Wren stands on the chalk circle with a protractor. "Comet, hold the string from the pin to this spoke mark."
Comet holds it. Wren marks two points on the rim and reads the central angle. "Eighty degrees."
"Now I stand over here," he says, walking to the far side of the rim. He sights both marks. "Forty."
"Half," Comet says. "Is that a coincidence? What can we make of it?"
Wren moves to a new spot on the far rim and measures again. "Still forty. Every spot on this side."
Nova projects the circle with the two angles drawn. "Would you like a hint? Both angles cut off the same arc."
"Center angle equals the arc. Rim angle is half the arc," Wren says. "Our designer, test it on the brace too."
A central angle has its vertex at the center. Its measure equals the measure of its arc.
An inscribed angle has its vertex on the circle and two chords as sides. Its measure is half its arc.
So every inscribed angle on the same arc is equal, and each is half the central angle on that arc.
Thursday is Proof Day, when you see why. Today you use it.
| Arc | Central angle | Inscribed angle |
|---|---|---|
| 60° | 60° | 30° |
| 80° | 80° | 40° |
| 120° | 120° | 60° |
| 180° | 180° | 90° |
A brace lies across the turntable 1.8 meters from the center. The radius is 3 meters. How long is the brace?
The radius perpendicular to the brace bisects it. Half the brace, the 1.8 meter distance and a 3 meter radius make a right triangle.
Half the brace = square root of 3² - 1.8² = square root of 5.76 = 2.4. The brace is 2 × 2.4 = 4.8 meters.
Why does this work? The two radii to the brace ends are equal, so the perpendicular from the center is a line of symmetry.
| Statement | True or false? |
|---|---|
| Every inscribed angle on the same arc has the same measure. | ? |
| An inscribed angle is twice its arc. | ? |
| A tangent is perpendicular to the radius at the touching point. | ? |
| A radius perpendicular to a chord bisects the chord. | ? |
| 40 × 2 = 80 | ? |
Strong work, designer. Tomorrow is Shop Lab: you measure inscribed angles yourself and fit a square corner into a semicircle.