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

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

PD = PC − CD = PC − AB = (50 − x) m

In ∆PAC,

\[\tan45^o = \frac{PC}{AC}\]
\[ \Rightarrow 1 = \frac{PC}{AC}\]
\[ \Rightarrow AC = PC = 50 m\]

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

\[\tan30^o= \frac{PD}{BD}\]
\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{PD}{BD}\]
\[ \Rightarrow BD = PD\sqrt{3}\]
\[ \Rightarrow BD = \left( 50 - x \right)\sqrt{3}\]

Now, BD = AC

\[\therefore \left( 50 - x \right)\sqrt{3} = 50\]
\[ \Rightarrow \sqrt{3}x = 50\left( \sqrt{3} - 1 \right)\]
\[ \Rightarrow x = \frac{50\left( \sqrt{3} - 1 \right)}{\sqrt{3}}\]
\[ \Rightarrow x = \frac{50\left( 3 - \sqrt{3} \right)}{3}\]
\[ \Rightarrow x = \frac{50\left( 3 - 1 . 732 \right)}{3}\]
\[ \Rightarrow x = 21 . 13 m \left( \text{approx} \right)\]

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