Posted by **Sonia** on Tuesday, July 20, 2010 at 4:22pm.

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

- Trig -
**Henry**, Tuesday, July 20, 2010 at 9:00pm
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.

- Trig -
**Sonia**, Wednesday, July 21, 2010 at 7:15am
My answer in the book was 32km, so how can I get this answer?

- Trig -
**Henry**, Wednesday, July 21, 2010 at 3:50pm
Sonia, check the angles and make sure

they are the same as those posted.

- Trig -
**Sonia**, Wednesday, July 21, 2010 at 4:14pm
Oh sorry this is right, I read the wrong answer, thank u for your help:)

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