Angles
Angles
Angles
Angles
If two rays.
1 and L
L
, and L2 , emanate
9
L
from If
thefrom
samethe
pointiii
plane,
1
andpoint
, inthethe
9
L
from the
If two
same
thensame poin
If two
rays.rays,
1 and1 L
L
,
9
from two
the rays.
same L
pointiii
the plane,
plane,
then their
angle is aismeasurement
of how
close
two two
lines
theangle
are to
other.
their
is alines
measurement
ofother.
how
close the two line
ofthen
how
close
the
lines
areeach
each
other.
thentheir
theirangle
angle is aa measurement
measurement of
how
close
two
the
are
totoeach
To findTo
thefind
angle
between
1
L
and
,
2
L
move
them
to
the
point
(0,
in
the
0)
To move
find
the
angle
between
1
L
,
2
move
the angle
between
L1 and
L
them
to point
the point
0)thein
the them to the
2 , move
To find the angle
between
1 and
L
,
2
L
them
to the
(0, and
inL
0)(0,
so
that
plane, and
the
ray
1
L
matches
with
positive
half
up
the
of
the
x-axis.
so
plane,
and
ray
1half
with the positive
plane,
so that
the L
raymatches
L1 matches
up with
thethe
positive
the up
x-axis.
so that
plane,
and and
the ray
1
positive
halfL
up with
thethat
ofmatches
theofx-axis.
After being
moved moved
to the point
will Lintersect
the
ray
2 moved
L
the unit
being
After
theintersect
point
the ray L
2 will
to point
the (0,0),
point
(0,the
0),
the
ray
the unit
beingbeing
AfterAfter
moved to the
will2 towill
intersect
(0,0),
ray
2
L
the (0,0),
unit
1
circle C’
in exactly
one point.
in exactly
point. circle C’ in exactly one point.
circlecircle
one one
point.
C’ inC exactly
betweenbetween
The angle
L, and LL
21 and
is the 2number
with
0 between
[0, 2ir)
thewith
property
and
2 property
L
is
isThe
theangle
number
✓ 22ir)
[0,L,with
2⇡)
thethe
property
number 0 [0, 2ir)
angleangle
between L, and
The The
2 isLthe
L
number
0 [0,
the
1
that wind(0)
is
the
point
where
2
L
intersects
the
unit
circle,
C’.
wind(0)
theunit
point
where
wind(✓)
the point
where
L2 intersects
the
circle,
CL
. intersects the unit circ
2
that that
wind(0)
is theis point
where
2 that
L
intersects
the isunit
circle,
C’.
L
i
2
t
it
2
L
wiya(e)
wiya(e)
c’ c’
L
i
2
t
wiya(e
c’
ee
e
I—I I—I
179
179
196
I—
179
this by
the
between
two rays
rays
equals
&, then
then we
wedenote
usuallythis
denote
by
If the
between
two
equals
✓,equals
then
usually
by this
this we
gle between two
raysangle
equals
&,angle
thenbetween
werays
usually
denote
by
IfIf the
angle
two
&,
usually
denote
labeling
the
rays as
shown
labeling
the
rays asbelow:
as showii
showii below:
below:
e rays as showii
below:
labeling
the
rays
.
Examples.
Examples.
Examples.
• A right
angle
is an
angle
resulting
from
raysfrom
thatrays
intersect
perpenresulting
from
rays
that intersect
intersect perpen
perpen
right
angle
tht
angle resulting
resulting
from
rays
that
intersect
perpen
right angle is tht angle
that
AA right
angle
isistht
angle
as
the angle
x- as
and
y-axes
do.
The angle
between
positive
of the
dicularly,
the
and
y-axes
The
angle
between
thehalf
positive
halfofofthe
the
asthe
do.half
y-axes do.dicularly,
The
between
they-axes
positive
ofangle
the the
s the x- and dicularly,
The
between
the
positive
half
x-x- and
do.
