# Help with Trig problem

• Feb 7th 2011, 02:11 PM
rokr32
Help with Trig problem
http://img819.imageshack.us/img819/622/semttulogz.png

• Feb 7th 2011, 02:20 PM
pickslides
Try using this ratio,

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

You now have one equation with one variable.
• Feb 7th 2011, 02:21 PM
rokr32
yea but then i get stuck @ 4y=root3 (y^2+1) and don't know how to solve from there.
• Feb 7th 2011, 02:33 PM
pickslides
Expand the RHS, then group all terms onto the one side.

After this apply the quadratic formula.
• Feb 7th 2011, 02:34 PM
rokr32
thanks :D
• Feb 7th 2011, 02:53 PM
Soroban
Hello, rokr32!

There are two solutions . . .

Quote:

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

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

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

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

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

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

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