# Thread: Sector Problem

1. Imagine we've take an equilateral triangle, bisected it and oriented one of the halves like so:

\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 $x$};
\node[right=5pt of {(5.196,1.5)}] {\large $\dfrac{1}{2}$};
\node[above=5pt of {(2.598,1.5)}] {\large $1$};
\end{tikzpicture}

Can you use the Pythagorean theorem to find $x$? Once you know $x$, then you can find the sine and cosine of 30 and 60 degrees...

I cant

3. Originally Posted by Ilikebugs
I cant
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 $r$.

I dont know how to solve for r

5. Originally Posted by Ilikebugs
I dont know how to solve for r
Well, as I said, the first step is to substitute for the sine functions:

$\displaystyle 3\cdot\frac{1}{2}Rr\cdot\frac{1}{2}=\frac{1}{2}R^2\cdot\frac{\sqrt{3}}{2}$

Next, we may multiply through by $\displaystyle \frac{4}{\sqrt{3}R}$ to obtain:

$\displaystyle \sqrt{3}r=R$

Hence:

$\displaystyle r=\frac{R}{\sqrt{3}}$

Now you need to plug in the given $R=6\text{ cm}$ and simplify to finish the problem.

r=2√3?

7. Originally Posted by Ilikebugs
r=2√3?
$r=2\sqrt{3}\text{ cm}$

8. Having found that the triangle is equilateral one can determine its height (altitude) with the Pythagorean theorem:

$$\sqrt{6^2-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt3$$

There is a point on a median (the line that emanates from a vertex and terminates on the midpoint of the opposite side) of the triangle, called the centroid, that is concurrent with the circumcenter in the case of an equilateral triangle and one of its properties is that it divides a median (in this case an altitude, which is concurrent with a median in the case of an equilateral triangle) in the ratio 2:1. Hence the radius we seek (the radius of the circumcircle of $\triangle{MON}$) is

$$\frac23\cdot3\sqrt3=2\sqrt3$$

The three medians of a triangle intersect at the centroid.

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