An important question from trigonometry (height and distance) in which we have to find the distance of the cliff from the ship and the height of the cliff if a man on the deck of a ship, 16 m above water level, observe that that angle of elevation and depression respectively of the top and bottom of a cliff are 60° and 30°.

Class 10, RS Aggarwal, Chapter 14, qquestion no 23

This answer was edited.Consider A as the man on the deck of a ship B and CE is the cliff,

Then AB=16m

Given

Angle of elevation from the top of the cliff = 45o

Angle of depression at the base of the cliff = 30o

Take CE=h and AD=x

Then, CD=h−16 and AD=BE=x

In right triangle CAD

tanθ=CD/AD

Substituting the values,

⇒tan45°=(h−16)/x

So, we get

⇒1=(h−16)/x

⇒x=h−16…………..(1)

In right △ADE

tanθ=DE/AD

Substituting the values,

⇒tan30°=16/x

So, we get

⇒31/√3=16/x

⇒x=16√3…………..(2)

Using both the equations

h−16=16√3

⇒h=16√3+16

⇒h=16(1.732+1)

⇒h=16(2.732)

⇒h=43.712=43.71 m

Substituting the value in equation (1)

x=h−16

⇒x=43.71−16

⇒x=27.71

Hence, the distance of cliff = 27.71m.

An the height of cliff = 43.71m.