Let we have triangle ABC, where AC is the distance between two people (7220 ft), AB is a distance between one of them and baloonist, and angle BAC is equal to 35.6 degrees. BC is a distance from another person to the baloonist, and angle BCA is equal to 58.2 degrees. Let point H be a projection of baloonist to the AC, and BH is a height of the balloonist above the ground.
(1)tan(BAH)=BH/AH⟹BH=tan(BAH)∗AH,
(2)tan(BCH)=BH/(AC−AH)⟹BH=tan(BCH)∗(AC−AH)
from these two equations we have
tan(BAH)∗AH=tan(BCH)∗(AC−AH), so
AH=AC∗tan(BCH)/(tan(BAH)+tan(BCH))
let use it for equation (1)
BH=AC∗tan(BAH)∗tan(BCH)/(tan(BAH)+tan(BCH))
BH=7220(ft)∗tan(35.6degrees)∗tan(58.2degrees)/(tan(35.6degrees)+tan(58.2degrees))
BH=7220(ft)∗0.7159∗1.6128/(0.7159+1.6128)
BH=3579.8(ft)
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