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Problem with Sines on both sides

  1. Aug 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve for ∅

    1.33 sin 25.0° = 1.50 sin ∅

    2. Relevant equations

    Law of Sines?

    3. The attempt at a solution

    I did this:

    ([STRIKE]1.33[/STRIKE] sin 25.0°)/[STRIKE]1.33[/STRIKE] = (1.50 sin ∅)/1.33
    sin 25.0° = (1.50 sin ∅)/1.33

    but I'm solving for ∅, so I modified a little:

    (1.33 sin 25.0°)/1.50 = ([STRIKE]1.50 [/STRIKE]sin ∅)/[STRIKE]1.50[/STRIKE]
    sin ∅ = (1.33 sin 25.0°)/1.50

    I have a basic understanding of trigonometry (SOH CAH TOA) but I'm not too sure how to do this. Can I use an inverse sine to figure? Does the law of sines apply?

    There is no triangle.
     
    Last edited: Aug 9, 2013
  2. jcsd
  3. Aug 9, 2013 #2

    SteamKing

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    The law of sines is irrelevant to solving for the unknown angle phi. If you want to solve for phi, then, yes, you must use an inverse sine in your calculations.
     
  4. Aug 9, 2013 #3

    Simon Bridge

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    You have/i] to use an inverse sign.
    1st solve for ##\sin\phi## and then take the inverse sine of both sides.
     
  5. Aug 9, 2013 #4


    So will this solve for ∅?

    [STRIKE]sin-1[/STRIKE]([STRIKE]sin[/STRIKE] ∅) = sin-1((1.33 sin 25.0°)/1.50)
    ∅ = sin-1((1.33 sin 25.0°)/1.50)
     
    Last edited: Aug 9, 2013
  6. Aug 9, 2013 #5

    Simon Bridge

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    Absolutely: ##\sin^{-1}(\sin\phi )=\phi## by definition, so if ##\sin\phi = x## then ##\phi=\sin^{-1}x## where ##x## stands for everything on the RHS.

    Welcome to PF BTW :)
     
  7. Aug 10, 2013 #6

    SammyS

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    Actually, ##\sin^{-1}(\sin\phi )=\phi## is only true for

    ##\displaystyle \ \ -\,\frac{\pi}{2}\le\phi\le \frac{\pi}{2}\ .##
     
  8. Aug 10, 2013 #7

    Simon Bridge

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    Yah well - that would be the rest of the definition ... the geometry here is Snell's law.
     
  9. Aug 10, 2013 #8

    SammyS

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    What you say makes perfect sense !

    It would have helped if OP had mentioned what he/she was applying the "Law of Sines" to.
     
  10. Aug 11, 2013 #9

    Simon Bridge

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    "Law of sines" is usually derived on a scalene triangle - so the angle range is 0-pi.
    The other clue is that this is "introductory physics homework"... where else would this structure come up?

    If this question were posted in an algebra, or signals processing (say), context, I'd have had to bring up the periodicity next. No need to provide everything in one go - one step at a time huh? ;)

    Nice heads up though.
     
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