Question

from the top of a 60 metre height building the angle of depression to the top and foot of a cell tower are 30 degree and 60 degree respectively find the height of the cell tower​

Answer 1

Answer:

In ∆abc we have,

Tan30° = ab/bc

So 1/√3= h/x

So x=h√3 -eq(1)

In∆dec we have,

Tan60° =CD/ED

√3=60/x

so x√3=60/x -eq(2)

From (1) and (2)

h√3(√3) =60

So 3h =60

And h = 20

So height of tower = 60+20 =80m

May it help you