Question

as observed from top of 100m high light house from sea level angle of depression of two ships 30° and 45° one ship exactly. behind other on same side of light house find distance between two ships

Answer 1

heya....!!!!!

IN ∆ ACD

tan45° = AD / DC

=> 1 = 100/DC

=> DC = 100M

in ∆ ABD,

tan30° = AD/DB

=> 1/√3 = 100/DB

=> DB = 100√3

=> DB = 100√3M


DB = DC + CD

=> 100√3= 100 + CD

=> (100√3 - 100) = CD

=> 100(1.732 - 1) = CD
[ √3 = 1.732 ]

=> CD = 100×0.732

=> CD = 73.2M

Answer 2

Find the distance BC:

tan θ = opp/adj

tan (45) = BC/100

BC = 100tan(45)

Find the distance BD:

tan θ = opp/adj

tan(60) = BD/100

BD = 100 tan (60)

Find the distance between the two ships:

Distance = 100 tan(60) - 100 tan (45)

Distance = 100√3 - 100

Distance =73.2 m

Answer: The distance between the two ships is 73.2 m