sine and cosine rule

Trigonometrical rules for finding sides and angles in triangles which are not right angled

A B

C

First, a word about labelling triangles……

The vertices (corners) of a triangle are usually labelled using capital letters, for example A, B, C

The sides of the triangle are usually labelled using lower case letters, in this case, a, b and c, and are positioned opposite the respective vertices. So

Side a will be opposite vertex A

a

Side b will be opposite vertex B

b

Side c will be opposite vertex C

c

IMPORTANT!!

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!

A B

C

ab

c

The Sine Rule states that in any triangle ABC….

C

c

B

b

A

a

sinsinsin

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

B

b

A

a

sinsin More on this later!

A B

C

ab

c

The Cosine Rule states that in any triangle ABC….

Cabbac cos2222

This formula has c2 as the subject, but the letters can be interchanged, so it can also be written as

Baccab cos2222

or

Abccba cos2222

Study the patterns and locations of the letters in the three formulae closely. More on the cosine rule later!

The Sine RuleB

b

A

a

sinsin

Proof of the Sine Rule:

Let ABC be any triangle with side lengths a, b, c respectivelyC

BAc

ab

Now draw AD perpendicular to BC, and let the length of AD equal h

h

In BDC

a

hB sin

Bah sin

In ACD

b

hA sin

Abh sin

and

As both expressions are equal to h, we can say a sin B = b sin A

Dividing through by (sinA)( sinB) this becomes B

b

A

a

sinsin

which is the Sine Rule

D

Example 1 – Use the Sine Rule to find the value of x in the triangle:

C A

B

88

12m

x m

54

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.

The formula for the sine rule requires

• three “givens” (in this case, 88, 54 and 12 m) and one unknown (x)

Note that the third angle C and its opposite side c are not used in this problem!

• 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.

B

b

A

a

sinsin

C A

B

88

12m

x

54

B

b

A

a

sinsin

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

88sin54sin

12 x

Cross-multiply

88sin1254sinxDivide through by sin 54

to make the subject

54sin

88sin12x

x = 14.82 (to 2 dec. pl)

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!

14.82m (looks OK)

These too!


95

35cm

x cm

22

Here, no vertices are labelled so we will have to create our own. But first…

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!

Step 3, Allocate letters A, a, B, b (or any other letters of your choice) to matching pairs.

A = 95a = 35B = 22b = x

A = 95a = 35B = 22b = x

B

b

A

a

sinsin

22sin95sin

35 x

95sin22sin35 x

95sin

22sin35x

x = 13.16 (2dec pl) Remember to check that the answer fits the context of the diagram.

95

35 cm

x cm

22

A

a

B

b

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.

62

4.7m

5.1m

62sin

1.5

sin

7.4

B

b

A

a

sinsin

sin1.562sin7.4

1.5

62sin7.4sin

8137.0sin

8137.0sin 1

'2854459.54 or Remember to check that the answer fits the context of the diagram.


68

33

x m

35.7mLooking 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.

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.

68sin

7.35

79sin

x

79

68sin

79sin7.35x

x = 37.80 (2 dec pl)

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.

A

B

14m

32C1

10m

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:

Now extend side AC1 past C1 to the new point C2 where the new length BC2 is the same as it was previously (10m)…..

A

B

14m

32C1

10m

C2

10m

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.

A

B

14m

32C1

10m

ANGLE C is obtuse

B

14m

32C1

C2

10m

A

TRIANGLE 1 TRIANGLE 2

ANGLE C is acute

How are the two C angles related? (if at all)

A

B

14m

32C1

10m

C2

10m

Let angle BC2C1 = .

angle BC1C2 = . (isos )

angle BC1A = 180 –

(straight line)180 –

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.

Applying the Sine Rule,

sin

14

32sin

10

B

14m

32C2

10m

A

10

32sin14sin

9.47

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.

• The Sine Rule formula is C

c

B

b

A

a

sinsinsin

• 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 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!

• 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

• 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

b c

C B

A

..sin

sin

sin

sin..

etcA

C

a

c

B

A

b

aei

These are just re-shaped versions of the original sine rule formulae.

The Cosine RuleThere are two variations of this….

To find a side use

c2 = a2 + b2 – 2ab cos C

To find an angle use

ab

cbaC

2cos

222

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 respectivelyA

BCa

cb

Now draw AD perpendicular to BC, and let the length of AD equal h

h

In ACD

b

xC cos

In ABDPythagoras gives

222 )( xahc

D

Let the length CD = x, and so length BD will be a – x.

x a – x

Cbx cos (1)2222 2 xaxahc

(2)

In ACDPythagoras gives

222 xhb 222 xbh (3)

2222 2 xaxach

The formulae (2) and (3) are both for h2 so we make them equal to each other.

NOTE!! The expansion

(a – x)2 = a2 – 2ax + x2

22222 2 xbxaxac

Now cancel the x2 on each side and make c 2

the subject…

axbac 2222

From the first box on the previous slide, taking result (1)

x = b cos C

(4)

and substituting this into (4), we get

Cabbac cos2222

which is a version of the Cosine Rule (for finding a side)

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

b 2 = a2 + c2 – 2ac cos B

a 2 = b2 + c2 – 2bc cos A

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.

ab

cbaC

2cos

222

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

bc

acbA

2cos

222

ac

bcaB

2cos

222

ab

cbaC

2cos

222

When do we use the Cosine Rule?

