Find the angle in degrees

0.342Dcosx + hcos(squared)x = 16D(squared) / V(squared)

V = 60

D = 80

h = 2

Printable View

- Aug 25th 2008, 03:47 PMneeTrigonometric Equaiton Help Please!
Find the angle in degrees

0.342Dcosx + hcos(squared)x = 16D(squared) / V(squared)

V = 60

D = 80

h = 2 - Aug 25th 2008, 04:08 PMlewisb13
- Aug 25th 2008, 04:11 PMnee
The way it's written in the text is 16D(squared).

- Aug 25th 2008, 04:23 PMlewisb13
- Aug 25th 2008, 04:26 PMJhevon
- Aug 25th 2008, 04:27 PMnee
- Aug 25th 2008, 04:32 PMlewisb13
So you've got 2x^2+27.36x^2-256/9=0

Here we are letting x = cos x

Use the quadratic equation. Youll get 2 numbers. Probably one positive and one negative.

FOR EXAMPLE you get 5 and -16

5 = cos x

-16 = cos x

Solve for x. - Aug 25th 2008, 04:34 PMlewisb13
x = -14.6508 or x = .97075

- Aug 25th 2008, 04:35 PMlewisb13
13.892 degrees

- Aug 25th 2008, 04:38 PMnee
Thanks alot, I'll give it a try!

- Aug 25th 2008, 04:52 PMnee
I keep getting -12.55 and -1.133 which don't give me an angle!?

- Aug 25th 2008, 04:54 PMlewisb13
Did you solve the quadratic? Or are those ^^^^ the numbers you got while solving the quadratic?

- Aug 25th 2008, 05:00 PMnee
Ya I used

http://upload.wikimedia.org/math/2/9...80bd1d670f.png

and put the values into:

http://upload.wikimedia.org/math/3/e...a3bb558916.png

I end up with

x = -27.36 +or- 22.83 / 4

which gives me

12.55 and -1.133