Cos^-1(1.61cos(theta))
\[\cos ^{-1}(1.61\cos(\theta))\]
That is a better look at it. I am trying to figure this out and I was wondering if the cos inverse can cancel out the cos on the inside
is there a context to it?
Well in all honesty I am trying to figure out a formula for a physics class...I have all of the equations, I just need to do the math part to figure it out. I will write the two equations I have
Equation #1 \[T_c*\cos(\phi) - T_D *\cos(\theta) = 0\] Equation #2 \[T_D*\sin(\theta) - T_c*\sin(\phi) - mg= 0\] T_D is known to be 3592 T_c is known to be 2230 and G is the acceleration due to gravity which is 9.81
What I am trying to solve for is m
Or at least m is what would lead me to the answer that I want
\(\bf T_D\cdot \sin(\theta) - T_c\cdot \sin(\phi) - mg= 0\implies \cfrac{T_D\cdot \sin(\theta) - T_c\cdot \sin(\phi)}{g}=m\quad ?\)
Ok, but I don't know what theta or phi are
I am an idiot
...I guess dunno
Sorry I said that because I just looked and at the smallest part of the page, it says that theta =62.5 and phi = 35
*facepalm*
heheh, there are you mystery angles =)
Thanks for the help even though I was an idiot haha
He who asks is a fool for five minutes, but he who does not ask remains a fool forever. ~~ Chinese proverb ~~
To see what is in front of one's nose needs a constant struggle. ~~ George Orwell, "In Front of Your Nose" ~~
=)
Lol thanks
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