Two ships are there in the sea on either side of a light houses in such a way that the ships and the lighthouse are in the same straight line. The angles of depression of two ships as observed from the top o the light house are 60 � and 45 �. If the height of the light house is 200 m, find the distance between the two ships. [Use √3 = 1.73]

Let AB be the light house and C and D be the two ships.

And the distance between the two ships is d. Let the distance of C from AB is x m, then the distance of D from AB is (d – x) m.

Consider right angled ΔABC,

⇒ tan45° = AB/BC

⇒ 1 = 200/x

⇒ x = BC = 200m … (1)

Now consider right angled ΔABD,

⇒ tan60° = AB/BD

⇒ √3 = 200/ (d – x)

⇒ (d – x) = 200/√3

By rationalizing the denominator,

⇒ (d – x) = 200√3 / √3(√3)

⇒ (d – x) = 200√3/ 3 m …(2)

Substituting (1) in (2), we get

⇒ d – 200 = 200√3/3

⇒ d = CD = (200√3/3) – 200

⇒ d = 200(3 + √3)/3

Given √3 = 1.73

⇒ d = 200(3 + 1.73)/3

⇒ d = 946/3

⇒ d = 315.333 m = m

Ans. The distance between the two ships is m.