A man standing on a deck of a ship which is 10 m above water level observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30° .find the distance of the hill from the ship and the height of the hill
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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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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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