The Sine Rule
The Sine Rule is used for cases in which the Cosine Rule cannot be applied. It is
used to find:
1. An unknown side, when we are given two angles and a side.
2. An unknown angle when we are given two sides and an angle that is not included.
Cosine Rule
12
19
12
a2 = b2 + c2 – 2bcCosA
75o
18
18
Sine Rule
12
50o
29/07/2017
18
?
12
75o
50o
1
The Sine Rule
Consider a general triangle ABC.
Deriving the rule
C
b
a
B
P
CP
 CP  aSinB
a
CP
also SinA 
 CP  bSinA
b
 aSinB  bSinA
aSinB

b
SinA
a
b


SinA SinB
SinB 
c
Draw CP perpendicular to BA
A
This can be extended to
a
b
c


29/07/2017
SinA SinB SinC
or equivalently
SinA SinB SinC


a
b
c 2
a
b
c


SinA SinB SinC
The Sine Rule
To find an unknown side we need 2 angles and a side.
1.
Not to
scale
5.1 cm
a
60o
15o
29/07/2017
145o
m
45 m
85o
m
12.7

Sin 63o
Sin 85o
Sin 63o x 12.7
m 
 11.4cm (1 dp)
Sin 85o
p
p
63o
12.7cm
45o
a
5.1

Sin 45o Sin 60o
5.1Sin 450
a 
 4.2 cm (1 dp)
Sin 60o
3.
2.

45
Sin15o
Sin145o
45Sin145o
p 
 99.7m (1 dp)
Sin15o
3
The Sine Rule
a
b
c


SinA SinB SinC
To find an unknown angle we need 2 sides and an angle not included.
2.
1.
5.1 cm
4.2 cm
60o
Sinx
4.2
x

Sin 60
SinA SinB SinC


a
b
c
o
12.7cm
yy
Siny
5.1
12.7
4.2Sin 60o
 Sinx 
5.1
29/07/2017
99.7 m
145o
11.4cm
Sin 63o
11.4
 y  83.0o (1 dp )
Sinz
z

Not
to
scale
12.7Sin 63o
 Siny 
11.4
 x  45.5o (1 dp )
3.
63oo
45 m
45

Sin145o
99.7
45Sin145o
 Sinz 
99.7
 z  15o (1 dp )
4
The Sine Rule
a
b
c


SinA SinB SinC
The angle of elevation of the
top of a building measured
from point A is 25o. At point
D which is 15m closer to the
building, the angle of
elevation is 35o Calculate the
height of the building.
Application Problems
T
10o
36.5
35o
B
Angle TDA = 180 – 35 = 145o
Angle DTA = 180 – 170 = 10o
TD
15

Sin 25o Sin10o
15Sin 25o
TD29/07/2017

 36.5 m
Sin10
145o
25o
D
15 m
Sin 35o 
A
TB
36.5
 TB  36.5Sin 35o  20.9 m
5
a
b
c


SinA SinB SinC
The Sine Rule
The angle of elevation of the top of a column measured from point A, is 20o.
The angle of elevation of the top of the statue is 25o. Find the height of the
statue when the measurements are taken 50 m from its base
Angle BCA = 180 – 110 = 70o
Cos 20o 
50
AC
50
Cos 20o
 53.2 m (1dp )
 AC 
Angle ACT = 180 – 70 = 110o
TC
53.2

