“Teach A Level Maths”
Vol. 1: AS Core Modules
35:The Sine Rule
© Christine Crisp
The Sine Rule
Module C2
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The Sine Rule
Triangles that aren’t Right Angled
To find unknown sides and angles in non-right angled
triangles we can use one or both of 2 rules:
•
the sine rule
• the cosine rule
The next few slides prove the sine rule. The
cosine rule is on the next presentation.
You do not need to learn the proof.
The Sine Rule
The Sine Rule
ABC is a scalene triangle
a, b and c are the sides opposite angles A, B and C
C
b
A
a
c
B
The Sine Rule
ABC is a scalene triangle
Draw the perpendicular, h, from C to BA.
In
C
ΔACN ,
b
a
h
A
c
N
B
The Sine Rule
ABC is a scalene triangle
In
ΔACN ,
C
sin A 
b
A
a
h
c
N
B
The Sine Rule
ABC is a scalene triangle
In
ΔACN , sin A  h
b
 b sin A  h
In
ΔBCN ,
C
b
a
h
A
c
N
B
The Sine Rule
ABC is a scalene triangle
In
ΔACN , sin A  h
b
 b sin A  h
In
ΔBCN ,
sin B 
A
C
b
a
h
c
N
B
The Sine Rule
ABC is a scalene triangle
ΔACN , sin A  h
b
 b sin A  h
h
In ΔBCN , sin B 
a
 a sin B  h
C
In
A
b
a
h
c
N
B
The Sine Rule
so,
b sin A  h
and a sin B  h
C
 b sin A  a sin B
b
A
h
c
a
B
The Sine Rule
so,
b sin A  h
and a sin B  h
C
 b sin A  a sin B

sin A
sin B

a
b
b
A
a
c
B
The Sine Rule
The triangle ABC . . .
. . . can be turned so that BC is the base.
We would then get h  c sin B  b sin C
A

C
b
sin B
sin C

b
c
c
a
b
h
A
c
B
B
a
C
The Sine Rule
So,
We now have
sin B
sin C

b
c
sin A
sin B

a
b
So,
and
sin B
sin C

b
c
sin A
sin B sin C


a
b
c
The Sine Rule
The sine rule can be used in a triangle when we know
•
One side and its opposite angle, plus
•
One more side or angle
e.g. Suppose we know p, q and angle Q in triangle
PQR
sin Q
sin P

p
q
Tip: We need one complete “pair” to use the sine rule.
The angle or side that we can find is the one that
completes another pair.
The Sine Rule
e.g. 1 In the triangle
ABC, find the size of
angles A and C.
A
C
12
10
62 
sin B
sin A
We don’t need the

a
3rd part of the rule
b
a sin B
10 sin 62
sin A 
 sin A 
12
b
Solution: Use


sin A  0.7358
A  47 4  (3 s.f.)

C  180  62  47  4   70  6 
B
The Sine Rule
e.g. 2 In the triangle XYZ,
find the length XY.
X
z
13
55 
29 
Y
Z
Solution: As the unknown is a side, we “flip” the sine
rule over. The unknown side is then at the “top”.
z
y

sin Z
sin Y
13 sin 55

z
sin 29


z 
y sin Z
sin Y
z  22 0 ( 3 s.f.)
The Sine Rule
SUMMARY
 The sine rule can be used in a triangle when we
know
• One side and its opposite angle, plus
• One more side or angle
 We write the sine rule so that the unknown angle
or side is on the left of the equation
 If 2 sides and 1 angle are known we use:
sin A
sin B

a
b
 If 1 side and 2 angles are known we use:
b
a

sin A
sin B
The Sine Rule
Exercises
1. In triangle ABC, b = 3 6 cm, c = 4 2 cm and angle
C = 110. Find the size of angles A and B.
2. In triangle PQR, PQ = 23 cm, angle R = 42 and
angle P = 17  . Find the size of side QR.
C
Solution:

110
3 6
A
a

4 2

B
B  53 7  
sin C
sin B

b
c
3  6 sin 110
sin B 
4 2
A  16 3 
The Sine Rule
Exercises
2. In triangle PQR, PQ = 23 cm, angle R = 42 and
angle P = 17  . Find the size of QR.
Solution:
R
p
r

sin P
sin R
42 
p
P
17 
23

Q

23 sin 17
p
sin 42
p  10 0 ( 3 s.f.)
The Sine Rule
angle, is
longest
side, 2
e.g.IfInan
a unknown
triangle PQR
p =opposite
5 cm, r the
= 7.2
cm and
triangles may be possible: one will have an angle
angle P = 37 . 
greater than 90
Drawing side r and angle P, we have:
R1
This is one possible
complete triangle.
5
37 
P
7.2
Q
The Sine Rule
e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and
angle P = 37  .
This is the other.
R1
5
R2
5
37 
P
7.2
Q
The Sine Rule
e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and
angle P = 37  .
This is the other. The 2 possible values of
R are connected since

