Results 1 to 6 of 6

Math Help - Help with Trig problem

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    5

    Help with Trig problem



    Uploaded with ImageShack.us


    please help me guys. im stuck.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Try using this ratio,

    \displaystyle \frac{2}{\sqrt{3}} = \frac{y^2+1}{2y}

    You now have one equation with one variable.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    5
    yea but then i get stuck @ 4y=root3 (y^2+1) and don't know how to solve from there.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Expand the RHS, then group all terms onto the one side.

    After this apply the quadratic formula.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    5
    thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, rokr32!

    There are two solutions . . .


    \text{Find }y.
    Code:
    
          *
          *60*
          *     *   y^2+1
          *        *
          *           *
          *           30 *
          *  *  *  *  *  *  *
                  2y

    \text{We have: }\;\cos30^o \:=\:\dfrac{2y}{y^2+1} \quad\Rightarrow\quad \dfrac{\sqrt{3}}{2} \:=\:\dfrac{2y}{y^2+1}

    . . \sqrt{3}\,y^2 + \sqrt{3} \:=\:4y \quad\Rightarrow\quad \sqrt{3}\,y^2 - 4y + \sqrt{3} \:=\:0


    \text{Quadratic Formula: }\;y \;=\;\dfrac{4 \pm\sqrt{4^2 - 4(\sqrt{3})(\sqrt{3})}}{2\sqrt{3}}

    . . . . . . . . . . . . . . . y \;=\;\dfrac{4 \pm 2}{2\sqrt{3}} \;=\;\dfrac{2 \pm1}{\sqrt{3}}


    \text{Therefore: }\;y \;=\;\begin{Bmatrix}\dfrac{2+1}{\sqrt{3}} &=& \dfrac{3}{\sqrt{3}} &=& \sqrt{3} \\ \\[-3mm]<br />
\dfrac{2 - 1}{\sqrt{3}} &=&  \dfrac{1}{\sqrt{3}} &=& \dfrac{\sqrt{3}}{3} \end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig word problem - solving a trig equation.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 14th 2011, 07:07 AM
  2. Replies: 3
    Last Post: January 2nd 2011, 08:20 PM
  3. Trig problem
    Posted in the Trigonometry Forum
    Replies: 13
    Last Post: April 21st 2009, 03:51 PM
  4. Trig problem...
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 15th 2009, 11:46 AM
  5. another trig problem
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 24th 2008, 10:16 AM

Search Tags


/mathhelpforum @mathhelpforum