Question

From the top of a tower of height 50 m, the angles of depression of the top and bottom of a pole are ${30}^{\circ }$ and ${45}^{\circ }$ respectively. Find:

i) how far the pole is from the bottom of the tower?

ii) the height of the pole. $[Use\sqrt{3}=1.732]$

Answer 1

In this fig, AB is the tower.

CD is the pole.

The distance between the tower and the pole is BC

In right angle triangle ABC, one angle is 45

Hence the other angle is 180-90-45=45

Hence it is a 45 -45- 90 triangle.

BC=AB=50m.

Hence the distance between the tower and the pole is 50m.

Now in rt.angled $△AED$, $ED=BC=50m$

AED is a 30-60-90 Triangle.

$\frac{AE}{ED}=\frac{1}{\surd 3}$

$AE=\frac{50}{\sqrt{3}}$

Hence $BE=AB-AE=$$50-\frac{50}{\surd 3}$

Hence $CD=BE=$ height of the pole$=50-\frac{50}{\surd 3}$