Question

From the top of a tower of height 50 m, the angles of depression of the top and bottom of a pole are 30° and 45° respectively. Find
(i) how far the pole is from the bottom of a tower,
(ii) the height of the pole. (Use $$\sqrt{3}=1.732$$)

 

Answer 1

Let PC and AB represents the tower and the pole, respectively.
Suppose AB = x m.
From the above figure, we have

$$\mathrm{\angle }PAC=\mathrm{\angle }QPA={45}^o\left(\text{Alternate angles}\right)$$
$$\mathrm{\angle }PBD=\mathrm{\angle }QPB={30}^o\left(\text{Alternate angles}\right)$$

PD = PC − CD = PC − AB = (50 − x) m

In ∆PAC,

$$\tan {45}^o=\frac{PC}{AC}$$
$$\Rightarrow 1=\frac{PC}{AC}$$
$$\Rightarrow AC=PC=50m$$

Therefore, the distance between the foot of the pole and the bottom of the tower is 50 m.
In ∆PBD,

$$\tan {30}^o=\frac{PD}{BD}$$
$$\Rightarrow \frac{1}{\sqrt{3}}=\frac{PD}{BD}$$
$$\Rightarrow BD=PD\sqrt{3}$$
$$\Rightarrow BD=(50-x)\sqrt{3}$$

Now, BD = AC

$$\therefore (50-x)\sqrt{3}=50$$
$$\Rightarrow \sqrt{3}x=50(\sqrt{3}-1)$$
$$\Rightarrow x=\frac{50(\sqrt{3}-1)}{\sqrt{3}}$$
$$\Rightarrow x=\frac{50(3-\sqrt{3})}{3}$$
$$\Rightarrow x=\frac{50(3-1.732)}{3}$$
$$\Rightarrow x=21.13m\left(\text{approx}\right)$$

Hence, the approximate height of the pole is 21.13 m.