1. The Cosine Rule A B C a c b
Pythagoras’ Theorem allows us to calculate unknown lengths in right-angled triangles using the relationship a2 = b2 + c2
It would be very useful to be able to calculate unknown sides for any value of the angle at A. Consider the square on the side opposite A when angle A is not a right-angle.
a2 = b2 + c2 A a2 1 A
Angle A obtuse a2 2 A
Angle A acute a2 3
a2 > b2 + c2
a2 < b2 + c2

2. *Since Cos A = x/c  x = cCosA
The Cosine Rule
The Cosine Rule generalises Pythagoras’ Theorem and takes care of the 3 possible cases for Angle A.
a2 > b2 + c2
a2 < b2 + c2
a2 = b2 + c2 A 1 2 3 A B C a b c
Consider a general triangle ABC. We require a in terms of b, c and A.
Deriving the rule
BP2 = a2 – (b – x)2
Also: BP2 = c2 – x2
a2 – (b – x)2 = c2 – x2
a2 – (b2 – 2bx + x2) = c2 – x2
a2 – b2 + 2bx – x2 = c2 – x2
a2 = b2 + c2 – 2bx*
a2 = b2 + c2 – 2bcCosA P x
b - x b
Draw BP perpendicular to AC
*Since Cos A = x/c  x = cCosA
When A = 90o, CosA = 0 and reduces to a2 = b2 + c2 1
Pythagoras
When A > 90o, CosA is negative,  a2 > b2 + c2 2
Pythagoras + a bit
When A < 90o, CosA is positive,  a2 > b2 + c2 3
Pythagoras - a bit

3. Finding an unknown side.B C a b c
The Cosine Rule
The Cosine rule can be used to find:
1. An unknown side when two sides of the triangle and the included angle are given.
2. An unknown angle when 3 sides are given.
Finding an unknown side.
a2 = b2 + c2 – 2bcCosA
Applying the same method as earlier to the other sides produce similar formulae for b and c. namely:
b2 = a2 + c2 – 2acCosB
c2 = a2 + b2 – 2abCosC

4. To find an unknown side we need 2 sides and the included angle.
a2 = b2 + c2 – 2bcCosA
The Cosine Rule
To find an unknown side we need 2 sides and the included angle.
Not to scale
8 cm
9.6 cm a 1.
40o 2.
7.7 cm
5.4 cm
65o m
85 m
100 m
15o 3. p
m2 = – 2 x 5.4 x 7.7 x Cos 65o
m = ( – 2 x 5.4 x 7.7 x Cos 65o)
m = 7.3 cm (1 dp)
a2 = – 2 x 8 x 9.6 x Cos 40o
a = ( – 2 x 8 x 9.6 x Cos 40o)
a = 6.2 cm (1 dp)
p2 = – 2 x 85 x 100 x Cos 15o
p = ( – 2 x 85 x 100 x Cos 15o)
p = 28.4 m (1 dp)

5. a2 = b2 + c2 – 2bcCosA The Cosine Rule Application Problem L H B
A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left onto a bearing of 035o and sails to a lighthouse (L) 24 miles away. It then returns to harbour.
Make a sketch of the journey
Find the total distance travelled by the boat. (nearest mile)
HL2 = – 2 x 40 x 24 x Cos 1250
HL = ( – 2 x 40 x 24 x Cos 1250)
= 57 miles
Total distance = = 121 miles. H
40 miles
24 miles B L
125o

6. The Cosine Rule a2 = b2 + c2 – 2bcCosA P Q W Not to Scale
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 430 miles North to a point P before turning left onto a bearing of 230o to a second point Q, 570 miles away. It then returns to base.
(a) Make a sketch of the flight.
(b) Find the total distance flown by the aircraft. (nearest mile)
The Cosine Rule
a2 = b2 + c2 – 2bcCosA
Not to Scale P
570 miles W
430 miles Q
50o
QW2 = – 2 x 430 x 570 x Cos 500
QW = ( – 2 x 430 x 570 x Cos 500)
= 441 miles
Total distance = = 1441 miles.

