Answer to Question #90164 in Trigonometry for alexis

Question

two people, 7220 ft. apart, observe a balloonist flying overhead between them. their measures of the angles of elevation at 3:30pm are 35.6° and 58.2° respectively. what is the height of the balloonist above the ground?

Answer 1

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)

$BH=AC*tan(BAH)*tan(BCH)/(tan(BAH)+tan(BCH))$

$BH=7220(ft)*tan(35.6 degrees)*tan(58.2 degrees)/(tan(35.6 degrees)+tan(58.2 degrees))$

$BH=7220(ft)*0.7159*1.6128/(0.7159+1.6128)$

$BH=3579.8(ft)$