Results 1 to 4 of 4

Math Help - Help on triogonometry. Test tomorrow!

  1. #1
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177

    Help on triogonometry. Test tomorrow!

    √ 29 cos (x + 68.2 )=3.1

    solve all angle of x. range is 0 to 360

    ans is 236.9 and 347.7 i think. ppls expalin to me how to do it. thank you. by the way, just afraid that it is not clear, the front part is square root 29.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by helloying View Post
    √ 29 cos (x + 68.2 )=3.1

    solve all angle of x. range is 0 to 360

    ans is 236.9 and 347.7 i think. ppls expalin to me how to do it. thank you. by the way, just afraid that it is not clear, the front part is square root 29.
    sqrt(29) *cos(x +68.2deg) = 3.1
    So,
    cos(x +68.2deg) = 3.1 / sqrt(29)
    cos(x +68.2deg) = 0.57566

    It is a positive cosine value, so angle (x +68.2) is in the 1st and 4th quadrants.

    (x +68.2deg) = arccos(0.57566) = 54.9 deg

    In the 1st quadrant:
    (x +68.2deg) = 54.9 deg
    x = 54.9 -68.2 = -13.3 deg
    That is in the 4th quadrant. If measuring by the usual counterclockwise way,
    x = 360 -13.3 = 346.7 degrees ------**

    In the 4th quadrant:
    (x +68.2deg) = (360 -54.9) deg
    x +68.2 = 305.1 deg
    x = 305.1 -68.2 = 236.9 deg ------------**

    Therefore, x = 236.9 deg and 346.7 deg. --------answer.
    Last edited by ticbol; August 5th 2008 at 01:28 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177

    oh i see

    but how did u know when to put the negative in the 4th quadrant or watever and when to reject it? i had mistaken the negative i had gotten to be out of the range so i reject the -13.3
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by helloying View Post
    but how did u know when to put the negative in the 4th quadrant or watever and when to reject it? i had mistaken the negative i had gotten to be out of the range so i reject the -13.3
    The -13.3 degrees?
    It is in the 4th quadrant because it is a negative acute angle.
    If it were 13.3 degrees, then it should be in the 1st quadrant because it is a positive acute angle.

    Our usual direction for measuring angles is counterclockwise starting from the right branch of the horizontal axis or of the x-axis. So if an angle is negative, then it is being meaured clockwise. So -13.3 degrees is in the 4th quadrant.
    The four quadrants are always "counted" counterclokwise whether the angles are measured counterclockwise or clockwise.

    Another example is -95 degrees. That should be in the 3rd quadrant. And if -95 degrees is to be measured as usual....counterclockwise.... then it will be
    -95 deg = (360 -95) = 265 deg. It is still in the 3rd quadrant.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help! Test tomorrow.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 2nd 2008, 06:47 PM
  2. Help: Test tomorrow!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 11th 2008, 05:50 PM
  3. Please i need help i have a test tomorrow
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 7th 2008, 08:02 PM
  4. Help w/ Problems for Test Tomorrow
    Posted in the Calculus Forum
    Replies: 11
    Last Post: July 4th 2006, 09:24 PM
  5. plz help!! test tomorrow!
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 11th 2006, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum