Finding the Angular Size of a Minor Arc
The question is as follows:
A 20-inch chord is drawn in a circle with a 12-inch radius. What is the angular size of the minor arc of the chord? What is the length of the arc, to the nearest tenth of an inch?
The two points of the chord, I labeled, is $A$ and $B$. The point in which the radius goes through the chord and touches the outside of the circle is labeled as point $D$, and the intersection point of the radius and the chord was $E$. I know that $E$ perpendicularly bisects the chord $AB$.
After that part, I was not able to make much progress. Any help will be greatly appreciated.
$\endgroup$1 Answer
$\begingroup$There is a theorem which states that $AB=2r\sin\beta$
Thus $\sin\beta=\dfrac{20}{24}=\dfrac{5}{6}$
$\beta=\sin^{-1}\left(\dfrac{5}{6}\right)\approx 56.44°$
$\widehat{AOB}=2\beta\approx 112.885°$
$ \stackrel{\frown}{AB}$ is proportional to circumference $C=2\pi r\approx 150.796$
$\dfrac{\stackrel{\frown}{AB}}{150.796}=\dfrac{112.885°}{360°}$
$\stackrel{\frown}{AB}\approx 47.3$inch
