Question

The angles of depression of the top and the foot of a 9 m tall building from the top of a multistoreyed building are 30° and 60°, respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use `sqrt(3)` = 1.73).

Answer 1

Let AD be multi-storeyed building and BC be building.

In ΔDCE,

tan 30° = `(DE)/(CE)`

tan 30° = `(h - 9)/x`

`1/sqrt(3) = (h - 9)/x`

x = `sqrt(3) (h - 9)`   ...(1)

In ΔABD,

tan 60° = `(AD)/(AB)`

tan 60° = `h/x`

`sqrt(3) = h/x`

h = `xsqrt(3)`   ...(2)

Substitute value of h in equation (1).

`sqrt(3)(xsqrt(3) - 9) = x`

`3x - 9sqrt(3) = x`

2x = `9sqrt(3)`

x = `(9sqrt(3))/2`

x = `4.5sqrt(3)`

x = 4.5 × 1.73

x = 7.785

x = 7.79 m

Now, substitute value of x in equation (2).

h = `4.5 sqrt(3) xx sqrt(3)`

h = 4.5 × 3

h = 13.5 m

Hence, the required height is 13.5 m and distance is 7.79 m.