1. society
2. crime
3. property crime
4. usury

Trigonometrical rules for finding sides and angles in
triangles which are not right angled
First, a word about labelling triangles……
A
c
B
a
b
The vertices (corners) of a triangle are usually
labelled using capital letters, for example A, B, C
C
The sides of the triangle are usually labelled using
lower case letters, in this case, a, b and c, and are IMPORTANT!!
positioned opposite the respective vertices. So
Side a will be opposite vertex A
Side b will be opposite vertex B
Side c will be opposite vertex C
Note also that side a could also be
called BC as it connects vertex B to
vertex C etc….
(We won’t be using this labelling
system in this unit of work)
We will look at the two rules very briefly before starting to use them!
c
A
The Sine Rule states that in any
triangle ABC….
B
a
b
a
b
c


sin A sin B sin C
C
This is the general formula for the sine rule. In reality however, you will use
only two of the three fractions at any one time. So the rule we will be using is
a
b

sin A sin B
More on this later!
c
A
The Cosine Rule states that in any
triangle ABC….
B
a
b
c 2  a 2  b 2  2ab cos C
This formula has c2 as the subject, but
the letters can be interchanged, so it
can also be written as
b 2  a 2  c 2  2ac cos B
or
a 2  b 2  c 2  2bc cos A
C
Study the patterns and
locations of the letters
in the three formulae
closely. More on the
cosine rule later!
The Sine Rule
a
b

sin A sin B
Proof of the Sine Rule:
Let ABC be any triangle with side lengths a, b, c respectively
C
b
h
a
D
A
B
c
Now draw AD perpendicular to BC, and let the length of AD equal h
In BDC
sin B 
In ACD
h
a
and
sin A 
h  a sin B
h  b sin A
As both expressions are equal to h, we can say
Dividing through by (sinA)( sinB) this
becomes
h
b
a sin B = b sin A
a
b

sin A sin B
which is the
Sine Rule
Example 1 – Use the Sine Rule to find the value of x in the
triangle:
xm
C
A
54
a
b

sin A sin B
12m
VERY IMPORTANT!! Take time to study the diagram.
Note the positions of the three “givens” (actual values you’re
told) – the 88, 54 and 12 m, and the one “unknown”, x.
88
B
The formula for the sine rule requires
• three “givens” (in this case, 88, 54 and 12 m) and one unknown (x)
• two of these givens must be an angle and its opposite side (in this case,
the 54 and the 12 m which we will make our A and a).
• the third given (88) and the unknown (x) must also be an angle and its
opposite side.
Note that the third angle C and its opposite side c are
not used in this problem!
x 14.82m (looks OK)
C
A
54
a
b

sin A sin B
12m
88
Now we substitute the 3 givens and the
unknown into this formula…..
A = 54
a = 12
Remember these two “givens” must
be an angle and its matching opposite
side
B = 88
b=x
These too!
12
x

sin 54 sin 88
Cross-multiply
x sin 54  12 sin 88
Divide through by sin 54
12 sin 88
to make the subject
x
sin 54
B
Substituting the values
into the formula
Finally, label the x as 14.82
on the diagram and check
that your answer fits with
the other numbers in the
problem!
x = 14.82 (to 2 dec. pl)

Example 2 – Use the Sine Rule to find the value of x in the
triangle:
x cm
Here, no vertices are labelled so we
will have to create our own. But
first…
95
35cm
Step 1, check that there are 4 “labels” – i.e. 3
givens and 1 unknown. There are a 95, 22,
35 cm and x cm so this fits our requirements.
Step 2, check that 2 of the 3 givens are a matching angle and
opposite side. 95 and 35 cm fit this. Also check that the remaining
given and the unknown form another matching angle and opposite
side (22 and x cm). They do! All our requirements are in place so we
can now use the Sine Rule!
A = 95
a = 35
Step 3, Allocate letters A, a, B, b (or any
B = 22
other letters of your choice) to matching
b=x
pairs.
22
A = 95
a = 35
B = 22
b=x
x cm
b
a
b

sin A sin B
35
x

sin 95 sin 22
35 sin 22  x sin 95
35 sin 22
x
sin 95
x = 13.16 (2dec pl)
A
95
a
35 cm
22
B
Remember to check that the answer
fits the context of the diagram.
Example 3 – Use the Sine Rule to find the value of  in the
triangle:
62
4.7m