• First, check to see if you can use the Sine Rule. It’s easier!

• You are told ALL THREE SIDES and asked to FIND ANY ANGLE

You can use the Cosine Rule when

OR8m

9m

10m

• You are told TWO SIDES and THEIR INCLUDED ANGLE (i.e. the angle between those two sides) and asked to FIND THE THIRD SIDE

20 cm

45

15 cm

x

Here, we use ab

cbaC

2cos

222

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:

Note that we have 2 given sides (3 cm and 4 cm) and their included angle (65)

65 4 cm

c

C

3 cm

A

B

c 2 = a2 + b2 – 2ab cos C

Leta = 3b = 4C = 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

Finally, check that c = 3.85 fits the diagram.

so we can use the Cosine Rule for finding a side…

Example 7 – Use the Cosine Rule to find the size of C in the triangle:

Note that we have 3 given sides and are asked to find angle at C (opposite 7.5)

7.5 m

8 m

?

B

9 mC

A

Leta = 8b = 9c = 7,5

Ans: Angle C is equal to 51.95 (to 2 dec pl)

or 5157’ (to nearest minute)

Finally, check that C = 51.95 fits the diagram.

so we can use the Cosine Rule for finding an angle…

ab

cbaC

2cos

222

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.

982

5.798cos

222

C

= 0.6163

NOTE !! Bracket

numerator and

denominator

when entering

into calculator.

)6163.0(cos 1C

95.51C

Example 8 – Use the Cosine Rule to find the value of x in the triangle:

Note that we have 2 given sides (10 m and 11 cm) and their included angle (100)

11 m

x

10 m

c 2 = a2 + b2 – 2ab cos C Leta = 10b = 11c = xC = 100

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

Finally, check that x = 16.10 fits the diagram. x is the longest side so this would seem reasonable.

so we use the Cosine Rule for finding a side…

Example 9 – Use the Cosine Rule to find the value of in the triangle:

29 mm

21 mm

40 mm

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…

ab

cbaC

2cos

222

Leta = 21b = 29c = 40C =

29212

402921cos

222

2611.0

)2611.0(cos 1

= 105.13Ans: is approx. equal to105.13 (to 2 dec pl) or1058’ (to nearest min)

remember the brackets

Note the negative cos. This means our angle is obtuse! ALL OBTUSE ANGLES HAVE A NEGATIVE COSINE!

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)

• 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

ab

cbaC

2cos

222 c 2 = a2 + b2 – 2ab cos C

To find a SIDE To find an ANGLE

a 2 = b2 + c2 – 2bc cos A

b2 = a2 + c2 – 2ac cos B bc

acbA

2cos

222

ac

bcaB

2cos

222

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

• 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).

• To use the Cosine Rule to find a side you must be given the other two sides and their included angle.

• 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.

A

35

71

x m

16 m

E

80

9 cm

6 cm

C

29

119

x cm

12 cmB

14 cm

10 cm

12 cm

D

67

x m

11 m

13 mF

33

9 cm

x cm12 cm

A

35

71

x m

16 m

Example 10

First check to see if we can use the Sine Rule.

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

B

b

A

a

sinsin

35sin

16

71sin

x

35sin

71sin16x

38.26x to two dec pl.

Ans: the length of side x is 26.38 mapproximately.

Remember to check

appropriateness of

your answer!

Example 11


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

Ans: the size of angle is 44.42 or 4425’ approx.

Remember to check

appropriateness of

your answer!

B

14 cm

10 cm

12 cm

ab

cbaC

2cos

222

Let….C = c = 10 a = 12b = 14

14122

101412cos

222

7143.0cos

7143.0cos 1

42.44

Example 12


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

B

b

A

a

sinsin

29sin

32sin12x

12.13x to two dec pl.

Ans: the length of side x is 13.12 cm approximately.

Remember to check

appropriateness of

your answer!

C

29

119

x cm

12 cm32

29sin

12

32sin

xLet….a = xA = 32 b = 12B = 29

Example 13


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

Ans: the size of side x is 13.35 m (to 2 dec places)

Remember to check

appropriateness of

your answer!

Let….C = 67c = xa = 11b = 13

D

67

x m

11 m

13 m

c 2 = a2 + b2 – 2ab cos C

x 2 = 112 + 132 – 2 × 11 × 13 × cos 67

x 2 = 178.251

x = 13.35

Example 14


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

B

b

A

a

sinsin

Ans: the size of angle is approx. 58.96 or 5858’

Remember to check

appropriateness of

your answer!

80sin

9

sin

6

Let….a = 6A = b = 9B = 80

E

80

9 cm

6 cm

9

80sin6sin

04.41

04.4180180

96.58

Example 15

F

33

9 cm

x cm12 cm


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)

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 1 (finding )

33sin

9

sin

12

9

33sin12sin

57.46

Finding

= 180 – 33 – 46.57

= 100.43

Part 2 (finding x)

33sin

9

43.100sin

x

25.16x

Note!! Here the diagram is quite out of scale. This becomes apparent on checking the reasonableness of your answer

sine and cosine rule

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