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
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.