Sine, Cosine
and Area Rules
NCS Mathematics
DVD Series
Outcomes for this DVD
In this DVD we will:
1. Calculate the area of a triangle given an angle and the two
adjacent sides.
Lesson 1
2. Apply the Sine Rule for triangles to calculate an unknown
side or an unknown angle of a given triangle.
Lesson 2
3. Apply the Cosine Rule for triangles to calculate an unknown
side or an unknown angle of a given triangle.
Lesson 3
4. Apply the Sine and the Cosine rules to solve
problems in 2-dimensions.
Lesson 4
Lesson 1
The Area Rule
NCS Mathematics
DVD Series
Some basic definitions – a reminder
Trigonometric Ratios
In a right angled triangle, the 3 trigonometric ratios for an angle
 are defined as follows:
hypotenuse
opposite

adjacent
sin  
opposite
hypotenuse
tan  
opposite
adjacent
cos  
adjacent
hypotenuse
The area formula of a triangle
Consider a non-right angled triangle ABC.
a, b and c are the sides opposite angles A, B and C respectively.
( This is the conventional way of labelling a triangle ).
Draw the perpendicular, h, from C to BA.
1
Area of  
base  height
2
 Area  12 c  h --- (1)
C
b
h
h
In ACN , sin A 
b
 b sin A  h
Substituting for h in (1)
Area 
1
2
c b sin A
A
Area 
a
c
1
2
N
bc sin A
B
Different forms of the area formula
A similar argument gives the same formula for the area
if B is obtuse i.e. B  90
The formula always uses
C
 2 sides and the
 angle formed by those sides (Included )
b
a
Any angle can be used as such in
area formula, so
 90
A
Area =
1
2
ab sin C =
1
2
bc sin A
h
c
=
1
2
B
ca sin B
N
Three possible approaches to find
the area of a triangle
Any angle can be used in the formula, so
C
C
Area  12 bc sin A
b
b
a
C
A
A
a
Area  12 ab sin C
A
B
Area  12 ca sin B
B
c
c
b
a
c
B
The area of a triangle – Example 1
Find the area of PQR.
Solution: We must use the angle formed by the
2 sides with the given lengths.
We know PQ and RQ so use the included angle Q
1
Area of PQR   QP  QR  sin Q
2
1
2
  8  7  sin 64 cm
2
64
 25, 2 cm
2
The area of a triangle – Example 2
Find the area of ABC .
A useful application of the area formula occurs when we
have a triangle formed by 2 radii and a chord of a circle.
A
r
C
1
Area   CA CB  sin C
2
But CA  CB  r

r
B
1 2
 Area  r sin 
2
Tutorial 1: Area of a Triangle
Find the areas of the triangles shown in the diagrams.
Give your answers accurate to 2 decimal digits
2)
1)
40
8 cm
30
10 cm
PAUSE
• Do Tutorial 1
• Then View Solutions
radius  6 cm
AOB  120
Tutorial 1: Problem 1: Area of a Triangle: Solution
Y  180   40  30   110
40
8 cm
30
10 cm
1
1) Area XYZ   XY  YZ  sin Y
2
1

 z  x  sin Y
2
1
2
  8 10  sin 110  cm
2
 37,59 cm
2
Tutorial 1: Problem 2: Area of a Triangle: Solution
Given:
radius  6 cm
AOB  120
1 2
2) Area AOB   r  sin O
2
1
2
   6 cm   sin120
2
 15,59 cm
2
Lesson 2
The Sine Rule
NCS Mathematics
DVD Series
The Sine Rule for Triangles
One way to find unknown sides and angles in non - right angled
triangles is by using the Sine Rule:
Suppose ABC is a scalene triangle Drop CN  AB
h
In ACN, sin A 
b
a
b
 h  b sin A
h
h
In BCN , sin B 
a
 h  a sin B
c
N
sin A sin B
or

b sin A  a sin B
a
b
The Complete Sine Rule for Triangles
ABC can be turned so that BC is the base.
We then get
C
b
Now h  c sin B  b sin C
sin B sin C


b
c
A
a
c
A
c
h
B
sin A sin B sin C
So


a
b
c
b
B
a
C
When do we use the Sine Rule?
The sine rule can be used in a triangle when:
 Two angles and a side are given

