• Aug 25th 2008, 03:47 PM
nee
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 PM
lewisb13
Quote:

Originally Posted by nee
Find the angle in degrees

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

V = 60
D = 80
h = 2

Is it (16*D)^2 or 16*(D^2)?
• Aug 25th 2008, 04:11 PM
nee
The way it's written in the text is 16D(squared).
• Aug 25th 2008, 04:23 PM
lewisb13
Quote:

Originally Posted by nee
The way it's written in the text is 16D(squared).

Ok, which one of my 2 ways would you interpret that?
• Aug 25th 2008, 04:26 PM
Jhevon
Quote:

Originally Posted by lewisb13
Ok, which one of my 2 ways would you interpret that?

@ nee: meaning, was it $16D^2$ or $(16D)^2$
• Aug 25th 2008, 04:27 PM
nee
• Aug 25th 2008, 04:32 PM
lewisb13
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 PM
lewisb13
x = -14.6508 or x = .97075
• Aug 25th 2008, 04:35 PM
lewisb13
13.892 degrees
• Aug 25th 2008, 04:38 PM
nee
Thanks alot, I'll give it a try!
• Aug 25th 2008, 04:52 PM
nee
I keep getting -12.55 and -1.133 which don't give me an angle!?
• Aug 25th 2008, 04:54 PM
lewisb13
Did you solve the quadratic? Or are those ^^^^ the numbers you got while solving the quadratic?
• Aug 25th 2008, 05:00 PM
nee
Ya I used