In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x⁰ where tan x = 2/5 and AF = 200 m. The elevation of T from B, where AB = 80 m, is y⁰. Calculate:

(i) the height of the tower TF.

(ii) the angle y, correct to the nearest degree.

Solution:

Consider the height of the tower TF = x

It is given that

tan x = 2/5, AF = 200 m, AB = 80 m

(i) In right triangle ATF

tan x⁰ = TF/AF

Substituting the values

2/5 = x/200

So we get

x = (2 × 200)/ 5

x = 400/5

x = 80 m

Hence, the height of tower is 80 m.

(ii) In right triangle TBF

tan y = TF/BF

Substituting the values

tan y = 80/ (200 – 80)

tan y = 80/120

tan y = 2/3 = 0.6667

So we get

y = 33⁰ 41’ = 34⁰