⇡
the positive
half
of the
y-axis
is half
, because
positive
of the
thehalf
positive
halfofofthe
the
x-axis
and
the
positive
half
the
is ,,the
because
the
positive
of is
the
the positive x-axis
half of and
thex-axis
y-axis
is ,
because
the
positive
half
2y-axis
and
the
positive
half
ofofthe
y-axis
because
⇡
y-axis
the unit
at
the
point
(0, point
1),
and(0,1),
(0, 1)aild
= wind
To
the
point
(0,1),
aild
(0,1)2==.windQ).
windQ).
To
y-axis
intersects
the
unit
circle
the point
(0,1),circle
aild
(0,1)
= windQ).
To
sects the unit
circleintersects
aty-axis
(0,1)
To
intersects
the
unit
circle
atat the
⇡
right
angles
areangles
anglesare
of angles
.
summarize,
right
angles
are
right anglessummarize,
are angles
of
2angles
summarize,
right
ofof
.
.
.
(oi)
‘iia()
(oi)
(oi)
‘iia()
‘iia()
CC
C
positive
of the
x-axis
and
the
negative
half
of the
x-axis
i-axis
Thehalf
positive
half
the
i-axis
and the
the negative
negative
half
the i-axis
i-axis
he positive half of •theThe
i-axis
and
the
negative
half
of
the
The
positive
half
ofof the
i-axis
and
half
ofof the
are two rays
each
emanate
from
the from
point
(0,
0).
The
between
point
The
anglethe
betweenthe
the
arethat
two
rays that
that
each
emanate
fromthe
thethe
(0,angle
0). The
point
The
angle
between
s that each emanate
from
the rays
(0, each
0).
point
angle
between
are
two
emanate
(0,
0).
positiveray
ray
the
ray
is ⇡raybecause
negative
intersects
the
the
negative
because
theray
negative
rayintersects
intersects
the
positive
raynegative
and
ray
is and
thethe
negative
intersects
the
y and the negative
ir because
negative
isisirirthe
because
the
negative
ray
the
positive
ray
and
ray
circle=
in(—1,0).
the
point
wind(⇡)
( 1, 0).== (—1,0).
unit
circle
thepoint
point=
wind(7r)
(—1,0).
in the point unit
wind(7r)
unit
circle
inin the
wind(7r)
n(
rn(
rn(
197
I
pose a right triangle has angles , , and
If the length
the angle has length 1, then the length of the sides
and are and , respectively.
⇡ ⇡
⇡
Lemma (5). Suppose
triangle has
angles
andangles
length
Lemmaa right
(5). Suppose
a right
triangle
, ,
and
If the len
6 , 3 , has
2 . If the
⇡
of the side opposite
the
angle
has
length
1,
then
the
length
of
the
sides
of the side opposite
thep angle has length 1, then the length of the si
2
3
⇡
⇡
1
Proof:
with
the triangle
We begin
Proof: We
withand
the
opposite
the begin
angles
are
opposite
the3 triangle
angles
are
and ,
respectively.
6
2 andand
2 , respectively.
.
.
‘Z.
I-
I
2.
coyly.
with the triangle
2
Now triangle,
let’s drawflipped
a second
of the same triangle,
Now
let’s We
draw
a second
of the same
overcopy
the vertical
Proof:
begin
with copy
the
Proof:
We triangle
begin
with
the
triangle
‘Z.
‘Z.
coyly.
I-
coyly.
econd copy of the same triangle, flipped over the vertical
N
Now let’s draw aNow
second
samecopy
triangle,
over theflipped
vertical
let’scopy
drawofa the
second
of theflipped
same triangle,
181 over the verti
181
side.
N
181
N
198
181
Lemma (5). Suppose a right triangle has angles , , and
of the side opposite the angle has length 1, then the length
.
opposite the angles and j are and , respectively.
This new triangle is twice as big as the one we started with, and all of
its angles equal ⇡3 . It’s what’s called a regular triangle. All three sides of
a regular triangle have the same length, so all three sides of this particular
regular triangle have length 1.
