# Thread: Help with Trig problem

1. ## Help with Trig problem

2. Try using this ratio,

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

You now have one equation with one variable.

3. yea but then i get stuck @ 4y=root3 (y^2+1) and don't know how to solve from there.

4. Expand the RHS, then group all terms onto the one side.

After this apply the quadratic formula.

5. thanks

6. Hello, rokr32!

There are two solutions . . .

$\displaystyle \text{Find }y.$
Code:

*
*60*
*     *   y^2+1
*        *
*           *
*           30 *
*  *  *  *  *  *  *
2y

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

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

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

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

$\displaystyle \text{Therefore: }\;y \;=\;\begin{Bmatrix}\dfrac{2+1}{\sqrt{3}} &=& \dfrac{3}{\sqrt{3}} &=& \sqrt{3} \\ \\[-3mm] \dfrac{2 - 1}{\sqrt{3}} &=& \dfrac{1}{\sqrt{3}} &=& \dfrac{\sqrt{3}}{3} \end{Bmatrix}$