Question #90164
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?
1
Expert's answer
2019-05-27T12:59:51-0400

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/AHBH=tan(BAH)AHtan(BAH)=BH/AH⟹BH=tan(BAH)∗AH,

(2)tan(BCH)=BH/(ACAH)BH=tan(BCH)(ACAH)tan(BCH)=BH/(AC−AH)⟹BH=tan(BCH)∗(AC−AH)


from these two equations we have


tan(BAH)AH=tan(BCH)(ACAH)tan(BAH)∗AH=tan(BCH)∗(AC−AH), so


AH=ACtan(BCH)/(tan(BAH)+tan(BCH))AH=AC*tan(BCH)/(tan(BAH)+tan(BCH))


let use it for equation (1)


BH=ACtan(BAH)tan(BCH)/(tan(BAH)+tan(BCH))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)*tan(35.6 degrees)*tan(58.2 degrees)/(tan(35.6 degrees)+tan(58.2 degrees))


BH=7220(ft)0.71591.6128/(0.7159+1.6128)BH=7220(ft)*0.7159*1.6128/(0.7159+1.6128)


BH=3579.8(ft)BH=3579.8(ft)

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