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2016-03-19T18:03:36+05:30

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Consider the given figure,

In ΔBDC
tan30∘=10x
13√=10x⇒x=103√=10×1.732=17.32 m
In ΔAEC
tan60∘=AECF
3√=AE103√⇒AE=30m
So, Height of the hill =30+10=40 m
Distance of the hill from the ship=17.32 m

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2016-03-19T19:54:51+05:30
Tan 30= h/x where h is triangular height and x is length of base... h/x = root 3.. h = x into root 3..------1 And tan 30 = 10/x.. 1 / root 3 = 10/x.. X = 10 root 3.. In 1 Then h = 10 root 3 into root 3 =30m.. Total distance = 10+30 = 40.. Hence, the distance of the hill from the ship is 10 3 m and the height of the hill is 40 m.
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