Question

The angle of elevation from a point P of the top of a tower QR, 50 m high is 60°, and that of the tower PT from a point Q is 30°. Find the height of the tower PT, correct to the nearest meter.

Answer 1

In ΔPQR

`tan 60^circ = (RQ)/(PQ)`

`sqrt(3) = 50/(PQ)`

`PQ = 50/sqrt(3)`

In ΔPQT

`tan 30^circ = (PT)/(PQ)`

`PT = PQ xx 1/sqrt(3)`

`PT = 1/sqrt(3) xx 50/sqrt(3)` 

= `50/3`

= 16.67 m

Answer 2

Given: The angle of elevation from A is 30°, from B (20 m closer) is 60°.

Let the tower height = h and the distance from A to the foot of the tower = x.

Step-wise calculation:

1. From point A:

`tan 30^circ = h/x` 

⇒ h = x × tan 30°

⇒ `h = x/sqrt(3)`

2. From point B:

Distance to tower = x – 20.

`tan 60^circ = h/(x - 20)` 

⇒ `h = sqrt(3) (x - 20)`

3. Equate the two expressions for h:

`x/sqrt(3) = sqrt(3) (x - 20)`

4. Multiply both sides by `sqrt(3)`:

x = 3(x – 20)

⇒ x = 3x – 60

⇒ 2x = 60

⇒ x = 30 m

5. Height:

`h = x/sqrt(3)`

= `30/sqrt(3)`

= `10sqrt(3)` m ≈ 17.32 m

Height of the tower = `10sqrt(3)` m (≈ 17.32 m).

Distance of the tower from point A = 30 m.