Question

Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships are observed from the top of the light house are 60° and 45° respectively. If the height of the light house is 200 m, find the distance between the two ships.

Answer 1

The distance between two ships  is 315.47 meters

Step-by-step explanation:

Given as :

Two ships are there in the sea on either side of a light house

The height of light house = H = 200 meters

The distance of ship 1 from base of house = y meters

The distance of ship 2 from base of house = x meters

The angle of depression from the top of house on ship 1 = 45°

The angle of depression from the top of house on ship 2 =  60°

The distance between two ships  = BD = (x + y) meters

According to question

From figure

In Δ ADC

Tan angle = $\dfrac{perpendicular}{base}$

Tan  45° = $\dfrac{AC}{CD}$

1   = $\dfrac{h}{y}$

Or, y = h = 200

∴  The distance of ship 1 from base of house = y = 200 meters       ...........1

Again

In Δ ABC

Tan angle = $\dfrac{perpendicular}{base}$

Tan  60° =  $\dfrac{AC}{CB}$

√3 = $\dfrac{200}{x}$

Or,  x =  $\dfrac{200}{\sqrt{3} }$                  ..............2

∴  The distance of ship 2 from base of house = x = $\dfrac{200}{\sqrt{3} }$ meters

From eq 1 and eq 2

Total distance between two ships  = BD = (x + y) meters

i.e BD =  ( $\dfrac{200}{\sqrt{3} }$ + 200 ) meters

Or , BD = 315.47 meters

So, The distance between two ships  = BD = 315.47 meters

Hence, The distance between two ships  is 315.47 meters , Answer

Answer 2

Let the ships be at A and B and the top of the light house be C as shown in the figure.

The distance between the ships id AB

AB = AO+OB

given that OC = 200m

tan45 = OC/OB

⇒ 1 = 200/OB

⇒OB = 200m

tan 60 = OC/AO

⇒√3 = 200/AO

⇒AO = 200/√3 = 115.47m

AB = AO+OB = 200+115.47 = 315.47m

Distance between ships is 315.47m