Az
Proof: We begin with the triangle
Lemma
Suppose triangle
hasthe
angles
, ,side
andof the
If the length
a right triangle
The bottom
side of (5).
our original
is half of
bottom
1
of thesoside
opposite
angle has length 1, then the length of the sides
larger triangle,
it has
length the
2.
opposite the angles and are and , respectively.
.
A
/
1/ I
Az/
I
31
A
of the same
triangle,
Now
let’sside
drawof aour
second
copytriangle
Let’s call the length of the
third
original
a. What
we flipped over
have left to do is find what a equals, and for this we can use the Pythagorean
Theorem which says that a2 + ( 12 )2 = 12 , which simplifies to a2 + 14 = 1.
1
2
Subtracting Proof:
= 34 , with
and since
a is a positive number,
we can take
4 gives us
the triangle
Wea begin
q
2
the square root of the equation a =
3
4
to find that a =
3
4
=
p
p3
4
=
p
3
2 .
a”.i
car,’.
2.
Co
181
⌅
Now let’s draw a second 199
copy of the same triangle, flipped over the vertical
2
p
3
⇡
1
Lemma (6).
windQ)
(6). wind
Lemma
=
(6). windQ) =
3 =
2, 2
Proof: Let’s
draw aa ray
from (0,
Let’s draw
0)0) through
point wind ⇡ and aa
Proof:
ray
through the
the
from (0, 0) through
Let’s draw a ray
the from
point (0,
a pointwind()XNJrnd
3
wind()XNJrnd
⇡ to the i-axis that creates a riht angle
vertical line
line segment
segment from
from wind
wind()
vertical
to
the
x-axis
that
creates
a
right
angle
segment from wind() to the i-axis that creates
a riht angle
3
with
the
i-axis.
with the x-axis.
axis.
C
C
(6). wind()
Let’s draw a ray from (0,0) through the point wind()xNynd a
The
ray, the
thetovertical
vertical
line segment,
segment, and
andathe
the
i-axis
form aa right
right triangle.
triangle. The
ray,
line
x-axis
form
The
ehesegment
wind
that
riht
angle
vertical from
lineThe
segment,
andthe
the i-axis
i-axis
form creates
a right triangle.
The
⇡ equals 1, because wind()
⇡ is
distance
between
the
points
and
wind()
(0,
0)
distance
the points
(0,1,0)because
and wind
axis.
tween the points
and wind()
wind()
is 1, because wind 3 is
3 equals
(0, 0)between
equals
⇡ the right angle is ⇡ and
a
point
on
the
unit
circle.
The
angle
at
is
(0,
0)
a point
theatunit
at (0, is0) is and
the unit circle.
The on
angle
the angle
right
angle
3 , the right angle is 2 , and
(0, circle.
0) is The
⇡
⇡
⇡
the third
third angle must
must then equal
equal , because
because +
= r, and the sum of
+ +
+⇡ =
the
⇡, and the sum of
6
ngle must then
equal angle
because then
and the 3sum2 of
+ + = 6r,
the
three
angles
of
a
triangle
ir.
the three angles of a triangle
ngles of a triangle
ir. must always equal ⇡.
WInCU
WInCU
()
C
,
,
, ,
,
,
he vertical line segment, and the xaxis form a right triangle. The
(0.0’
tween the points (0, 0) and windQ) equals
(0.0’ 1, because wind() is
the unit circle. The angle at (0, 0) is
the right angle is
and
Now
Lemma
tells
that
the
length
of
the
other
sides of
of the
the triangle
triangle
(5) tells +
us that
two sides
Lemma
(5)
the
of
the
other
ngle
must
equal
the
sum
of two
because
= length
ir, and
+ two
ma (5)
tellsthen
that
the,plength
of the us
other
sides
of the
triangle
us Now
3
are 1 and
and
ngles of a triangle
yay)fEequa1
71.
ti
are
2
2 .
