1. © T Madas

2. The Cosine Rule

3. © T Madas A B C a b c a2a2 = b2b2 + c2c2 – 2 b ccosA b2b2 = a2a2 + c2c2 – 2 a ccosB c2c2 = a2a2 + b2b2 – 2 a bcosC The cosine rule in any triangle

4. © T Madas R P Q r p q p2p2 = r2r2 + q2q2 – 2 r qcosP The cosine rule

5. © T Madas x 9 45° 65° 70° d 9 8 d 2d 2 = 9292 + 8282 – 2 x 8 x cos45° The cosine rule

6. © T Madas x 4 41° 68° 71° x 4 7 x 2x 2 = 4242 + 7272 – 2 x 7 x cos68° The cosine rule

7. © T Madas 45° 72° 63° x 6 cm 8 cm Calculate the missing length in the triangle below: Using The Cosine Rule

8. © T Madas Calculate the missing length in the triangle below: 45° 72° 63° x 6 cm 8 cm x 6 x 2x 2 = 6262 + 8282 – 2 x 8 x cos72 ° x 2x 2 = 36 + 64 – 96cos72 ° x 2x 2 = 100 – 96cos72 ° x = x ≈ 8.4 cm

9. © T Madas Calculate the missing angle in the triangle below: θ 5 cm 6 cm 8 cm

10. © T Madas The Sine Rule

11. © T Madas A B a b c The Sine Rule C in any triangle

12. © T Madas Calculate the missing length in the triangle below: 45° 72° 63° x 6 cm Using The Sine Rule

13. © T Madas Calculate the missing length in the triangle below: 45° 72° 63° x 6 cm

14. © T Madas Calculate the missing length in the triangle below: 35° 75° 70° x 8 cm

15. © T Madas Calculate the missing angle in the triangle below: 35° x 7 cm 5 cm

16. © T Madas

17. A soldier walked from his base for 4 km on a bearing of 060° to a point A. He then walked a further 5 km due east to a point B. Find: 1. The distance of point B from the base. 2. The bearing of B as measured from the base. 3. The bearing of the base as measured from B. Base N 4 060° A5B C 30° d 150°

18. © T Madas Base N 060° A5B C 30° d 150° – 40 x 5 d 2d 2 = 5252 + 4242 – 2x 4x cos150° By the cosine rule on ABC c d 2d 2 = 25 + 16 cos150° c d 2d 2 ≈ 75.64 c d ≈ 8.7 km 4 8.7

19. © T Madas A soldier walked from his base for 4 km on a bearing of 060° to a point A. He then walked a further 5 km due east to a point B. Find: 1. The distance of point B from the base. 2. The bearing of B as measured from the base. 3. The bearing of the base as measured from B. Base N 4 060° A5B C 30° 150° 8.7 8.7 km θ

20. © T Madas Base N 4 060° A5B C 30° 150° 8.7 θ By the sine rule on ABC : sinθ 5 sin150° 8.7 = c 5sin150° 8.7 sinθ = c x 5 5 x 0.287 sinθ ≈ c sin -1 (0.287) θ ≈ c 17° θ ≈

21. © T Madas Base N 4 060° A5 C 30° 150° 8.7 A soldier walked from his base for 4 km on a bearing of 060° to a point A. He then walked a further 5 km due east to a point B. Find: 1. The distance of point B from the base. 2. The bearing of B as measured from the base. 3. The bearing of the base as measured from B. 8.7 km 077° 17° B is at a bearing of 077° from the base 257° B 77°

22. © T Madas