Trig
posted by Sonia .
From two different tracking stations, a weather balloon is spotted from two angles of elevation, 57degrees and 83 degrees, respectively. The tracking stations are 15 km apart. Find that altitude of the balloon.
Any help given will be greatly appriciated

First, I'm going to give you
instructions and INFO for drawing
a picture of the problem. The picture
is the key to understanding the
problem. So it has to be drawn correctly.
1. Draw a horizontal line. 2. At the left end, draw a vertical line upward
to form a rt. angle. 3. Draw a hypotenuse from the top of vertical
line to rt. end of the horizontal
line. 4. Label the acute angle between
hyp. and the horizontal 57 deg.
5. Draw a 2nd hyp. from the top of
vertical line to horizontal line at
a point greater than half way to left
end. Label the acute angle formed
83 deg. 6. The distance between the
2 hyp. should be labeled 15km on the
hor. line.Label the remaining distance
X on hor. line. The total hor. dist.
= X + 15. 7. Label vertical line h.
Tan57 = h / (X+15), h = ( x + 15)7an57.
Tan 83 = h/X, h = X Tan83.
h = (X +15) Tan 57 = X Tan 83.
Solve for X:
(X +15)*1.54 = 8.14X
1.54X + 23.1 = 8.14X
X = 3,5km
h = X Tan 83 = 3.5 * 8.14 = 28.5km.=
Altitude. 
My answer in the book was 32km, so how can I get this answer?

Sonia, check the angles and make sure
they are the same as those posted. 
Oh sorry this is right, I read the wrong answer, thank u for your help:)

I didn't get the labeling! Like what is h ?
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