Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

Cos^-1(1.61cos(theta))

OpenStudy (anonymous):

\[\cos ^{-1}(1.61\cos(\theta))\]

OpenStudy (anonymous):

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

OpenStudy (jdoe0001):

is there a context to it?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

What I am trying to solve for is m

OpenStudy (anonymous):

Or at least m is what would lead me to the answer that I want

OpenStudy (jdoe0001):

\(\bf T_D\cdot \sin(\theta) - T_c\cdot \sin(\phi) - mg= 0\implies \cfrac{T_D\cdot \sin(\theta) - T_c\cdot \sin(\phi)}{g}=m\quad ?\)

OpenStudy (anonymous):

Ok, but I don't know what theta or phi are

OpenStudy (anonymous):

I am an idiot

OpenStudy (jdoe0001):

...I guess dunno

OpenStudy (anonymous):

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

OpenStudy (anonymous):

*facepalm*

OpenStudy (jdoe0001):

heheh, there are you mystery angles =)

OpenStudy (anonymous):

Thanks for the help even though I was an idiot haha

OpenStudy (jdoe0001):

He who asks is a fool for five minutes, but he who does not ask remains a fool forever. ~~ Chinese proverb ~~

OpenStudy (jdoe0001):

To see what is in front of one's nose needs a constant struggle. ~~ George Orwell, "In Front of Your Nose" ~~

OpenStudy (jdoe0001):

=)

OpenStudy (anonymous):

Lol thanks

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!