Question

As observed from top of a 100m high light house from sea level the angles of depression of two ships are 30degree and 45degree if one ship is exactly behind the other on the same side of the light house find the distance between two ships

Answer 1

Let AB be the height of the light house & two ships be at C and D .
In ∆ABC,
tan 45° = AB/BC = P/B
1 = 100 /BC
[tan 45° = 1]
BC = 100 m
In ∆ABD
tan 30° = AB/BD
1/√3 = 100/(BC + CD)
1/√3 = 100/(100 + CD)
100 + CD = 100√3
CD = 100√3 - 100
CD = 100(√3 - 1)
CD = 100 (1.732 - 1)
[Given √3 = 1.732]
CD = 100 × 0.732
CD = 73.2 m
Hence, the distance between the two ships is 73.2 m

HOPE THIS WILL HELP YOU...

Answer 2

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