trig. precalc. word problem?

2017-12-27 1:45 pm
A hot air balloon is between two spotters who are 1.2 miles apart. One spotter reports that the angle of elevation of the balloon is 76 degrees and the other spotter reports it is 65 degrees. What is the altitude of the balloon?

I have the answer I need someone to guide me through solving this.
answer is 1.7 miles

回答 (4)

2017-12-27 2:28 pm
✔ 最佳答案
.
let height = h


h/tan65° + h/tan76° = 1.2
h ( 1/tan65° + 1/tan76° ) = 1.2
h = 1.2 tan(76°) tan(65°) / ( tan65° + tan76° )
h = 1.677
h = 1.7 miles
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2017-12-27 3:06 pm
Refer to the diagram below.
B is the balloon, P and Q are the two spotters, and AH is the altitude of the balloon.

In ΔBPH:
tan76° = BH/PH
PH = BH/tan76°

In ΔBQH:
tan65° = BH/HQ
HQ = BH/tan65°

PH + HQ = 1.2 miles
(BH/tan76°) + (BH/tan65°) = 1.2 miles
BH [(1/tan76°) + (1/tan65°)] = 1.2 miles
BH = 1.2 / [(1/tan76°) + (1/tan65°)] miles
BH = 1.7 miles (to 2 sig. fig.)
tan(76) = h / (1.2 - x)
tan(65) = h / x

x * tan(65) = h

tan(76) = x * tan(65) / (1.2 - x)
1.2 * tan(76) - x * tan(76) = x * tan(65)
1.2 * tan(76) = x * (tan(65) + tan(76))
x = 1.2 * tan(76) / (tan(65) + tan(76))

h = x * tan(65)
h = 1.2 * tan(76) * tan(65) / (tan(65) + tan(76))
h = 1.2 * (sin(76) * sin(65) / (cos(76) * cos(65))) / ((sin(65) * cos(76) + sin(76) * cos(65)) / (cos(65) * cos(76)))
h = 1.2 * sin(76) * sin(65) / (sin(65) * cos(76) + sin(76) * cos(65))
h = (6/5) * sin(76) * sin(65) / sin(65 + 76)
h = 6 * sin(76) * sin(65) / (5 * sin(141))
h = 1.6768309146671395853123514255406...
2017-12-27 3:48 pm
Ans:


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