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

Answer:

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

Step-by-step 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$ABD$,

         ⇒                    $tan {40\si {\degree}} = \frac{AB}{BD}$

         ⇒                      $AB = tan (40) \times BD$          ....................(1)

Now, In triangle Δ$ABE$

         ⇒                  $tan (10)= \frac{AB}{200+BD}$

         ⇒                 $AB = tan(10)\times (200+BD)$

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

                        $tan(10)\times (200+BD)= tan(40)\times BD$

                      $tan(10)\times 200+tan(10)\times BD= tan(40)\times BD$

        ⇒         $BD\times (tan(40)-tan(10))=tan(10)\times 20$        

        ⇒                   $BD\times 0.6628 = 35.2654$

        ⇒                       $BD = \frac{35.2654}{0.6628} =53.2067 ft$        

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

                                 $AB= tan(40)\times 53.2067=0.8391\times 53.2067$

                                 $AB=44.6457 ft$              

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  $AC=50.1457 ft$.