9.2 Define General Angles and
Use Radian Measure
What are angles in standard position?
What is radian measure?
Angles in Standard Position
In a coordinate plane, an angle
90°
can be formed by fixing one ray
called the initial side and
rotating the other ray called the 180°
vertex
terminal side, about the vertex.
An angle is in standard position
if its vertex is at the origin and
270°
its initial side lies on the positive
x-axis.
The measure of an angle is positive if the rotation of its
terminal side is counterclockwise and negative if the
rotation is clockwise.
The terminal side of an angle can make more than one
complete rotation.
0°
Draw an angle with the given measure in standard
position.
a. 240º
SOLUTION
a.
Because 240º is 60º more
than 180º, the terminal side
is 60º counterclockwise past
the negative x-axis.
Draw an angle with the given measure in standard
position.
b.
500º
SOLUTION
b.
Because 500º is 140º more
than 360º, the terminal side
makes one whole revolution
counterclockwise plus 140º
more.
Draw an angle with the given measure in standard
position.
c.
–50º
SOLUTION
c.
Because –50º is negative, the
terminal side is 50º clockwise
from the positive x-axis.
Coterminal Angles
Coterminal angles are angles whose terminal sides
coincide.
An angle coterminal with a given angle can be found
by adding or subtracting multiples of 360°
The angles 500° and 140°
are coterminal because their
terminal sides coincide.
Find one positive angle and one negative angle that are
coterminal with (a) –45º
SOLUTION
There are many such angles, depending on what
multiple of 360º is added or subtracted.
a.
–45º + 360º = 315º
–45º – 360º = – 405º
Find one positive angle and one negative angle that are
coterminal with (b) 395º.
b. 395º – 360º = 35º
395º – 2(360º) = –325º
Draw an angle with the given measure in standard
position. Then find one positive coterminal angle and
one negative coterminal angle.
1.
65°
65º + 360º = 425º
65º – 360º = –295º
2.
230°
230º + 360º = 590º
230º – 360º = –130º
3.
300°
300º + 360º = 660º
300º –
360º
4.
= –60º
740°
740º –
2(360º)
740º –
3(360º)
= 20º
= –340º
Radian Measure
One radian is the measure of an
angle in standard position whose
terminal side intercepts an arc of
length r.
Because the circumference of a
circle is 2𝜋𝑟, there are 2𝜋 radians
in a full circle.
Degree measure and radian
measure are related by the
equation 360° = 2𝜋 radians or
180°=𝜋 radians.
Converting Between Degrees and Radians
Degrees to radians
Radians to degrees
Multiply degree measure
Multiply radian
180°
measure by
by
𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
180°
𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
Degree and Radian Measures of Special Angles
The diagram shows equivalent
degree and radian measures for
special angles for 0° to 360°
( 0 radians to 2𝜋 radians).
It will be helpful to memorize
the equivalent degree and
radian measures of special
angles in the 1st quadrant and
𝜋
for 90° = radians.
2
All other special angles are
multiples of these angles.
Convert (a) 125º to radians and (b) – π radians to
12
degrees.
( π radians )
a. 125º = 125º
180º
=
25π
36
radians
(
)(
)
π
180º
π
–
radians
b. –
=
12
π radians
12
= –15º
Convert the degree measure to radians or the radian
measure to degrees.
5. 135°
135º = 135º
=
(
3π
4
π radians )
180º
radians
(
–50° = –50°
6. –50°
=
7.
5π
4
(
5π
4 =
– 5π
18
π radians )
180º
radians
)(
)
5π
180º
radians
π radians
4
= 225º
8.
π
10
π
10
=
(
= 18º
π
10
)(
radians
)
180º
π radians
Sectors of Circles
A sector is a region of a circle
that is bounded by two radii
and an arc of the circle.
The central angle 𝜃 of a sector
is the angle formed by the two
radii.
Arc Length and Area of a Sector
The arc length s and
area A of a sector with
radius r and central
angle 𝜃 (measures in
radians) are as follows:
Arc length: 𝒔 = 𝒓𝜽
Area: 𝑨 =
𝟏 𝟐
𝒓 𝜽
𝟐
Softball A softball field forms a sector with the dimensions
shown. Find the length of the outfield fence and the
area of the field.
SOLUTION
STEP 1 Convert the measure of
the central angle to
radians.
π
π radians )
90º = 90º
radians
=
2
180º
STEP 2 Find the arc length and the area of the sector.
Arc length: s = r θ= 180 ( π )= 90π ≈ 283 feet
2
1
Area: A = 1 r2θ =
(180)2 ( π ) = 8100π ≈ 25,400 ft2
2
2
2
(
Arc length: s = r θ= 180 ( π )= 90π ≈ 283 feet
2
1
Area: A = 1 r2θ =
(180)2 ( π ) = 8100π ≈ 25,400 ft2
2
2
2
ANSWER
The length of the outfield fence is about 283 feet.
The area of the field is about 25,400 square feet.
9.2 Assignment
Page 566, 3-37 odd