Wednesday

March 4, 2015

March 4, 2015

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

Any help given will be greatly appriciated

- Trig -
**Henry**, Tuesday, July 20, 2010 at 9:00pmFirst, 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 7:15amMy answer in the book was 32km, so how can I get this answer?

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

they are the same as those posted.

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

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