A hot 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.
+128475 I assume that you probably want to know the height of the balloon???
If so...call the ground distance from the balloon to the soccer field on the right side = x
And on the other side ....let the grouhd distance to he football field = 4 - x
So we have that
tan (20° 15') = height / x
Note that 15' = (15/60)° = (1/4)° = .25°
So
tan (20.25°) = height / x
x tan (20.25) = height
And on the other side [ 30' = .5 degrees ]
tan ( 62.5°) = height / (4 - x)
(4 - x) tan (62.5) = height
Set the heighs equal
(4 - x) tan (62.5) = xt an(20.25)
4tan(62.5) - x tan(62.5) = x tan (20.25)
4tan(62.5) = x [ tan (20.25 + tan (62.5) ]
x = 4tan (62.5) [ tan(20.25) + tan (62.5) ] ≈ 3.36 miles
So...the height is
(3.36)tan(20.25) = height ≈ 1.24 miles
