Pessimist Singularitarian

#13
December 3rd, 2016,
22:20
Originally Posted by

**Ilikebugs**
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:

$ \displaystyle x^2+\left(\frac{1}{2}\right)^2=1^2$

$ \displaystyle x^2+\frac{1}{4}=1$

$ \displaystyle x^2=1-\frac{1}{4}=\frac{3}{4}$

Since $x$ represents a linear measure, we take the positive root:

$ \displaystyle 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:

$ \displaystyle \sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\dfrac{1}{2}}{1}=\frac{1}{2}$

$ \displaystyle \sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\dfrac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}$

So, returning to the equation:

$ \displaystyle 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 $r$.