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November 28, 2014

November 28, 2014

Posted by **William320** on Thursday, August 1, 2013 at 11:42am.

- Math -
**Reiny**, Thursday, August 1, 2013 at 12:53pmmake a sketch, labeling the balloon P and the point directly below it on the ground as Q

(We have to find PQ)

Label the first observer as A and the second as B

AB = 3

angle PAB = 15° , angle PBQ = 24°

In triangle PAB,

angle A = 15, angle PBA = 156° , so angle APB = 9°

by the sine law:

AP/sin15 = 3/sin9

AP = 3sin15/sin9 = 4.963465... ( I stored in calculator's memory)

In the right-angled triangle, PBQ

sin24 = PQ/AP

PQ = APsin24 = 2.0188 miles high

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