Practice question for Circles, Radians, and a Little More Trigonometry, #5, p. 272, 4th Ed.

Here’s the question:

So the first thing you’ll want to do is draw in a strategically placed radius (or two) and label the lengths you know.

Because we know that M is the midpoint of \overline{AB}, we can actually cut that 5\sqrt{3} in half: AM=\dfrac{5\sqrt{3}}{2} andBM=\dfrac{5\sqrt{3}}{2}.

We also know that we can always draw a perpendicular bisector from the center of a circle to the midpoint of a chord, so you know that by drawing in \overline{OM}, you’re creating two right triangles: AOM and BOM.

Let’s just look at triangle AOM.

At this point, we can certainly use the Pythagorean Theorem to solve, but the presence of the \sqrt{3} should have activated our Spidey-senses from the beginning: this is a 30°-60°-90° triangle! So we can also just see how the numbers we know fit into the known 1:\sqrt{3}:2 ratio. The hypotenuse in a 30°-60°-90° is double the length of the short leg. Our hypotenuse is 5, so the short leg we seek must be half that: \dfrac{5}{2}. We know we’re right because the long leg in a30°-60°-90° is the short leg times \sqrt{3}, and sure enough: \dfrac{5}{2}\times\sqrt{3}=\dfrac{5\sqrt{3}}{2}. So we have our answer. OM=\dfrac{5}{2}.

Gettin’ Pythaggy wit it:

    \begin{align*}OM^2+\left(\dfrac{5\sqrt{3}}{2}\right)^2&=5^2\\\\OM^2+\dfrac{25(3)}{4}&=25\\\\OM^2+\dfrac{75}{4}&=\dfrac{100}{4}\\\\OM^2&=\dfrac{25}{4}\\\\OM&=\dfrac{5}{2}\end{align*}