Sin 5o Sin 65o
 TC 
53.2 Sin 5
 5.1 m
Sin 65o
Angle ATC =
T
180 – 115 = 65o
65o
110o
C
70o
5o
A
29/07/2017
20o
25o
50 m
B
6
The Sine Rule. (Used for Non-Right Angled Triangles)
Calculating a Length
a
=
b
Sin A
Sin B
b
7 cm
0
58
A
The angles are
OPPOSITE
their sides.
B0
39
x
a
Working Out
a
=
0
Sin 58
9.43
7Let ‘a’
stand
0
Sin for
39 the
unknown
length.
a
=
0
Sin 58
7
0
Sin 39
0
a = 7 Sin 58
0
Sin 39
a
= 9.43 cm
7
Menu
Calculate the Missing Lengths. (Answers to 2 d.p.)
3)
1)
0
x = 7.36 cm110
0
70
0
50
0
30
x
x = 4.06 m
2)
4)
0
80
x
0
40
0
60
x
x= 21.70 m
0
50
x= 4.57 m
20 m
8
Answers
Menu
The Sine Rule. (Used for Non-Right Angled Triangles)
Calculating an Angle.
b
6.5 cm
a
=
b
Sin A
Sin B
a
13.7 cm
B
0
x
0
A
27
Working Out
13.7
=
6.5
0
Sin A
Sin 27
0
-1
0
Sin0.9569
730.9569
13.7
Sin A
=
13.7 Sin 27 = Sin A
A 6.5
6.5
0
Sin 27
0.9569 = Sin A
-1
Sin 0.9569 = A
0
73 = A
9
Menu
Calculate The Missing Angles. (Answers to 1 d.p.)
1)
0
35.3
2)
80
48
31.7
0
x
0
15 cm
3)
x
4)
0
48.9
70
x
0
40
0
x
0
78.2
0
25 cm
10
Answers
Menu
The Sine Rule
The Sine Rule is used for cases in which the Cosine Rule cannot be applied. It is
used to find:
1. An unknown side, when we are given two angles and a side.
2. An unknown angle when we are given two sides and an angle that is not included.
Cosine Rule
12
19
12
a2 = b2 + c2 – 2bcCosA
75o
18
18
Sine Rule
12
50o
18
?
12
75o
50o
The Sine Rule
Consider a general triangle ABC.
Deriving the rule
C
b
a
B
P
CP
 CP  aSinB
a
CP
also SinA 
 CP  bSinA
b
 aSinB  bSinA
aSinB

b
SinA
a
b


SinA SinB
SinB 
c
Draw CP perpendicular to BA
A
This can be extended to
a
b
c


SinA SinB SinC
or equivalently
SinA SinB SinC


a
b
c
a
b
c


SinA SinB SinC
The Sine Rule
To find an unknown side we need 2 angles and a side.
1.
Not to
scale
5.1 cm
a
60o
15o
145o
m
45 m
85o
m
12.7

Sin 63o
Sin 85o
Sin 63o x 12.7
m 
 11.4cm (1 dp)
Sin 85o
p
p
63o
12.7cm
45o
a
5.1

Sin 45o Sin 60o
5.1Sin 450
a 
 4.2 cm (1 dp)
Sin 60o
3.
2.

45
Sin15o
Sin145o
45Sin145o
p 
 99.7m (1 dp)
Sin15o
The Sine Rule
a
b
c


SinA SinB SinC
To find an unknown angle we need 2 sides and an angle not included.
2.
1.
5.1 cm
4.2 cm
60o
Sinx
4.2
x

Sin 60
SinA SinB SinC


a
b
c
o
yy
12.7
4.2Sin 60o
 Sinx 
5.1
99.7 m
145o
11.4
 y  83.0o (1 dp )
Sinz
z

11.4cm
Sin 63o
12.7Sin 63o
 Siny 
11.4
 x  45.5o (1 dp )
3.
12.7cm
Siny
5.1
63oo
45 m
45

Sin145o
99.7
45Sin145o
 Sinz 
99.7
 z  15o (1 dp )
Not
to
scale
The Sine Rule
a
b
c


SinA SinB SinC
The angle of elevation of the
top of a building measured
from point A is 25o. At point
D which is 15m closer to the
building, the angle of
elevation is 35o Calculate the
height of the building.
Application Problems
T
10o
36.5
35o
B
Angle TDA = 180 – 35 = 145o
Angle DTA = 180 – 170 = 10o
TD
15

Sin 25o Sin10o
15Sin 25o
TD 
 36.5 m
Sin10
145o
25o
D
15 m
Sin 35o 
A
TB
36.5
 TB  36.5Sin 35o  20.9 m
a
b
c


