With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x°, where tan x° = $\dfrac{2}{5}$. Calculate:

(i) the distance AB in metres;

(ii) angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.

Let's take AD to be the height of the man, AD = 2 m.

From figure, BE = AD = 2 m.

Also,

CE = BC - BE = (10 - 2) = 8 m.

(i) In ∆CED,

$\Rightarrow \text{tan x} \degree = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan x} \degree = \dfrac{CE}{DE} \\[1em] \Rightarrow \dfrac{2}{5} = \dfrac{8}{DE} \\[1em] \Rightarrow DE = \dfrac{8 \times 5}{2} \\[1em] \Rightarrow DE = \dfrac{40}{2} = 20 \text{ m}.$

From figure,

AB = DE = 20 m.

Hence, AB = 20 m.

(ii) Let A'D' be the new position of the man and θ be the angle of elevation of the top of the tower.

Such that, D'E = 15 m

In ∆CED',

$\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan θ} = \dfrac{CE}{D'E} \\[1em] \Rightarrow \text{tan θ} = \dfrac{8}{15} \\[1em] \Rightarrow \text{tan θ} = 0.533 \\[1em] \Rightarrow \text{tan θ} = \text{tan } 28° \\[1em] \Rightarrow \text{θ} = 28°.$

Hence, angle of elevation of the top of the pole when the man is standing 15 metres from the pole is 28°.