# A man standing on

In the fig. AB is the height of hill AB = AE+EB

h+10

In ∆AEB

tan 60° =

√3 =

h = √3 DE ------- (1)

In ∆DEB

tan 30° =

DE = 10√3 ------(2)

the distance of the hill from the ship = DE

10√3 m

From eqn. (1) and eqn. (2) we get,

h = √3× 10√3
= 10(3) m
= 30 m

Height of hill = 30 + 10
= 40 m

Therefore the distance between ship and hill is 10√3 m and the height of the hill is 40 m.

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