SinA SinB SinC
The Sine Rule
The angle of elevation of the top of a column measured from point A, is 20o.
The angle of elevation of the top of the statue is 25o. Find the height of the
statue when the measurements are taken 50 m from its base
Angle BCA = 180 – 110 = 70o
Cos 20o 
50
AC
50
Cos 20o
 53.2 m (1dp )
 AC 
Angle ACT = 180 – 70 = 110o
TC
53.2

Sin 5o Sin 65o
 TC 
53.2 Sin 5
 5.1 m
Sin 65o
Angle ATC =
T
180 – 115 = 65o
65o
110o
C
70o
5o
A
20o
25o
50 m
B
The Sine Rule
The Sine Rule is used for cases in which the Cosine Rule cannot be applied. It is
used to find:
1. An unknown side, when we are given two angles and a side.
2. An unknown angle when we are given two sides and an angle that is not included.
Cosine Rule
12
19
12
a2 = b2 + c2 – 2bcCosA
75o
18
18
Sine Rule
12
50o
29/07/2017
18
?
12
75o
50o
17
The Sine Rule
Consider a general triangle ABC.
Deriving the rule
C
b
a
B
P
CP
 CP  aSinB
a
CP
also SinA 
 CP  bSinA
b
 aSinB  bSinA
aSinB

b
SinA
a
b


SinA SinB
SinB 
c
Draw CP perpendicular to BA
A
This can be extended to
a
b
c


29/07/2017
SinA SinB SinC
or equivalently
SinA SinB SinC


a
b
c 18
a
b
c


SinA SinB SinC
The Sine Rule
To find an unknown side we need 2 angles and a side.
1.
Not to
scale
5.1 cm
a
60o
15o
29/07/2017
145o
m
45 m
85o
m
12.7

Sin 63o
Sin 85o
Sin 63o x 12.7
m 
 11.4cm (1 dp)
Sin 85o
p
p
63o
12.7cm
45o
a
5.1

Sin 45o Sin 60o
5.1Sin 450
a 
 4.2 cm (1 dp)
Sin 60o
3.
2.

45
Sin15o
Sin145o
45Sin145o
p 
 99.7m (1 dp)
Sin15o
19
The Sine Rule
a
b
c


SinA SinB SinC
To find an unknown angle we need 2 sides and an angle not included.
2.
1.
5.1 cm
4.2 cm
60o
Sinx
4.2
x

Sin 60
SinA SinB SinC


a
b
c
o
12.7cm
yy
Siny
5.1
12.7
4.2Sin 60o
 Sinx 
5.1
29/07/2017
99.7 m
145o
11.4cm
Sin 63o
11.4
 y  83.0o (1 dp )
Sinz
z

Not
to
scale
12.7Sin 63o
 Siny 
11.4
 x  45.5o (1 dp )
3.
63oo
45 m
45

Sin145o
99.7
45Sin145o
 Sinz 
99.7
 z  15o (1 dp )
20
The Sine Rule
a
b
c


SinA SinB SinC
The angle of elevation of the
top of a building measured
from point A is 25o. At point
D which is 15m closer to the
building, the angle of
elevation is 35o Calculate the
height of the building.
Application Problems
T
10o
36.5
35o
B
Angle TDA = 180 – 35 = 145o
Angle DTA = 180 – 170 = 10o
TD
15

Sin 25o Sin10o
15Sin 25o
TD29/07/2017

 36.5 m
Sin10
145o
25o
D
15 m
Sin 35o 
A
TB
36.5
 TB  36.5Sin 35o  20.9 m
21
a
b
c


SinA SinB SinC
The Sine Rule
The angle of elevation of the top of a column measured from point A, is 20o.
The angle of elevation of the top of the statue is 25o. Find the height of the
statue when the measurements are taken 50 m from its base
Angle BCA = 180 – 110 = 70o
Cos 20o 
50
AC
50
Cos 20o
 53.2 m (1dp )
 AC 
Angle ACT = 180 – 70 = 110o
TC
53.2

