{B}}}$

Where :

• P = Height
• B = Base

Using tan θ and substituting the values in it, we get :

$:\implies \bf{tan\:\theta = \dfrac{P}{B}}$

$:\implies \bf{tan60^{\circ} = \dfrac{550\sqrt{3}}{x}}$

$:\implies \bf{\sqrt{3} = \dfrac{550\sqrt{3}}{x}}$

$:\implies \bf{x = \dfrac{550\sqrt{3}}{\sqrt{3}}}$

$:\implies \bf{x = \dfrac{550\not{\sqrt{3}}}{\not{\sqrt{3}}}}$

$:\implies \bf{x = 550}$

$\underline{\boxed{\therefore \bf{x = 550\:m}}}$

Hence the distance Between B and D is 550 m.

Distance between C and D :

Let the distance between C and D be y m.

According to the figure and given information , ADC is a Right-angled triangle .

Here :

• AD is the Height (P)
• CD is the Base (B)
• AC is the Hypotenuse (H)

We are provided with the height and to find the base , hence we will use tan θ.

Since,

$\boxed{\bf{tan\:\theta = \dfrac{P}{B}}}$

Where :

• P = Height
• B = Base

Using tan θ and substituting the values in it, we get :

$:\implies \bf{tan\:\theta = \dfrac{P}{B}}$

$:\implies \bf{tan30^{\circ} = \dfrac{550\sqrt{3}}{y}}$

$:\implies \bf{\dfrac{1}{\sqrt{3}} = \dfrac{550\sqrt{3}}{y}}$

$:\implies \bf{y = 550\sqrt{3} \times \sqrt{3}}$

$:\implies \bf{y = 550 \times \sqrt{3 \times 3}}$

$:\implies \bf{y = 550 \times 3}$

$:\implies \bf{y = 1650}$

$\underline{\boxed{\therefore \bf{x = 1650\:m}}}$

Hence the distance Between A and D is 1650 m.

Distance between the lighthouse and the tower :

⠀⠀⠀⠀⠀⠀⠀⠀⠀==> AC = BD + DC

⠀⠀⠀⠀⠀⠀⠀⠀⠀==> AC = 550 + 1650

⠀⠀⠀⠀⠀⠀⠀⠀⠀==> AC = 2200 m

Hence the distance between the lighthouse and the tower is 2200 m.