To calculate second side
 Two sides and the non-included angle are given
To calculate second angle

Application of the Sine Rule - Example 1
In ABC, find the size of angles A and C.
sin A
sin B
Solution: Use

a
b
a sin B
 sin A 
b
10 sin 62
 sin A 
12
 A  47,4
A is opposite the shorter of the 2 given sides.
A  62  A must be an acute angle.
(Only one possibility as can be seen from sketch)
Thus C  180  62  47, 4  70, 6
Application of the Sine Rule - Example 2
In PQR it is given that:
QR  5, PR  4 and Q  48.
Determine P.
P is opposite the longer of the 2 given sides.
P  48  P can be an acute or obtuse angle.
(  Two possibilities as can be seen from sketch below)
sin Q
sin P
Solution : Use

q
p
 sin P 

p sin Q
5  sin 48

P1
q
4
P2  68,3 or
P1  180  68,3  111,7
P2
Application of the Sine Rule - Example 3
In XYZ , find the length XY .
Solution : As the unknown is a side, we use the sine rule in
its reciprocal form. The unknown side is then at the top.
z
sin Z
y

sin Y
y sin Z
z 
sin Y


13sin 55
z
sin 29
z  22, 0
Tutorial 2: Sine Rule
1. In ABC , find B.
(Correct to two decimal places)
2. In PQR, find QR and the area of PQR
PAUSE DVD
• Do Tutorial 2
• Then View Solutions
Tutorial 2: Problem 1: Sine rule: Solution
Find B.
(2 decimal places)
Given:
B  35  B acute or obtuse
sin A sin B
sin 35 sin B



a
b
7
10
10sin 35
 sin B 
7
 B1  55,02
or B2  180  55,02  124,98
Tutorial 2: Problem 1: Why two solutions?
Obtained: B1  55, 02
Given:
or B2  124,98
B  35  B acute or obtuse
B1 (Obtuse)
B2 (Acute)
Tutorial 2: Problem 2: Sine rule: Solution
2. Find QR and the area of PQR.
QR
67

sin 64 sin 80
Given:
67sin 64
 QR 
 61,15 cm
sin 80

R

80

1
Area of PQR   QP  QR  sin 36
2
1
  67  61,15  sin 36
2
2
 1 204, 09 cm
Lesson 3
The Cosine Rule
NCS Mathematics
DVD Series
The Cosine Rule for Triangles
The Cosine Rule for ABC is given by:
a  b  c  2bc cos A
2
2
2
Symmetry also implies that:
b  a  c  2ac cos B
or
A
2
2
2
C
b
a
c
B
c  a  b  2ab sin C
2
2
2
We use this form to find the third side when
two sides and included angle are given.
Proof of the Cosine Rule
In CAD :
x
2
2
2
cos A  and b  x  h
b
b
x
a
h
In BCD :
2
2
2
2
2
2
a  h   c  x   h  c  2cx  x
cx
 a   b  x   c  2c  b cos A  x
2
2
2
2
2
 b  c  2bc cos A
2
2
Proofs for symmetrical results are similar.
A second form of the Cosine Rule
Know: a  b  c  2bc cos A
2
2
2
 2bc cos A  b  c  a
2
2
2
b c a
 cos A 
2bc
2
2
2
We use this form to find any angle of
a triangle when we know all 3 sides.
Applications of the Cosine Rule - Example 1
Find p in the PQR
Apply the Cosine Rule
Q
p
6
120
P
7
R
p  q  r  2qr cos P
2
2
2
 p  7  6  2  7  6  cos120
2
2
2
 127
 p  11, 3 1 decimal accuracy 
Applications of the Cosine Rule - Example 2
Find X in the XYZ
Solution: Use the Cosine Rule
y z x
cos X 
2 yz
2
2
2
8
X
4
6
82  62  42
 cos X 
2(8)(6)
 X  29,0 ( 1 dec )
Z
Y
Applications of the Cosine Rule  Example 3
Find side c and B in the given ABC.
A
Cosine rule:
b  15
c  b  a  2ba cos C
2
2
2
C
c
30 
 c  15  19  2(15)(19) cos30
 c  9,61 ( 2 decimal places )
2
2
Sine rule:
2
a  19
B
sin B sin C  sin B  15 sin 30

