Ahot air balloon is flying above groveburg. to the left side of the balloon, the balloonist measure the angle of depression to the groveburg soccer fields to be 20° 15'. to the right side of the balloon, the balloonist measures the angle of depression to the high school football field to be 62° 30'. the distance between the two athletic complexes is 4 miles.

what is the distance from the balloon to the football field?

The answer is 1.4 miles ⇒ answer (d)

Step-by-step explanation:

* Let the balloon is at the vertex A, and the soccer fields

  at vertex B and the football field at vertex C

∴ m∠B = 20° 15' = 20.25°

∴ m∠C = 62° 30' = 62.5°

∴ m∠A = 180 - 20.25 - 62.5 = 97.25°

∵ BC = 4 miles

* By using the sin Rule:

∵ AC/sin(B) = BC/sin(A)

∴ AC = (BC)(sinB)/sin(A)

∴ AC = [4 × sin(20.25)] ÷ sin(97.25) = 1.395 ≅ 1.4 miles

∴ The distance from the balloon to the football filed = 1.4 miles

or

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof