Mathematics 3200
Angles and Angle Measure
Unit 4 Lesson 2
Coterminal Angles in General Form:
By adding or subtracting multiples of one full rotation, you can write an infinite number
of angles that are coterminal with any given angle.
For example, some angles that are coterminal with 50 are
50  3601  410
50  3602   770
50  3601  310
50  7202   670
In general, the angles coterminal with 50 are 50  360n , where n is any natural
number.
Some angles coterminal with
3
3 8
 2 1 

4
4
4
11

4
3
3 16
 2 2 

4
4
4
19

4
3
are
4
3
3 8
 2 1 

4
4
4

5
4
3
3 16
 2 2 

4
4
4
13

4
3
3
 2n , where n is any natural
In general, the angles coterminala with
are
4
4
number.
General Form:
 An expression containing parameters that can be given specific values to generate
any answer that satisfies the given information or situation.
 Represents all possible cases.
Any given angle has an infinite number of angles coterminal with it, since each time you
make one full rotation from the terminal arm, you arrive back at the same terminal arm.
Angles coterminal with any angle  can be described using the expression
  360n and   2 n ,
Where n is a natural number. This way of expressing an answer is called the general
form.
Example 1:
Express Coterminal Angles in General Form
Write an expression for all possible angles coterminal with each given
angle. Identify the angles that are coterminal that satisfy
 360    360 or  2    2 .
9
4
An expression for all possible angles coterminal with  500 is  500  360n .
A)
A)
 500
B)
650
C)
We would have to add 360 or 720 to find an angle between  360    360 .
 500  360  140 or  500  720  220
An expression for all possible angles coterminal with 650 is 650  360n .
B)
We would have to subtract 360 or 720 to find an angle between
 360    360 .
650  360  290 or 650  720  70
C)
An expression for all possible angles coterminal with
We would have to subtract
9
9
 2n .
is
4
4
8
16
or
to find an angle between  2    2 .
4
4
9 8 
9 16
7

 or


4
4
4
4
4
4
Arc Length of a Circle
 
All arcs that subtend a right angle   have the same central angle, but they have
2
different arc length depending on the radius of the circle. The arc length is proportional to
the radius. This is true for any central angle and related arc length.
Consider two concentric circles with centre O .
The radius of the smaller circle is 1 , and the
radius of the larger circle is r . A central angle of
 radians is subtended by arc AB on the
circle and arc CD on the larger one. You can
write the following proportion, where x
represents the arc length of the smaller circle
and a is the arc length of the larger circle.
r
1
O

a r

x 1
a  xr
Consider the circle with radius 1 and the sector with central angle  . The ratio of the arc
length to the circumference is equal to the ratio of the central angle to one full rotation.

2r 2
 
x
2 1
 2 
x 
x

Substitute x   in a  xr .
a  r
This formula, a  r , works for any circle, provided that  is measured in radians and
both a and r are measured in the same units.
A
x
a
B
Example 2:
Determine Arc Length in a Circle
If a represents the length of an arc of a circle with radius r , subtended by
a central angle of  , determine the missing quantity. Give you answers to
the nearest tenth of a unit.
A)
r  8.7cm ,   75 , a  ?cm
B)
r  ?mm ,   1.8 , a  4.7mm
C)
r  5m , a  13m ,   ?
A)
  
75  75

 180 
5

12
a  r
 5 
  8.7 
 12 
 1.318.7 
 11.4
The arc length is 11.4cm .
B)
a  r
4.7  1.8r 
4 .7 1.8r

1 .8 1 .8
2 .6  r
The radius is 2.6mm .
C)
a  r
13   5
13 5

5
5
2 .6  
x
2 .6

180 
180  x   180 2.6 
 180 
  
x  180 0.82 
x  148
The degree measure is 148 .
Page 175-178 #1-23