Question

One observer estimates the angle of elevation to the basket of a hot air balloon to be 60°, while another observer 100 m away estimates the angle of elevation to be 30°. Find:

1. The height of the basket from the ground. 
2. The distance of the basket from the first observer's eye. 
3. The horizontal distance of the second observer from the basket.

Answer 1

Given: One observer A measures angle of elevation 60°; another observer B (100 m away from A along the same straight line) measures 30°. Let the vertical projection (foot) of the basket on the ground be point C. Let AC = d (horizontal distance from first observer to the basket) and the basket height be h.

Step-wise calculation:

1. From observer A (angle 60°):

`tan 60^circ = h/d = sqrt(3)` 

⇒ `h = dsqrt(3)`

2. From observer B (angle 30°). If B is 100 m farther from A on the same line away from the basket, horizontal distance BC = d + 100.

So `tan 30^circ = h/(d + 100) = 1/sqrt(3)` 

⇒ `h = (d + 100)/sqrt(3)`

3. Equate the two expressions for h:

`dsqrt(3) = (d + 100)/sqrt(3)`

⇒ 3d = d + 100

⇒ 2d = 100

⇒ d = 50 m

4. Height: `h = dsqrt(3)`

= `50sqrt(3)  m ≈ 86.60  m`

5. Distance of the basket from the first observer's eye (slant distance):

`s_1 = sqrt(d^2 + h^2)` 

= `sqrt(50^2 + (50sqrt(3))^2)` 

= `sqrt(2500 + 7500)`

= `sqrt(10000)`

= 100 m

6. Horizontal distance of the second observer from the basket:

BC = d + 100

= 50 + 100

= 150 m

i. Height of the basket = `50sqrt(3)` m (≈ 86.60 m).

ii. Distance from the basket to the first observer's eye = 100 m.

iii. Horizontal distance of the second observer from the basket = 150 m.