Question

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

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