Question

The angle of elevation of the top of a tower 24 m high from the foot of another tower in the same plane is 60°. The angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.

Answer 1

Given:

Height of first tower (A) = 24 m.

Let the other tower (B) have height h (m).

Let the distance between the foot of the two towers = d (m).

Angle of elevation of top of A from foot of B = 60°.

Angle of elevation of top of B from foot of A = 30°.

Step-wise calculation:

1. From foot of B looking at top of A: 

`tan 60^circ = ("Height of" A)/d` 

⇒ `sqrt(3) = 24/d` 

So `d = 24/sqrt(3)` 

= `(24sqrt(3))/3` 

= `8sqrt(3)` m

2. From foot of A looking at top of B:

`tan 30^circ = ("Height of" B)/d` 

⇒ `1/sqrt(3) = h/d` 

So `h = d/sqrt(3)` 

= `(8sqrt(3))/sqrt(3)`

= 8 m

3. Length of wire joining the tops = Distance between the top points:

Vertical difference = 24 – 8 = 16 m.

Horizontal separation = d = `8sqrt(3)` m. 

Wire length L = `sqrt(d^2 + ("vertical difference")^2)`

= `sqrt((8sqrt(3))^2 + 16^2)` 

= `sqrt(64 xx 3 + 256)` 

= `sqrt(192 + 256)` 

= `sqrt(448)`

= `sqrt(64 xx 7)` 

= `8sqrt(7)` m ≈ 8 × 2.6458 

= 21.166 m (approx)

Distance between the two towers = `8sqrt(3)` m (≈ 13.856 m).

Height of the other tower = 8 m.

Length of the wire between the tops = `8sqrt(7)` m (≈ 21.166 m).