5.1m
A quick check indicates everything is in place to use the Sine Rule….
• 3 givens and one unknown
• One pair of givens (5.1 and 62) form a matching angle and opposite side; and
• The other pair (4.7 and ) form the second matching angle and opposite side.
Note the third side and angle are unmarked – we don’t use these.
a
b

sin A sin B
4.7
5.1

sin   sin 62
62
4.7m

5.1m
4.7 sin 62  5.1sin 
4.7 sin 62
5 .1
sin   0.8137
sin  
  sin 1 0.8137 
  54.459 or 5428'
Remember to check that the answer
fits the context of the diagram.
Example 4 – Use the Sine Rule to find the value of x in the
triangle:
Looking at the diagram, it seems we have a
problem! 
Although the 68 and 35.7 form a matching angle
and opposite side, the 33 and x do not.
35.7m
79
But…remembering the angle sum of a triangle is 180,
we can work out the 3rd angle to be 180 – 33 – 68 = 79.
So now we use the 79 as the matching angle for the x and
proceed as usual, ignoring the 33 which plays no further part.
x
35.7

sin 79 sin 68
x
35.7 sin 79
sin 68
x = 37.80 (2 dec pl)
33
xm
68
Example 5 – The “Ambiguous Case”. Draw two different shaped
triangles ABC in which c = 14m, a = 10m and A = 32. Hence
find the size(s) of angle C.
This process (drawing triangles from verbal data and no diagram) takes time and
practice. You need to access these types of problems and practise them
thoroughly. Below is one possible diagram:
B
14m
32
A
10m
Now extend side AC1 past C1 to
the new point C2 where the
new length BC2 is the same as
it was previously (10m)…..
B
C1
14m
32
A
10m
10m
C1
C2
The new ABC2 has the
same given properties as the
original ABC1 . Both
triangles have c = 14, a = 10
and A = 32 . But note the
angles at C are different!
One is acute and the other
obtuse.
B
TRIANGLE 1
14m
32
TRIANGLE 2
14m
10m
10m
32
C1
A
B
C1
A
ANGLE C is obtuse
C2
ANGLE C is acute
How are the two C angles related? (if at all)
Let angle BC2C1 = .
B
14m
A
32
 angle BC1C2 = . (isos )
10m
180 –  
C1
10m

 angle BC1A = 180 – 
(straight line)
C2
Conclusion: The (green) acute angle at C2 and the (blue) obtuse angle at C1 are
supplementary. Thus, for example if one solution is 73 then the other solution is
180 – 73 = 107
Back to the question!
Draw the triangle with the acute, rather than the obtuse, angle
at C.
B
14m
32
A
10m

C2
Applying the Sine Rule,
10
14

sin 32 sin  
14 sin 32
sin  
10
  47.9
One solution (the acute angle which is
the only one given by the calculator) is
therefore 47.9 and the second solution
(the obtuse angle) is 180 – 47.9 = 132.1
Ans:  = 47.9 or 132.1
• The Sine Rule can be used to find unknown sides or angles in triangles.
a
b
c