Sin 5o Sin 65o
 TC 
53.2 Sin 5
 5.1 m
Sin 65o
Angle ATC =
T
180 – 115 = 65o
65o
110o
C
70o
5o
A
29/07/2017
20o
25o
50 m
B
22
Trigonometry 5: Sine, cosine and tangent for any angle
y
P (x,y)
1

O
x
y
A
x
As the Point P moves in an anti-clockwise
direction around the circumference of the
circle, the angle  changes from 0o to 360o.
Consider the right-angled triangle
formed by the vertical line PA.
In this triangle the distance OA = x.
The distance OP = y.
y
So point P has co-ordinates (x,y).
P(cos ,sin )
1
O
sin 

x
cos  A
In triangle OAP, cos  
x
y
and sin  
1
1
Therefore x = cos  and y = sin .
So the co-ordinates of P are (cos , sin ).
29/07/2017
23
The Sine Rule
C. McMinn
SOH/CAH/TOA can only be used for right-angled triangles.
The Sine Rule can be used for any triangle:
C
b
The sides are labelled to match
their opposite angles
A
The Sine Rule:
a
=
sinA
a
B
c
b
=
sinB
c
sinC
A
Example 1:
Find the length of BC
76º
c
7cm
b
63º
C
a
a
B
c
=
sinA
sin76º ×
x
sinC
7 sides to the
Drawxarrows from the
=
× sin76º
opposite
help decide
sin76º angles tosin63º
which parts of the sine rule to use.
7
x =
× sin76º
sin63º
x
=
7.6 cm
P
Example 2:
Find the length of PR
82º
x
r
43º
q
55º
Q
p
p
15cm
R
q
=
sinP
sinQ
Draw15
arrows from thexsides to the
sin43º × opposite angles
= to help decide× sin43º
sin43º
sin82º
which parts of the sine rule to use.
15
= x
sin43º ×
sin82º
x
= 10.33 cm
B
1.
3. G
2.
F
53º
A
41º
x
5.5
62º
28º
D
130º
E
x
8.0
63º
76º
H
26 mm
C
4.
P
10.7
x
5.
37º
66º
35.3
x
61º
R
6.
57º
x
5.2
77º
35º
62º
Q
85º
7.
65º
x
6.6
86º
x
6.9
I
Remember:
•
•
•
•
Draw a diagram
Label the sides
Set out your working exactly as you have been
shown
Check your answers regularly and ask for help if you
need it
Finding an Angle
The Sine Rule can also be used to find an
angle, but it is easier to use if the rule is
written upside-down!
Alternative form
of the Sine Rule:
sinA
a
=
sinB
=
b
sinC
c
C
Example 1:
b
Find the size of angle ABC
6cm a
4cm
xº
72º
A
B
c
sinA
sinB
=
a
4×
4×
b
sin72º
sin xº
Draw arrows
from the sides
=
× 4 to the
6
opposite
angles 4to help decide
which parts of the sine rule to use.
sin72º
=
sin xº
6
sin xº = 0.634
x
= sin-1 0.634 =
39.3º
P
Example 2:
Find the size of angle PRQ
85º
q
r
xº
R
p 8.2cm
sinP
sinR
=
p
sin85º
7×
Q
sin xº
=
8.2
r
7
×7
sin85º
7×
=
8.2
sin xº
sin xº = 0.850
x
= sin-1 0.850 =
58.3º
1.
2.
3.
82º
47º
105º
66.6
xº
37.6
xº
45.5
xº
6 cm
5.
4.
xº
31.0
27º
3.5 km
xº
51.1
33º
6.
7.
xº
57.7
74º
70º
92.1
xº
52.3º (←Be careful!→)
22.9º
Remember:
•
•
•
•
Draw a diagram
Label the sides
Set out your working exactly as you have been
shown
Check your answers regularly and ask for help if you
need it