### Instruction: A sector with acute central angle $\theta$ is cut from a circle of radius 6. The radius of the circle circumscribed about the sector is
$\textbf{(A)}\ 3\cos\theta \qquad \textbf{(B)}\ 3\sec\theta \qquad \textbf{(C)}\ 3 \cos \frac12 \theta \qquad \textbf{(D)}\ 3 \sec \frac12 \theta \qquad \textbf{(E)}\ 3$.### Response: To find the radius of the circumscribed circle, I need to find the distance from the center of the sector to the midpoint of the arc that forms the sector.
I can draw a perpendicular bisector of the arc and form a right triangle with the center of the sector, the midpoint of the arc, and one of the endpoints of the arc.
The hypotenuse of this right triangle is the radius of the circumscribed circle, which I'll call $r$.
The adjacent side of this right triangle is half of the radius of the original circle, which is 3.
The angle opposite the adjacent side is half of the central angle $\theta$.
Using the definition of cosine, I can write an equation relating $r$, 3, and $\frac12 \theta$.
$\cos \frac12 \theta = \frac{3}{r}$
Solving for $r$, I get $r = \frac{3}{\cos \frac12 \theta}$
This matches choice D, so that is the answer.