• The Sine Rule formula is sin A sin B sin C
• To use the Sine Rule, you must have
 A matching angle and opposite side pair (two givens)
 A third given and an unknown, which also make an angle
and opposite side pair
• When asked to find the size of an ANGLE, first check whether the problem
could involve the ambiguous case (see Example 5). In that case, the two
answers are supplementary – i.e. add to 180
• When confronted with a problem where you have to decide whether to use
the Sine Rule or the Cosine Rule, always try for the Sine Rule first, as it is easier.
We will have this discussion later!
• In every triangle, the largest side is always opposite the largest angle. The
side lengths are in the ratio of the sines of their opposite angles.
In every triangle,
 The largest side is always opposite the largest angle.
 The middle sized side is always opposite the middle sized angle, and
 The smallest side is always opposite the smallest angle
• The ratio of any two side lengths is always equal to the ratio of the sines of their
respective opposite angles.
a sin A

b sin B
c sin C

etc..
a sin A
A
i.e.
b
C
c
a
B
These are just re-shaped versions of
the original sine rule formulae.
The Cosine Rule
There are two variations of this….
To find a side use
To find an angle use
c2 = a2 + b2 – 2ab cos C
a 2  b2  c2
cos C 
2ab
These formulae are just rearrangements of each
other. Verify this as an exercise.
Proof of the Cosine Rule:
Let ABC be any triangle with side lengths a, b, c respectively
A
b
C
c
h
x
D
a
NOTE!! The expansion
(a – x)2 = a2 – 2ax + x2
a–x
B
Now draw AD perpendicular to BC, and let the length of AD equal h
Let the length CD = x, and so length BD will be a – x.
In ACD
cos C 
x
b
x  b cos C
(1)
In ABD
Pythagoras gives
In ACD
Pythagoras gives
c 2  h 2  ( a  x) 2
c 2  h 2  a 2  2ax  x 2
h 2  c 2  a 2  2ax  x 2 (2)
b2  h2  x2
h 2  b 2  x 2 (3)
The formulae (2) and (3) are both for h2 so we make them equal to each other.
c  a  2ax  x  b  x
2
2
2
2
2
Now cancel the x2 on each side and make c 2
the subject…
c  a  b  2ax
2
2
2
From the first box on the previous slide, taking result (1)
x = b cos C
and substituting this into (4), we get
c 2  a 2  b 2  2ab cos C
which is a version of the Cosine Rule (for finding a side)
(4)
c 2 = a2 + b2 – 2ab cos C
(1) Note the positions of the letters. If the 2ab cos C were missing, this would just
be Pythagoras’ Theorem, c 2 = a2 + b2 . If the triangle were right angled, then C
would be 90 and as cos 90 = 0, it becomes Pythagoras’ Theorem!
(2) When c2 is the subject, the only angle in the formula is C (the angle opposite
to side c). Note A and B are absent from the formula.
(3) The above formula is to find a side length. The letters can be swapped
around and the same formula can be written
a 2 = b2 + c2 – 2bc cos A
b 2 = a2 + c2 – 2ac cos B
c 2 = a2 + b2 – 2ab cos C
Here are the three
variations of the
formula shown
together. Study
them closely and
note the patterns!
c 2 = a2 + b2 – 2ab cos C
(4) This formula can be rearranged to make cos C the subject, i.e.
a 2  b2  c2
cos C 
2ab
This is the version of the
Cosine Rule to use when
FINDING AN ANGLE.
(5) Again, the letters can be swapped around and the same formula can be written
a 2  b2  c2
cos C 
2ab
b2  c2  a 2
cos A 
2bc
a2  c2  b2
cos B 
2ac
When do we use the Cosine Rule?
• First, check to see if you can use the Sine Rule. It’s easier!
You can use the Cosine Rule when
• You are told TWO SIDES and
THEIR INCLUDED ANGLE (i.e.
the angle between those two
sides) and asked to
FIND THE THIRD SIDE
x
• You are told ALL THREE
SIDES and asked to
FIND ANY ANGLE
8m
10m

