1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find all the angles problem
Loading...