Question

A man is standing on the deck of a ship, which is 40 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 `(sqrt(3) = 1.732)`

Answer 1

Let the height of the hill BE be h m and the distance of the hill from the ship be x m

In the right ∆ABD

tan 30° = `"AD"/"DB"`

`1/sqrt(3) = 40/x`

x = `40sqrt(3)`  ...(1)

In the right ∆CDE

tan 60° = `"CE"/"DC"`

`sqrt(3) = ("h" - 40)/x`

x = `("h" - 40)/sqrt(3)`  ...(2)

From (1) and (2) we get

`("h" - 40)/sqrt(3) = 40sqrt(3)`

h – 40 = 40 × 3

h = 120 + 40 = 160 m

Height of the hill = 160 m

Distance of the hill from the ship = `40 xx sqrt(3)`

= 40 × 1.732

= 69.28 m