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October 22, 2014

October 22, 2014

Posted by **Adreeanna** on Sunday, January 20, 2013 at 8:58pm.

- Math -
**Reiny**, Sunday, January 20, 2013 at 10:26pmThe usual question for the above would be to find the height of the balloon, but this is easier

Hope you made a sketch.

Mine has a horizontal line PAB where the balloon is above P at a point Q

AB = 800

angle B = 37°

angle QBP = 65° , so angle QBA = 115°

which makes angle BQA = 180-115-37 = 28

by the sine law:

AQ/sin115 = 800/sin28°

AQ = 800sin115/sin28 = 1544.39 or 1544 to the nearest foot

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