Ilikebugs said:
The Pythagorrean theorem tells us that the square of the hypotenuse (the longest side in a right triangle, the one opposite the $90^{\circ}$ angle) is equal to the sum of the squares of the other two sides (the two shorter sides are called "legs"). So, using that with the triangle I gave in post #11, stated mathematically, this is:
$$x^2+\left(\frac{1}{2}\right)^2=1^2$$
$$x^2+\frac{1}{4}=1$$
$$x^2=1-\frac{1}{4}=\frac{3}{4}$$
Since $x$ represents a linear measure, we take the positive root:
$$x=\frac{\sqrt{3}}{2}$$
Okay, so we now have:
\begin{tikzpicture}
\draw[blue,thick] (0,0) -- (5.196,0);
\draw[blue,thick] (5.196,0) -- (5.196,3);
\draw[blue,thick] (5.196,3) -- (0,0);
\draw[gray,thin] (5.196,0.25) -- (4.946,0.25);
\draw[gray,thin] (4.946,0.25) -- (4.946,0);
\node[below=5pt of {(2.598,0)}] {\large $\dfrac{\sqrt{3}}{2}$};
\node[right=5pt of {(5.196,1.5)}] {\large $\dfrac{1}{2}$};
\node[above=5pt of {(2.598,1.5)}] {\large $1$};
\node[below=10pt of {(5.196,3)},xshift=-10pt] {\large $60^{\circ}$};
\node[right=18pt of {(0,0)},yshift=7pt] {\large $30^{\circ}$};
\end{tikzpicture}
Now, the sine of an angle in a right triangle is defined as the ratio of the side opposite the angle to the hypotenuse. Hence we have:
$$\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\dfrac{1}{2}}{1}=\frac{1}{2}$$
$$\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\dfrac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}$$
So, returning to the equation:
$$3\cdot\frac{1}{2}Rr\sin\left(30^{\circ}\right)=\frac{1}{2}R^2\sin\left(60^{\circ}\right)$$
All you need to do now is plug in the values for the sine functions, and solve for $r$. :D