SOLUTION: I need help with this word problem please. Felita decided to take a ride in a hot air balooon.as the air ballon rises vertically the angle of elevation from a point P on the gro

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Question 756463: I need help with this word problem please.
Felita decided to take a ride in a hot air balooon.as the air ballon rises vertically the angle of elevation from a point P on the ground level 120 km from the point Q directly underneath the balloon changes from 19 degree 20 seconds to 31 degrees 50 seconds.Approximatly how far does the ballon rise during this period?
Answer by KMST(5328) (Show Source):
You can put this solution on YOUR website!
NOTE # 1:
A hot air balloon cannot get that high and would be hard to see from 120 km away. A helium balloon can rise higher than a hot air balloon, and Felix Baumgartner supposedly got to 39km on a helim balloon for his skydiving record. (He was wearing some sort of spacesuit, because there is very little air pressure at that altitude). The problem would be more believable if the distance PQ was 12.0 km instead of 120 km. However, I'll calculate based on PQ=120km.

The ballon rises from point R to point S, both directly above point Q.
angle QPR = 19 degree 20 seconds
angle QPS = 31 degrees 50 seconds
QR = PQtan(QPR)
QS = PQtan(QPS) and
balloon rise = QS - QP = PQtan(QPS) - PQtan(QPR) = PQ(tan(QPS) - tan(QPR))
Using approximate values for those tangents,
QR = PQtan(QPR) = 120km X 0.344436 = 41.33232km
QS = PQtan(QPS) = 120km X 0.601191 = 72.14292km, and
balloon rise = 120km (0.601191 - 0.344436) = 120km X 0.256755 = 30.8106km

NOTE:
If the calculator cannot handle minute and second fractions of a degree, you will have to convert to degrees.
There are 60 minutes in 1 degree and 60 seconds in one minute, so there are
seconds in 1 degree.
19 degree 20 seconds = =19.00555555....
31 degree 50 seconds = =31.01388888888....