As observed from the top of a light house, 100m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30 degree to 45 degree. Determine the distance traveled by the ship during the period of observation.
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Answered by
112
height = AB= 100
BC= x
CD=y
triangle ABC
tan60= AB/BC
√3= 100/x
x= 100/√3
In triangle ABD
tan30= AB/CD
1/√3= 100/y
y= 100√3
as we know
BD=BC+ CD
y= x+ CD
CD= y-x
putting the values we get
CD= 100 x (3-1) /√3
=100 x 2/√3
=200/√3
BC= x
CD=y
triangle ABC
tan60= AB/BC
√3= 100/x
x= 100/√3
In triangle ABD
tan30= AB/CD
1/√3= 100/y
y= 100√3
as we know
BD=BC+ CD
y= x+ CD
CD= y-x
putting the values we get
CD= 100 x (3-1) /√3
=100 x 2/√3
=200/√3
Answered by
45
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