# Thread: Sector Problem

1. Is there a way other than guess and check to figure it out?

2. What kind of triangle is $\triangle MON$?

3. Hi Ilikebugs,

Yes, you can solve this systematically. Let $X$ be the center of the circle which passes through points $M$, $O$, and $P$. Show that $\triangle MXO\cong \triangle OXN$. Argue that $m\angle XON = 30º$. Draw an altitude of $\triangle XON$ from vertex $X$ to side $ON$. Say $Y$ is the point of intersection of the altidtude with side $ON$. Then consider the right triangle $\triangle OXY$ and solve the hypotenuse $OX$, which is the radius you want.

4. Let's begin by constructing the line segment $\overline{MN}$. since $\overline{OM}=\overline{ON}=R$, we know $\triangle MON$ is isosceles, and so we can label $\angle OMN=\angle ONM=\theta$:

\begin{tikzpicture}
\draw [blue,thick,domain=60:120] plot ({6*cos(\x)}, {6*sin(\x)});
\draw [blue,thick](0,0) -- ++(60:6cm)
(0,0) -- ++(120:6cm);
\draw[blue,thick] (-3,5.196) -- (3,5.196);
\node[below=5pt of {(0,0)}] {\large O};
\node[left=5pt of {(-3,5.196)}] {\large M};
\node[right=5pt of {(3,5.196)}] {\large N};
\node[above=15pt of {(0,0)}] {\large 60$^{\circ}$};
\node[left=5pt of {(-1.5,2.598)}] {\large $R$};
\node[right=5pt of {(1.5,2.598)}] {\large $R$};
\node[right=8pt of {(-3,5.196)},yshift=-8pt] {\large $\theta$};
\node[left=8pt of {(3,5.196)},yshift=-8pt] {\large $\theta$};
\end{tikzpicture}

Given that the sum of the interior angles in a triangle is $180^{\circ}$, we may state:

$\displaystyle 2\theta+60^{\circ}=180^{\circ}$

Solving this, we find:

$\displaystyle \theta=60^{\circ}$

And so we know $\triangle MON$ is equilateral:

\begin{tikzpicture}
\draw [blue,thick,domain=60:120] plot ({6*cos(\x)}, {6*sin(\x)});
\draw [blue,thick](0,0) -- ++(60:6cm)
(0,0) -- ++(120:6cm);
\draw[blue,thick] (-3,5.196) -- (3,5.196);
\node[below=5pt of {(0,0)}] {\large O};
\node[left=5pt of {(-3,5.196)}] {\large M};
\node[right=5pt of {(3,5.196)}] {\large N};
\node[left=5pt of {(-1.5,2.598)}] {\large $R$};
\node[right=5pt of {(1.5,2.598)}] {\large $R$};
\node[above=5pt of {(0,5.196)}] {\large $R$};
\end{tikzpicture}

Next, let's decompose $\triangle MON$ into 3 congruent isosceles triangles sharing a central common vertex:

\begin{tikzpicture}
\draw [blue,thick,domain=60:120] plot ({6*cos(\x)}, {6*sin(\x)});
\draw [blue,thick](0,0) -- ++(60:6cm)
(0,0) -- ++(120:6cm);
\draw[blue,thick] (-3,5.196) -- (3,5.196);
\draw[red,thick] (0,0) -- (0,3.464);
\draw[red,thick] (-3,5.196) -- (0,3.464);
\draw[red,thick] (3,5.196) -- (0,3.464);
\node[below=5pt of {(0,0)}] {\large O};
\node[left=5pt of {(-3,5.196)}] {\large M};
\node[right=5pt of {(3,5.196)}] {\large N};
\node[left=5pt of {(-1.5,2.598)}] {\large $R$};
\node[right=5pt of {(1.5,2.598)}] {\large $R$};
\node[above=5pt of {(0,5.196)}] {\large $R$};
\node[right=3pt of {(0,1.732)}] {\large $r$};
\node[above=5pt of {(-1.5,3.464)}] {\large $r$};
\node[above=5pt of {(1.5,3.464)}] {\large $r$};
\end{tikzpicture}

Now, using the formula for the area of a triangle, we may state:

$\displaystyle 3\cdot\frac{1}{2}Rr\sin\left(30^{\circ}\right)=\frac{1}{2}R^2\sin\left(60^{\circ}\right)$

Can you now solve for $r$?

5. Thread Author
I'm not sure

6. Well, what are $\sin\left(30^{\circ}\right)$ and $\sin\left(60^{\circ}\right)$?

7. Thread Author
0.5 and 0.86602540378

8. Originally Posted by Ilikebugs
0.5 and 0.86602540378
Hint: What are the exact values...

-Dan

9. Thread Author
1/2 and 43301270189/50000000000?

10. Originally Posted by Ilikebugs
1/2 and 43301270189/50000000000?
Put away the calculator! This will help in the long run... Memorize the sine and cosine functions for the angles 0, 30, 45, 60, and 90 degrees. It will save you a lot of time and grief.

sin(30) = 1/2 and sin(60) = sqrt(3) / 2.

Put these into your equation for r. Can you solve it now?

-Dan

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