7. A B C a b c
The Cosine Rule
To find unknown angles the 3 formula for sides need to be re-arranged in terms of CosA, B or C.
a2 = b2 + c2 – 2bcCosA
b2 = a2 + c2 – 2acCosB
c2 = a2 + b2 – 2abCosC
Similarly

8. To find an unknown angle we need 3 given sides.
The Cosine Rule
To find an unknown angle we need 3 given sides.
Not to scale
8 cm
9.6 cm
6.2 1. A 2.
7.7 cm
5.4 cm P
7.3 cm
85 m
100 m 3. R
28.4 m

9. The Cosine Rule Application Problems L H B 57 miles 24 miles A
A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left and sails for 24 miles to a lighthouse (L). It then returns to harbour, a distance of 57 miles.
Make a sketch of the journey.
Find the bearing of the lighthouse from the harbour. (nearest degree) H
40 miles
24 miles B L
57 miles A

10. The Cosine Rule a2 = b2 + c2 – 2bcCosA P Q W Not to Scale
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 530 miles North to a point (P) as shown, It then turns left and flies to a point (Q), 670 miles away. Finally it flies back to base, a distance of 520 miles.
Find the bearing of Q from point P. P
670 miles W
530 miles
Not to Scale Q
520 miles

11. The Cosine Rule. (Used for Non-Right Angled Triangles)B
a2 = b2 + c2 – 2bc Cos A
Why can’t you use the Sine Rule to calculate x ?
3 cm
In order to use the
Sine rule you need
an angle and its corresponding side. c a x
x2 = – 2 x 4 x 3 x Cos 600
Because you will
have 2 unknowns !
x2 = 25 – 24 Cos 600
600 b
x2 = 13 C A
4 cm b a =
x = 3.6 cm to 1 d.p.
Sin A
Sin B
‘A’ is the given angle 11
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12. We can re-arrange the formula :
a2 = b2 + c2 – 2bc Cos A
We can re-arrange the formula :
Cos x = 82 – 52 – 42
Cos A = a2 – b2 – c2
-2 x 5 x 4
5 cm
8 cm
-2bc
etc. x0
4 cm 12
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13. 1) 2) 3) 6) 4) 5) Calculate The Missing Angles (to 1 d.p.)
And Missing Lengths ( to 3 sig.figs).
10.5 cm
8.95 cm 1) 2) 3)
10.9 cm
700
1120
14 cm x
6 cm
8 cm
9 cm
5 cm
300 x
20 cm x
88.2 cm
100.30
93.80 6) 4) 5) x
25 cm
32 cm
6 cm
10 cm
6 cm
3 cm x x
5 cm
40 cm
7 cm 13
Answers
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14. know the Perpendicular Height ?
Area of A Triangle.
But what if we don’t
know the Perpendicular
Height ?
A = 1 x Base x Perpendicular Height 2
15 cm2
5 cm
6 cm
4 cm
24 cm2
12 cm 14
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15. The Cosine Rule A B C a c b
Pythagoras’ Theorem allows us to calculate unknown lengths in right-angled triangles using the relationship a2 = b2 + c2
It would be very useful to be able to calculate unknown sides for any value of the angle at A. Consider the square on the side opposite A when angle A is not a right-angle.
a2 = b2 + c2 A a2 1 A
Angle A obtuse a2 2 A
Angle A acute a2 3
01/06/2018
a2 > b2 + c2 15
a2 < b2 + c2

16. *Since Cos A = x/c  x = cCosA
The Cosine Rule
The Cosine Rule generalises Pythagoras’ Theorem and takes care of the 3 possible cases for Angle A.
a2 > b2 + c2
a2 < b2 + c2
a2 = b2 + c2 A 1 2 3 A B C a b c
Consider a general triangle ABC. We require a in terms of b, c and A.
Deriving the rule
BP2 = a2 – (b – x)2
Also: BP2 = c2 – x2
a2 – (b – x)2 = c2 – x2
a2 – (b2 – 2bx + x2) = c2 – x2
a2 – b2 + 2bx – x2 = c2 – x2
a2 = b2 + c2 – 2bx*
a2 = b2 + c2 – 2bcCosA P x
b - x b
Draw BP perpendicular to AC
*Since Cos A = x/c  x = cCosA
When A = 90o, CosA = 0 and reduces to a2 = b2 + c2 1
Pythagoras
When A > 90o, CosA is negative,  a2 > b2 + c2 2
Pythagoras + a bit
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When A < 90o, CosA is positive,  a2 > b2 + c2 3
Pythagoras - a bit

