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I can walk you through it if you are on... send me a message

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Hmm let me try your solution.

No problem!

Okay, I'm confused now....

WAIT, NO IT'S CORRECT!!!!!

Thank You, @PlasmaFuzer!!!!

I'm just confused about the arcsin part.

Wait a minute try did liquids B and C come out to be 1.35 and 1.40 respectively (rounded)?

That's fine! I should be okay with the rest, and it's just that one part that I don't get....

Yeah, I was wondering that, too.

One last thing, how did you get arcsin? :)

I think I'm going to ask my professor about the indices.... He might agree with us.

Like this? \[1.52\sin(\theta_a) = \sin(52)\]?

because I know we're solving for n, so we have two unknowns, yes?

Yes that is correct... and take the arcsin of both sides to get the angle itself

After dividing by 1.52 that is

Just wondering where that went, in \[\theta = \sin^{-1} (\frac{ \sin(52) }{ 1.52 })\]

Oh yes, okay. Wow.

Yes, that's right. It was there. It was just "one" so you didn't put it anymore.

sorry

When you do please message me with the answer I am really curious to know :D

Yes, of course! :)

Thank you :D :D

WAIT WAIT!!!!

???

Hold on my chat box is screwing up I have to refresh :/

No problem!

Get some rest! Hope my problem didn't wear you out that much. LOL.
Thanks! See you around! :)

I'll send you a message when I get to ask my prof.!