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: