Find the angle in degrees 0.342Dcosx + hcos(squared)x = 16D(squared) / V(squared) V = 60 D = 80 h = 2
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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)?
The way it's written in the text is 16D(squared).
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?
Originally Posted by lewisb13 Ok, which one of my 2 ways would you interpret that? @ nee: meaning, was it $\displaystyle 16D^2$ or $\displaystyle (16D)^2$
exactly
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.
x = -14.6508 or x = .97075
13.892 degrees
Thanks alot, I'll give it a try!
I keep getting -12.55 and -1.133 which don't give me an angle!?
Did you solve the quadratic? Or are those ^^^^ the numbers you got while solving the quadratic?
Ya I used and put the values into: I end up with x = -27.36 +or- 22.83 / 4 which gives me 12.55 and -1.133
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