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

the answer is 13/48

step-by-step explanation:

sas (side angle side)

aa(angle angle)

sss(side side side)

sas you have to find two sides with an angle in between

aa you find two angles because it will always add up to 180

sss find all sides

step-by-step explanation: