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An observer flying in an aeroplane at an altitude of 900m observes 2 ships in front of him, which are in the same direction at an angles of depression of

60 and 30 respectively. find the distance between the two ships
final ans
i have left it in cjoice cause of lack of time
completed physics


One at angle of depression 60 degree will be at distance of 900/ \sqrt{3}  
second at angle of depression of 30 degree will be at a distance 900 \sqrt{3}
0 0 0
u r wrng
that is 300 root3
yes if you take 3x300/ root 3 but haven't s solved it further
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See the attachment.
We need to find the distance between the ships which is CD (=y-x).

tan 60 = AB/BC
⇒ √3 = 900/x
⇒ x = 900/√3 = 300√3m

tan 30 = AB/BD
⇒ 1/√3 = 900/y
⇒ y = 900×√3 = 900√3m

CD = y-x = 900√3 - 300√3 = 600√3 m 

So the distance between the two ships is 600√3 m.
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