OR
20 cm
9m
15 cm
45
Here, we use
a 2  b2  c2
cos C 
2ab
Here, we use
c 2 = a2 + b2 – 2ab cos C
Example 6 – Use the Cosine Rule to find the value of c in
the triangle:
C
65
Finally, check
that c = 3.85 fits
the diagram.
4 cm
3 cm
A
B
c
Note that we have 2 given sides (3 cm and 4 cm) and their included angle (65)
so we can use the Cosine Rule for finding a side…
c 2 = a2 + b2 – 2ab cos C
Let
a=3
b=4
C = 65
c 2 = 32 + 42 – 2 × 3× 4 × cos 65
c 2 = 14.857 (do in one step on calculator)
c = 3.85 (to two dec pl)
Ans: The length of the required side is 3.85 cm
Example 7 – Use the Cosine Rule to find the size of C in the
triangle:
B
Finally, check
that C = 51.95 fits
the diagram.
8m
7.5 m
?
C
9m
A
Note that we have 3 given sides and are asked to find angle at C (opposite 7.5)
a 2  b2  c2
cos C 
2ab
so we can use the Cosine Rule for finding an angle…
Let
a=8
b=9
c = 7,5
Caution! Here we MUST make c =
7.5 as it is the side opposite the
angle we’re finding, i.e. C, whereas a
and b are interchangeable.
8
cos C 
 92  7.52
2  8  9
= 0.6163
2

C  cos 1 (0.6163)
Ans: Angle C is equal to 51.95 (to 2 dec pl) C  51.95
or 5157’ (to nearest minute)
NOTE !! Bracket
numerator and
denominator
when entering
into calculator.
Example 8 – Use the Cosine Rule to find the value of x in
the triangle:
10 m
11 m
x
Finally, check
that x = 16.10 fits
the diagram. x is
the longest side
so this would
seem reasonable.
Note that we have 2 given sides (10 m and 11 cm) and their included angle (100)
so we use the Cosine Rule for finding a side…
Let
a = 10
b = 11
c=x
C = 100
c 2 = a2 + b2 – 2ab cos C
x 2 = 102 + 112 – 2 × 10 × 11 × cos 100
x 2 = 259.2 (do in one step on calculator)
x = 16.10 (to two dec pl)
Ans: The length of the required side is 16.10 m
Example 9 – Use the Cosine Rule to find the value of  in
the triangle:
40 mm
29 mm

21 mm
Finally, check that  = 105 fits the diagram. 
LOOKS obtuse so this would seem
reasonable. Beware – you can’t always
presume the drawings are to scale, so be
careful when judging the appropriateness of
your answers (in all problems)
Note that we have 3 given sides and are asked to find angle opposite to 40 mm
so we use the Cosine Rule for finding an angle…
21
cos  
 292  402
2  21 29
2
Let
a = 21
b = 29
c = 40
C=

a 2  b2  c2
cos C 
2ab
remember the brackets
Note the negative cos. This means our angle is obtuse!
 0.2611 ALL OBTUSE ANGLES HAVE A NEGATIVE COSINE!
  cos 1 (0.2611)
= 105.13
Ans:  is approx. equal to
105.13 (to 2 dec pl) or
1058’ (to nearest min)
• The Cosine Rule can be used to find unknown sides or angles in triangles.
• There are two versions of the Cosine Rule formula and three variations within
each of these, depending on what is required as the subject
To find a SIDE
2
c =
a2
+
b2
– 2ab cos C
a 2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
To find an ANGLE
a 2  b2  c2
cos C 
2ab
b2  c2  a 2
cos A 
2bc
a2  c2  b2
cos B 
2ac
Make sure you familiarise yourself with how the PATTERNS in these
configurations work. Also remember each formula on the left is just a
rearrangement of its corresponding formula on the right.
• To use the Cosine Rule to find an angle you must be given all three sides
• To use the Cosine Rule to find a side you must be given the other two sides
and their included angle.
• When deciding whether to use the Sine Rule or the Cosine Rule, always try
the Sine Rule first, as it is easier (only one formula to deal with).
• When dealing with angles in the range 90 <  < 180, i.e. OBTUSE
ANGLES, remember that their cosines are negative. This does not apply to
their sines – they are still positive.
Mixed examples – which rule to use?
Study each of these diagrams and determine which rule to use – Sine Rule or
Cosine Rule? If Cosine Rule, which version? Answers & working on next slides.
10 cm
12 cm
16 m
xm
A
14 cm
71
12 cm
119
C

