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

*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

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

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 = 5.42 + 7.72 – 2 x 5.4 x 7.7 x Cos 65o m = (5.42 + 7.72 – 2 x 5.4 x 7.7 x Cos 65o) m = 7.3 cm (1 dp) a2 = 82 + 9.62 – 2 x 8 x 9.6 x Cos 40o a = (82 + 9.62 – 2 x 8 x 9.6 x Cos 40o) a = 6.2 cm (1 dp) p2 = 852 + 1002 – 2 x 85 x 100 x Cos 15o p = (852 + 1002 – 2 x 85 x 100 x Cos 15o) p = 28.4 m (1 dp)

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 = 402 + 242 – 2 x 40 x 24 x Cos 1250 HL = (402 + 242 – 2 x 40 x 24 x Cos 1250) = 57 miles Total distance = 57 + 64 = 121 miles. H 40 miles 24 miles B L 125o

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 = 4302 + 5702 – 2 x 430 x 570 x Cos 500 QW = (4302 + 5702 – 2 x 430 x 570 x Cos 500) = 441 miles Total distance = 1000 + 441 = 1441 miles.

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

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

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

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

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 = 42 + 32 – 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 Menu

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 Menu

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 Menu

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 Menu

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

*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 01/06/2018 16 When A < 90o, CosA is positive,  a2 > b2 + c2 3 Pythagoras - a bit

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

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 = 5.42 + 7.72 – 2 x 5.4 x 7.7 x Cos 65o m = (5.42 + 7.72 – 2 x 5.4 x 7.7 x Cos 65o) m = 7.3 cm (1 dp) a2 = 82 + 9.62 – 2 x 8 x 9.6 x Cos 40o a = (82 + 9.62 – 2 x 8 x 9.6 x Cos 40o) a = 6.2 cm (1 dp) p2 = 852 + 1002 – 2 x 85 x 100 x Cos 15o p = (852 + 1002 – 2 x 85 x 100 x Cos 15o) p = 28.4 m (1 dp) 01/06/2018 18

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 = 402 + 242 – 2 x 40 x 24 x Cos 1250 HL = (402 + 242 – 2 x 40 x 24 x Cos 1250) = 57 miles Total distance = 57 + 64 = 121 miles. H 40 miles 24 miles B L 125o 01/06/2018 19

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 = 4302 + 5702 – 2 x 430 x 570 x Cos 500 QW = (4302 + 5702 – 2 x 430 x 570 x Cos 500) = 441 miles Total distance = 1000 + 441 = 1441 miles. 01/06/2018 20

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 01/06/2018 21

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 01/06/2018 22

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 01/06/2018 23

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 01/06/2018 24