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Circles

Question
CBSEENMA10008339

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

Solution

Let C be position of the man. AB be the water level, and BH be the hill. The angles of elevation of the top and depression of foot from the deck of the ship be 60° and 30° respectively.
i.e., ∠DCH = 60° and ∠BCD = 30°

Let    HD = x m
In right triangle CDH, we have
tan space 60 degree equals HD over CD
rightwards double arrow space space square root of 3 space equals space straight x over CD
rightwards double arrow space space CD space equals space fraction numerator straight x over denominator square root of 3 end fraction space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
In right triangle CDB, we have
tan space 30 degree space equals space BD over CD
rightwards double arrow space space fraction numerator 1 over denominator square root of 3 end fraction equals 10 over CD
rightwards double arrow space space space CD space equals space 10 square root of 3 space space space space space space space space space space space space... left parenthesis ii right parenthesis
Hence, distance of the ship from the hill equals 10 square root of 3 space straight m.
Comparing (i) and (ii), we get
fraction numerator straight x over denominator square root of 3 end fraction equals 10 square root of 3 space straight m
rightwards double arrow space space straight x space equals space 10 square root of 3 space straight m space straight x space square root of 3 space equals space 30 space straight m.

Now, total height of the hill
= BD + DH = 10 + x
= 10 + 30 = 40 m
Hence, height of the hill = 40 m.