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If the point (x,square root 3/3) is on the unit circle, what is x?
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recall the Unit circle has a radius of " 1 " thus \(\bf \begin{array}{llll} (x&,&\frac{\sqrt{3}}{3})\\ \uparrow &&\uparrow \\ x&&y\\ \downarrow &&\downarrow \\ a&&b \end{array} \\ \quad \\ radius=1=c^2=a^2-b^2\implies c=\sqrt{a^2-b^2}\implies 1=\sqrt{x^2-\left(\frac{\sqrt{3}}{3}\right)^2}\)
Assume the center is (0,0). Unit circle has radius 1. now substituting it in the equation x^2+y^2=r^2 gives the following x^2=(1)^2-(sqrt 3/3)2 x^2= 1 - (3/9) x^2 = 1-(1/3) x^2 = 2/3 x = sqrt(2/3) Is my answer correct???
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