Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find all the angles problem

  1. May 21, 2005 #1
    Find all the angles from [tex]0^{\circ}[/tex] to [tex]360^{\circ}[/tex] inclusive which satisfy the equation
    [tex]$ \tan(x-30^{\circ}) - \tan 50^{\circ} = 0 [/tex]
     
  2. jcsd
  3. May 21, 2005 #2
    If you want help, you need to show us what you've done so far, or what your thoughts are on how to go about solving it.
     
  4. May 21, 2005 #3
    I haven't done anything. I don't have a clue what to do.
     
  5. May 21, 2005 #4
    Here's a hint: write tan (x - 30) in terms of tan (50). What can you see then?
     
  6. May 21, 2005 #5
    [tex]\tan(\alpha + \beta) = \frac{tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}[/tex]

    [tex]\tan(x - 30) = \frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}[/tex]

    [tex]\frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}=\tan(50)[/tex]

    [tex]\tan(x)+\tan(-30)=\tan(50) - \tan(x)\tan(-30)\tan(50)[/tex]

    [tex]\tan(x) + \tan(x) \tan(-30)\tan(50)=\tan(50) - \tan(-30)[/tex]

    [tex]\tan(x)(1 +\tan(-30)\tan(50))=\tan(50) - \tan(-30)[/tex]

    [tex]\tan(x)=\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}[/tex]

    :devil: :biggrin:
     
    Last edited: May 21, 2005
  7. May 21, 2005 #6
    Don't know :uhh:
     
  8. May 21, 2005 #7
    [tex]\tan(x-30^{\circ}) = \tan 50^{\circ}[/tex]

    Can you go from there?
     
  9. May 21, 2005 #8
    How did you get all that?? :confused:
     
  10. May 21, 2005 #9

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper



    :rofl: It's not that complex:

    [tex]\tan (x - 30) = \tan 50[/tex]

    Hence to work out an initial value just apply arctan on both sides to get:

    [tex]x - 30 = 50[/tex]
     
  11. May 21, 2005 #10
    There are two solutions to the problem, that is one of them.
     
  12. May 21, 2005 #11
    since the Tan curve goes in a period of 180 degrees, you take the value that you got as one of the solutions and add or subtract 180 to/from it, and every time the result is within the rang of 0 -360, so:

    you do
    [tex]\tan (x - 30) = \tan 50[/tex]
    [tex]x = 80[/tex]

    then

    [tex]80 \pm 180n = x [/tex]

    and the only other value that fits into the range is when

    [tex]n = 1 [/tex]
    [tex]80 + 180 = 260 [/tex]

    Therefor the 2 answers are

    [tex]x = 80, 260 [/tex]
     
  13. May 22, 2005 #12
    Yup. Thanks!
     
  14. May 22, 2005 #13
    Hence my smilies
    :wink:
     
  15. May 22, 2005 #14

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    Which if you expand is:

    [tex]\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}= \frac{8 \cos^7 (10) \sin (10) - 56 \cos^5 (10) \sin^3 (10) + 56 \cos^3 (10) \sin^5 (10) - 8 \cos (10) \sin^7 (10)}{\cos^8(10) - 28 \cos^6(10) \sin^2(10) + 70 \cos^4(10) \sin^4(10) - 28 \cos^2 (10) \sin^6(10) + \sin^8 (10)}[/tex]

    And it just so happens that nicely simplifies down to:

    [tex]\frac{8 \cos^7 (10) \sin (10) - 56 \cos^5 (10) \sin^3 (10) + 56 \cos^3 (10) \sin^5 (10) - 8 \cos (10) \sin^7 (10)}{\cos^8(10) - 28 \cos^6(10) \sin^2(10) + 70 \cos^4(10) \sin^4(10) - 28 \cos^2 (10) \sin^6(10) + \sin^8 (10)} = \tan (80)[/tex]

    :rolleyes:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook