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 is the midpoint of , we can actually cut that in half: and .

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 , you’re creating two right triangles: and .

Let’s just look at triangle . At this point, we can certainly use the Pythagorean Theorem to solve, but the presence of the 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 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: . We know we’re right because the long leg in a30°-60°-90° is the short leg times , and sure enough: . So we have our answer. .

Gettin’ Pythaggy wit it: 