Results 1 to 9 of 9

Math Help - Coterminal Angles

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    15

    Coterminal Angles

    Hello

    someone help me with these problems please, ASAP

    Which angle is coterminal with 38 degree 50' ?
    a)-398 degrees 50'
    b)-321 degrees 50'
    c) 321 degrees 10'
    d) 398 degrees 10'
    c) none of these


    ---------------------------------

    Find the exact values of 0 degrees and 360 degrees that make the following equation a true statement:
    tan theta= -square root of 3 divided by3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    A and B are coterminal angles if A-B=360.

    Thus it has to be 321 or -321. According to the definition, it'll be -321 degrees 50', since 38-(-321) ~ 360

    The problem is that when we are talking about degrees ', the ' corresponds to minutes of degrees. 1 degree makes 60' (1 degree is like an hour). So it can't be -32150', since 50'+50' \neq 1=60'.

    So it's none of the solutions.

    ~~~~~~~~~~~~

    \tan(\theta)=-\frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=-\frac{1}{\sqrt{3}}

    \tan(\theta)={\color{red}-\frac 12} \cdot \frac{2}{\sqrt{3}}=\frac 12 \cdot {\color{red}-\frac{2}{\sqrt{3}}}

    Can you continue ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2007
    Posts
    15
    Quote Originally Posted by Moo View Post
    Hello,

    A and B are coterminal angles if A-B=360.

    Thus it has to be 321 or -321. According to the definition, it'll be -321 degrees 50', since 38-(-321) ~ 360

    The problem is that when we are talking about degrees ', the ' corresponds to minutes of degrees. 1 degree makes 60' (1 degree is like an hour). So it can't be -32150', since 50'+50' \neq 1=60'.

    So it's none of the solutions.

    ~~~~~~~~~~~~

    \tan(\theta)=-\frac{\sqrt{3}}{3}=-\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=-\frac{1}{\sqrt{3}}

    \tan(\theta)={\color{red}-\frac 12} \cdot \frac{2}{\sqrt{3}}=\frac 12 \cdot {\color{red}-\frac{2}{\sqrt{3}}}

    Can you continue ?
    umm no i dont get the second problem
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1
    Quote Originally Posted by subzero06 View Post
    umm no i dont get the second problem
    Look at the result that Moo gave you:

    \tan(\theta)=\frac12\left(-\frac{2}{\sqrt{3}}\right)

    This is really the same as saying

    \displaystyle{\tan(\theta)=\frac{-\frac12}{\frac{\sqrt{3}}2}}

    You should know, from the basic trigonometric identities, that \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}, so what value for \theta gives a sine of -\frac12 and a cosine of \frac{\sqrt{3}}2?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2007
    Posts
    15
    Quote Originally Posted by Reckoner View Post
    Look at the result that Moo gave you:

    \tan(\theta)=\frac12\left(-\frac{2}{\sqrt{3}}\right)

    This is really the same as saying

    \displaystyle{\tan(\theta)=\frac{-\frac12}{\frac{\sqrt{3}}2}}

    You should know, from the basic trigonometric identities, that \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}, so what value for \theta gives a sine of -\frac12 and a cosine of \frac{\sqrt{3}}2?
    Ok,
    so the values for cosine is 30 degrees and sine -30 because is -(1/2)???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by subzero06 View Post
    Ok,
    so the values for cosine is 30 degrees and sine -30 because is -(1/2)???
    Hm...
    Let's take the first situation, OK ?

    \displaystyle{\tan(\theta)=\frac{-\frac12}{\frac{\sqrt{3}}2}}

    So it means that \sin(\theta)=-\frac 12 and \cos(\theta)=\frac{\sqrt{3}}{2}

    What is the angle for which these two conditions are available ?



    The other one is :

    \displaystyle{\tan(\theta)=\frac{\frac12}{-\frac{\sqrt{3}}2}}

    So it means that \sin(\theta)=\frac 12 and \cos(\theta)=-\frac{\sqrt{3}}{2}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1
    Quote Originally Posted by Moo View Post
    The other one is :
    There are indeed two possible values on the given interval. That must have slipped my mind; thanks.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Reckoner View Post
    There are indeed two possible values on the given interval. That must have slipped my mind; thanks.
    Actually, that's why I put two equalities with two red things ^^
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Lol, if you want the answers, I've just found it in google :

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coterminal Angles Work How Again?
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: March 14th 2011, 08:16 AM
  2. Coterminal Angles
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: June 4th 2009, 04:08 AM
  3. Understanding coterminal angles
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 7th 2008, 08:31 AM
  4. Quick question: Coterminal angles
    Posted in the Trigonometry Forum
    Replies: 8
    Last Post: October 17th 2007, 07:27 AM
  5. coterminal angles
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 2nd 2007, 05:03 PM

Search Tags


/mathhelpforum @mathhelpforum