17. Finding an unknown side.B C a b c
The Cosine Rule
The Cosine rule can be used to find:
1. An unknown side when two sides of the triangle and the included angle are given.
2. An unknown angle when 3 sides are given.
Finding an unknown side.
a2 = b2 + c2 – 2bcCosA
Applying the same method as earlier to the other sides produce similar formulae for b and c. namely:
b2 = a2 + c2 – 2acCosB
c2 = a2 + b2 – 2abCosC
01/06/2018 17

18. To find an unknown side we need 2 sides and the included angle.
a2 = b2 + c2 – 2bcCosA
The Cosine Rule
To find an unknown side we need 2 sides and the included angle.
Not to scale
8 cm
9.6 cm a 1.
40o 2.
7.7 cm
5.4 cm
65o m
85 m
100 m
15o 3. p
m2 = – 2 x 5.4 x 7.7 x Cos 65o
m = ( – 2 x 5.4 x 7.7 x Cos 65o)
m = 7.3 cm (1 dp)
a2 = – 2 x 8 x 9.6 x Cos 40o
a = ( – 2 x 8 x 9.6 x Cos 40o)
a = 6.2 cm (1 dp)
p2 = – 2 x 85 x 100 x Cos 15o
p = ( – 2 x 85 x 100 x Cos 15o)
p = 28.4 m (1 dp)
01/06/2018 18

19. a2 = b2 + c2 – 2bcCosA The Cosine Rule Application Problem L H B
A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left onto a bearing of 035o and sails to a lighthouse (L) 24 miles away. It then returns to harbour.
Make a sketch of the journey
Find the total distance travelled by the boat. (nearest mile)
HL2 = – 2 x 40 x 24 x Cos 1250
HL = ( – 2 x 40 x 24 x Cos 1250)
= 57 miles
Total distance = = 121 miles. H
40 miles
24 miles B L
125o
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20. The Cosine Rule a2 = b2 + c2 – 2bcCosA P Q W Not to Scale
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 430 miles North to a point P before turning left onto a bearing of 230o to a second point Q, 570 miles away. It then returns to base.
(a) Make a sketch of the flight.
(b) Find the total distance flown by the aircraft. (nearest mile)
The Cosine Rule
a2 = b2 + c2 – 2bcCosA
Not to Scale P
570 miles W
430 miles Q
50o
QW2 = – 2 x 430 x 570 x Cos 500
QW = ( – 2 x 430 x 570 x Cos 500)
= 441 miles
Total distance = = 1441 miles.
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21. A B C a b c
The Cosine Rule
To find unknown angles the 3 formula for sides need to be re-arranged in terms of CosA, B or C.
a2 = b2 + c2 – 2bcCosA
b2 = a2 + c2 – 2acCosB
c2 = a2 + b2 – 2abCosC
Similarly
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22. To find an unknown angle we need 3 given sides.
The Cosine Rule
To find an unknown angle we need 3 given sides.
Not to scale
8 cm
9.6 cm
6.2 1. A 2.
7.7 cm
5.4 cm P
7.3 cm
85 m
100 m 3. R
28.4 m
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23. The Cosine Rule Application Problems L H B 57 miles 24 miles A
A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left and sails for 24 miles to a lighthouse (L). It then returns to harbour, a distance of 57 miles.
Make a sketch of the journey.
Find the bearing of the lighthouse from the harbour. (nearest degree) H
40 miles
24 miles B L
57 miles A
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24. The Cosine Rule a2 = b2 + c2 – 2bcCosA P Q W Not to Scale
An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 530 miles North to a point (P) as shown, It then turns left and flies to a point (Q), 670 miles away. Finally it flies back to base, a distance of 520 miles.
Find the bearing of Q from point P. P
670 miles W
530 miles
Not to Scale Q
520 miles
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