DATE
PERIOD
10-2 NTM Measuring Angles and Arcs
Angles and Arcs A central angle is an angle whose vertex is
at the center of a circle and whose sides are radii. A central angle
separates a circle into two arcs, a major arc and a minor arc.
GF is a minor arc.
Here are some properties of central angles and arcs.
Z GEF is a central angle.
CHG is a major arc.
• The sum of the measures of the central angles of a circle
with no interior points in common is 360.
• The measure of a minor arc is less than 180 and equal to the
measure of its central angle.
• The measure of a major arc is 360 minus the measure of the
minor arc.
\
mZHEC + mZCEF + mZFEG + mZGEH = 360
mCF = mACEF
•
nriCGF - 360 - mCF
• The measure of a semicircle is 180.
• Two minor arcs are congruent i f and only i f their
corresponding central angles are congruent.
CF^FG
• The measure of an arc formed by two adjacent arcs is the sum
mCF + mFG = mCG
of the measures of the two arcs. (Arc Addition Postulate)
il and only if Z GEF s
ZFEG.
Example: AC is a diameter of QR. Find mAB and mACB. ^
-ARB is a central angle and m A ARB = 42, so mAB
Thus mACB = 360 - ^^or 3 / 6
>•
X..
....
. • .
,™
Exercises
Find the value of JC.
1.
pqo_
.
X
120
3UD - Zio-
I'So
iSo\
BD and AC are diameters of QO. Identify each arc as a major arc, minor arc, or semicircle of the circle.
Then find its measure.
3.mBA=l^
4.mBC
5. mCD
6. mACB =
l.mBCD
-V^'
=/SO-W=^2£l
JCO'HH
=(3^fj
DATE
NAME
PERIOD
\ r c Length An arc is part of a circle and its length is a part of the circumference of the circle,
f he length of arc I can be found using the following equation:
f =•
360
cf t ^
<2/^vc5 ace A-
-/•AA.-/
Example: Find the length of AR. Round to the nearest hundredth.
The length of arc AB, can be found using the following equation: AB =
360
27cr
Arc Length Equation
AB=—-2Tzr
360
Substitution
AB='^-2.(8)
AB ~ 18.85 in.
...
Use a calculator.
Exercises
Use 0 O to find the length of each arc. Round to the nearest hundredth.
AA
D
3Q0
0£
BE i f the radius is 2 meters
120°
B1
2. DEA i f the diameter is 7 inches D E A
3. B C i f 5 £ = 2 4 f e e t
5?t:^,rrci
"3^0'
j§C
jp^Fi
=
iZ S2
V
ihj
^-1T<Z^
4. CSA i f T)0 = 3 millimeters
Use 0 P to find the length of each arc. Round to the nearest hundredth.
5. RT, i f M r = 7 y a r d s
6. MR, if PR = 13 feet
• TT m
r^R -
3
7. M5T, i f MP = 2 inches
77
Co
. - ^ . ^ ' T .
^ ? T = ^ . ^ ' ^ < - ^ f
A/
/A?r
(
T MBS, i f PS = 10 centimeters
^^.&S
R
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