Question

NGC is a hot air balloon in the sky from her spot on the ground. The angle of elevation from Angie to the balloon is 40°. If she steps back 200 feet, the new angle of elevation is 10°. If Angie is 5.5 feet tall, how far off the ground is a hot air balloon?

Answer 1

The distance from hot air balloon to the ground
`AC=50.1457 ft`.

Explanation:

Labelled diagram of given scenario is shown below.

Given that,

An angle of elevation of Hot air balloon by Angie is
`40`°.

When she stepped back
`200 ft` then angle of elevation was
`10`°.

Height of Angie is
`5.5 ft`.

To find: How far off the ground is a hot air balloon.

So, from figure

Height of Angie
`(BC) = 5.5ft`

In triangle Δ
`ABD`,

⇒
`tan {40\si {\degree}} = (AB)/(BD)`

⇒
`AB = tan (40) * BD` ....................(1)

Now, In triangle Δ
`ABE`

⇒
`tan (10)= (AB)/(200+BD)`

⇒
`AB = tan(10)* (200+BD)`

Here, substituting the value
`AB` from Equation (1) we get,

`tan(10)* (200+BD)= tan(40)* BD`

`tan(10)* 200+tan(10)* BD= tan(40)* BD`

⇒
`BD* (tan(40)-tan(10))=tan(10)* 20`

⇒
`BD* 0.6628 = 35.2654`

⇒
`BD = (35.2654)/(0.6628) =53.2067 ft`

Now, finding the value of
`AB` from equation (1)

`AB= tan(40)* 53.2067=0.8391* 53.2067`

`AB=44.6457 ft`

Therefore Length of
`AC= AB+BC` =
`(44.6457 + 5.5) ft`

`AC=50.1457 ft`.

Hence,

The distance from hot air balloon to the ground
`AC=50.1457 ft`.