Question

The angles of depression of two ships from the top of a light house are 60° and 45° towards east. If the ships are 300 m apart, the height of the light house is

A) 200(3+√3) meter B) 250(3+√3) meter C) 150(3+√3) meter D) 160(3+√3) meter

Answer 1

Taking ∆ABD

In ∆ABD points C & D are the ships and the AB is the height i.e. lighthouse

Given ,

Angle of ship C = 45°

Angle of ship D = 60°

Now, in ∆ABC

tanC = tan45° = AB/BC

AB/BC=1

AB = BC

HERE, AB = BC , AB = height of lighthouse (h)

That means

BC = h = AB ……… eq1

In ∆ABD

tanD = tan60° = AB/BD

AB/BD = √3

AB = BD√3

h = BD√3

HERE, BD = BC + CD

h = {h + CD}√3

Here, distance between the ships that means CD = 300m

h/√3 = h + 300

(h/√3) - h = 300

(h - √3 h)/√3 = 300

h(1 - √3) = 300√3

h = 300√3/(1-√3)

h = 300√3/1-√3

Rationalising the denominator

h = -150√3(1 + √3)

h = -150√3 - 450

h = -150(√3 + 3)

Hence, distance cannot be negative so,

H = 150(√3 + 3)

Therefore, Option c). 150(√3 + 3) is the correct option.