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trig

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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.

Mar 5, 2019

#1
+106532
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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

Mar 5, 2019