Help with Trig problem

**rokr32** · February 7th 2011, 02:11 PM

Uploaded with ImageShack.us

please help me guys. im stuck.

**pickslides** · February 7th 2011, 02:20 PM

Try using this ratio,

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

You now have one equation with one variable.

**rokr32** · February 7th 2011, 02:21 PM

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

**pickslides** · February 7th 2011, 02:33 PM

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

After this apply the quadratic formula.

**rokr32** · February 7th 2011, 02:34 PM

thanks

**Soroban** · February 7th 2011, 02:53 PM

Hello, rokr32!

There are two solutions . . .

$\text{Find }y.$

Code:


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

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

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

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

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

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

Thread: Help with Trig problem

LinkBack

Thread Tools

Display

Help with Trig problem

Similar Math Help Forum Discussions

Trig word problem - solving a trig equation.

Weird Trig Angle of Elevation problem + Azimuth Word Problem

Trig problem

Trig problem...

another trig problem

Search Tags