35
x cm
29

xm
9 cm
D
11 m
B
67
13 m
E
33
6 cm
x cm
12 cm
80
F
9 cm
Example 10
First check to see if we can use the Sine Rule.
16 m
xm
35
A
71
We have a given angle and opposite side (35 and 16m),
and the unknown x and the other given (71) also form
a matching angle and opposite pair. So we can use the
SINE RULE
a
b

sin A sin B
x
16

sin 71 sin 35
Ans: the length of
side x is 26.38 m
approximately.
16 sin 71
x
sin 35
x  26.38 to two dec pl.
Remember to check
appropriateness of
your answer!
Example 11
First check to see if we can use the Sine Rule.
10 cm
14 cm
B

Let….
C=
c = 10
a = 12
b = 14
12 cm
We are not given any angle so we can’t use the Sine Rule
so we have to use the COSINE RULE – the angle
version
a 2  b2  c2
cos C 
2ab
12
cos  
 142  102
2 12 14
2
cos  0.7143
  cos 1 0.7143
  44.42

Ans: the size of
angle  is 44.42 or
4425’ approx.
Remember to check
appropriateness of
your answer!
Example 12
12 cm
32
119
C
29
First check to see if we can use the Sine Rule.
x cm
We have a given angle and opposite side (29 and 12cm),
but the unknown x and the other given (119) are NOT a
matching angle and opposite pair. BUT…the third angle is
180 – 119 – 29 = 32 so we can use the SINE RULE
a
b

sin A sin B
Let….
a=x
A = 32
b = 12
B = 29
x
12

sin 32 sin 29
Ans: the length of
side x is 13.12 cm
approximately.
12 sin 32
x
sin 29
x  13.12 to two dec pl.
Remember to check
appropriateness of
your answer!
Example 13
First check to see if we can use the Sine Rule.
xm
D
11 m
13 m
We are not given any angle and matching opposite side
so we can’t use the Sine Rule, so we have to use the
COSINE RULE – the side version
67
c 2 = a2 + b2 – 2ab cos C
Let….
C = 67
c=x
a = 11
b = 13
x 2 = 112 + 132 – 2 × 11 × 13 × cos 67
x 2 = 178.251
Ans: the size of side x is
x = 13.35
13.35 m (to 2 dec places)
Remember to check
appropriateness of
your answer!
Example 14

9 cm
E

Let….
a=6
A=
b=9
B = 80
First check to see if we can use the Sine Rule.
6 cm
80
We have a given angle and opposite side (80 and 9 cm), but the
unknown  and the other given (6 cm) are NOT a matching
angle and opposite side. HOWEVER…we can use the SINE
RULE to find the third angle  (which forms a matching pair
with the 6cm) then use the 180 rule to find 
a
b

sin A sin B
6
9

sin   sin 80
6 sin 80
sin   
9
  41.04
  180  80  41.04
  58.96
Ans: the size of
angle  is approx.
58.96 or 5858’
Remember to check
appropriateness of
your answer!
Example 15
First check to see if we can use the Sine Rule.
33
x cm
12 cm
F


We have a given angle and opposite side (33 and 9 cm), but the
unknown x and the other given (12 cm) are insufficient data for
Sine Rule. The Cosine Rule won’t work either as the triangle’s
data does not match either of the two configurations for the
Cosine Rule. HOWEVER…if we let  be the angle opposite the
12cm we then have a second matching pair and can begin with
using the SINE RULE to find angle . (This is PART 1)
9 cm
NOW FOR PART 2 …..Once we know  we can then find the
third angle  (which is opposite to x) and then apply the Sine
Rule a second time to find x.
Part 2 (finding x)
Part 1 (finding )
12
9

sin   sin 33
12 sin 33
sin  
9
  46.57
Finding 
 = 180 – 33 – 46.57
 = 100.43
x
9

sin 100.43 sin 33
x  16.25
Note!! Here the diagram is quite out of
scale. This becomes apparent on checking
the reasonableness of your answer