‘fri
,
(:(f
ti
‘fri
,
‘A
.1/-Jr
W’ncL =
Or
a
(:(f
%21
MojL.
Or
MojL.
“ir
(o.ó
.f
That
means
thatofthe
the
of wind
windQ)
equals 1 and
and the
the y-coordinate
y-coordinate
ma (5) tells us That
that the
of the ⇡triangle
length
the i-coordinate
other two sides
means
equals
2
ns that the i-coordinate
of that
windQ) x-coordinate
equals and of
the y-coordinate
p
p3
3 which is what we wanted to
1 J’
equals 3 . In
In other
other words,
words, wind
windQ)
(i ⇡ =
equals
which
=
2
3
2 ,wanted
2
In other words,
windQ)
which
is
what
we
to is what we wanted to
=
.
see.
see.
⌅
.
.
(, )
.
183
(f (, )
183
200
___________________
argument to
Lemma
(6)
to
a similar
argument
to the
the
one
from
Lemma
(6)show
to show
show
that
We We
can can
use use
a similar
argument
to the
one one
fromfrom
Lemma
(6) to
thatthat
p
3 1
⇡
wind()
wind
wind()
6 =
2 ,2 .
-
-
\
ff\
.
2
r-ç
r-ç
.
2
,),)
}4\7r
}4\7r
⇡ as shown
12 segments
each of
circle
be divided
into
12
segments
of length
length
as shown
The The
unit unit
circle
can can
be divided
into into
12 segments
eacheach
of length
as 6shown
below.
below.
(a”)
(a”)
4 4
1 1
I—-I—-2,2
2,2
frfr —I
—I
I—I—
‘2)‘2)
\\
(i,o)
(i,o)
If__ If__
“4.
“4.
//4i
//4i
(
(
—
2.) 2.)
—
-
cc
2, 2,
4 4
fi fi q”q”
T)T)
(o,H)
(o,H)
‘
‘
184
184
201
—
—
—
I
—
I
Exercises
Exercises
Exercises
Exercises
The
sum
ofangles
the
angles
oftriangle
a triangle
equal
⇡. That
That
is,c,,B,
↵, ,B,R
, Rare
2are
Rthe
are the
the
The
sum
ir.
is,
if,B,c,
R
are
The
ofofa atriangle
equal
ir.ir.That
is,is,ififc,
Thesum
sumof
ofthe
the
angles
equal
That
the
three
angles
oftriangle,
a triangle,
then
↵
+
+
= ⇡.
The
sum
of
the
of athen
triangle
That
three
angles
R are the
c+y
three
angles
ofofa angles
++y=ir.
+y
c c++ equal
three
angles
atriangle,
then
=7t.
7t.= 7t.is, if c, ,B,
three angles of a triangle, then c + +y = 7t.
‘‘ ‘‘
‘‘
For
#1-4,
find
the
unknown
angle
For
0.
For
find
#1-4,
angle
0.0. ✓.
For#1-4,
findthe
theunknown
unknown
angle
#1-4,
For #1-4, find the unknown angle 0.
1.)
3.)
1.)1.) 1.)
3.)3.) 3.)
1.)
3.)
2.)
2.)2.) 2.)
2.)
4.)
4.),
4.),4.),
4.),
185185
185
202
185
‘‘
For #5-12, write the given point in the plane in the form (a, b) for some
numbers a and b. The circle on page 201 will help with these questions.
5.) wind
5⇡
6
9.) wind(0)
6.) wind
7⇡
6
10.) wind
⇡
2
7.) wind
⇡
6
11.) wind
103⇡
3
8.) wind
⇡
3
12.) wind
13⇡
6
For #13-16, find the set of solutions of the given equations.
13.) log5 (2 + x)2 = 4
14.) 53x+2 = 7
15.)
1
x
+x=3
16.) (x + 1)(x + 2) = 1
203