9 , 61
b
c
 B  51,3 ( 1 dec. )
Tutorial 3: Cosine Rule
1. Given ABC with AB  6 cm; BC  4 cm and
ABC  60. Find AC correct to 2 decimal digits.
2. Find all the angles in XYZ , giving your
answers to one decimal place accuracy.
PAUSE DVD
• Do Tutorial 3
• Then View Solutions
Tutorial 3: Problem 1: Cosine Rule: Solution
Find AC (2 dec accuracy): Given:
AC  BC  AB  2  BC  AB  cos ABC
2
2
2
 4  6  2  4  6  cos 60
2
2
 28
 AC 
28  5, 29 cm
Tutorial 3: Problem 2: Cosine Rule: Solution
Determine all angle measures of XYZ.
y z x
cos X 
2 yz
2
2
2
42  92  7 2
Hence cos X 
2  4  9 
Given:
 X  48, 2
sin X sin Y
y sin X 4sin 48.2
Now

 sin Y 

x
y
x
7
 Y  25, 2
Then Z  180   X  Y   106,6
Lesson 4
Basic Applications:
Problems in 2-D
NCS Mathematics
DVD Series
Problems in 2  dimensions: Example 1
1. Points A and B are in the same horizontal plane as C ,
the foot of a vertical tower PC. B  42; PAC  65
and AB  25 m. Calculate PC.
BPA  65  42  23
Sine rule:
AP
25

sin 42
sin 23
23
25sin 42
 AP 
 42,81 m
sin 23
42
B
P
65
A
25 m
PC
sin 65 
AP
 PC  AP sin 65  42,81sin 65  38,8 m
C
Problems in 2  dimensions: Example 2
2. In the figure QR represents a proposed tunnel.
Q and R are visible from a point P.
The three points are in the same plane.
Given:
QP  100 m; PR  60 m
and QPR  110
R
Q
Calculate the length of tunnel.
100
60
110
P
QR  100  60  2 100 60 cos110
2
2
QR  133 m
2
Tutorial 4: Part 1: Problems in 2  D
1. From the ends of a bridge AB, 101 metres long,
the angles of depression of a point P on the ground
directly under the bridge is 42,2 and 70,1.
Find the height, h, of the bridge under this point.
A
101 m
70,1
42, 2
h
P
B
PAUSE DVD
• Do Tutorial 4 Part 1
• Then View Solutions
Tutorial 4: Part 1: Suggested Solution
A
101 m
70,1
42, 2
B
Question: Find h
h
P
APB  180   42, 2  70,1   67, 7
AP
101
101 m  sin 70,1

 AP 
sin 70,1 sin 67, 7
sin 67, 7
 AP  102, 65 m
h
But sin 42, 2 
102,65
 h  102, 65  sin 42, 2  68,95 m
Tutorial 4: Part 2: Problems in 2-D
2. ABCD is a wall of a room, AD being the
line of the ceiling. EF is a picture rail,
with E being directly below A.
AE = 2 metres; ACB  x and ECB  y
2cos x
2m
(a) Prove that EC 
sin( x  y )
y x
(b) Find the length and height
of the wall if x  33 and y  20. PAUSE DVD
• Do Tutorial 4 Part 2
• Then View Solutions
Tutorial 4: Part 2(a) Solution
2(a) Prove that
2 cos x
EC 
sin( x  y )
2m
x
y
x
Now ACE  x  y and CAD  x
Hence, CAE  90  x
From AEC :
EC
2

sin  90  x 
sin  x  y 
2 cos x
2sin  90  x 

 EC 
sin  x  y 
sin  x  y 
Tutorial 4: Part 2: Length of room
Know:
2 cos x
EC 
sin  x  y 
x  33 and y  20
2m
y
x
2 cos 33
EC 
 7, 46 m
sin13
BC
cos y 
 BC  EC  cos y
EC
 Length of room  BC  7, 46 m  cos 20
 7,01 m
End of the DVD on
Sine, Cosine and Area Rules
REMEMBER!
• Consult text-books for additional examples.
• Attempt as many as possible other similar examples
on your own.
• Compare your methods with those that were
discussed in the DVD.
• Repeat this procedure until you are confident.
• Do not forget:
Practice makes perfect!