• x  y  180
•
Triangle QR1 R2 is isosceles
R2
 R 1 x
37 
P
y
x
R1
x
5
5
7.2
Q
The Sine Rule
e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and
angle P = 37  .
This is the other. The 2 possible values of
R are connected since

• x  y  180
•
Triangle QR1 R2 is isosceles
R2
 R 1 x
so, R 1  R 2  180

37 
P
y
x
R1
x
5
5
7.2

90
The calculator will give the acute angle ( <
).
We subtract from 180to find the other possibility.
Q
The Sine Rule
e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and

angle P = 37 . Find 2 possible values of angle R and
the corresponding values of angle Q. Give the answers
correct to the nearest degree.
sin P
r sin P
Solution: sin R
r

p
 sin R 
p
7  2 sin 37
 sin R 
5
 sin R  0  8666
 R 1  60 or R 2  180  60  120
R 1  60
 Q 1  180  37  60  83
R 2  120  Q 2  180  37  120  23
The Sine Rule
We have either: R  60  and Q  83 
or:
 and
R  120
Q  23
R1
60 
5
R2
120
37 
P
7.2
83 
Q1
37 
P
5
7.2
23 
Q2
The Sine Rule
SUMMARY
 If the sine rule is used to find the angle opposite
the longest side of a triangle, 2 values may be
possible.
 Use the sine rule and a calculator to find 1 value.
This will be an acute angle ( less than 90  ).
 Subtract from 180 to find the other possibility.
 Use each value to find 2 possible values
for the 3rd angle.
The Sine Rule
Exercise
Find 2 possible values of angle ACB in triangle ABC if
AB = 15 cm, AC = 12 cm and angle B = 47 . Sketch
the triangles obtained.
Solution: sin C
c

15 sin 47
sin B
 sin C 
 b
12
 sin C  0  9142
C  66  1
C  66 1

C  113 9

or
C  113  9 
 A  180  66  1  47  66  9 

A  180  113  9  47  19  1
The Sine Rule
Exercise
AB = 15 cm, AC = 12 cm, angle B = 47 
(i) C  66 1
(ii) C  113 9 
A  66 9 
A  19 1
C
12
C
12
66 1 
19 1
A
66 9 
A
47 
15
B
113 9 
47 

15
B
The Sine Rule
The Sine Rule
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
The Sine Rule
ABC is a scalene triangle
a, b and c are the sides opposite angles A, B and C
The Sine Rule
C
sin A
sin B sin C


a
b
c
A
b
a
c
B
The Sine Rule
 The sine rule can be used in a triangle when we
know
• One side and its opposite angle, plus
• One more side or angle
 We write the sine rule so that the unknown angle
or side is on the left of the equation
 If 2 sides and 1 angle are known ( a, b and B )
we use:
sin A
sin B

a
b
 If 1 side and 2 angles are known ( A, b and B )
we use:
b
a

sin A
sin B
The Sine Rule
e.g. 1 In the triangle
ABC, find the size of
angles A and C.
C
12
10
62 
A
sin B
sin A
We don’t need the

3rd part of the rule
a
b
a sin B
10 sin 62
sin A 
 sin A 
12
b
Solution: Use


sin A  0.7358
A  47 4  (3 s.f.)

C  180  62  47  4   70  6 
B
The Sine Rule
e.g. 2 In the triangle XYZ,
find the length XY.
X
z
13
55 
29 
Y
Z
Solution: As the unknown is a side, we “flip” the sine
rule over. The unknown side is then at the “top”.
z
y

sin Z
sin Y
13 sin 55

z
sin 29


z 
y sin Z
sin Y
z  22 0 ( 3 s.f.)
TheisSine
Rule the longest side, 2
If an unknown angle
opposite
triangles may be possible: one will have an angle
greater than 90 
e.g. In a triangle PQR, p = 5 cm, r = 7.2 cm and

angle P = 37 . Find 2 possible values of angle R and
the corresponding values of angle Q. Give the answers
correct to the nearest degree.
Solution: sin R
sin P
r sin P


sin
R

r
p
p
7  2 sin 37
 sin R 
5
 sin R  0  8666
 R 1  60 or R 2  180  60  120
R 1  60
 Q 1  180  37  60  83
R 2  120  Q 2  180  37  120  23