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