1. Topic 2 Periodic Functions and Applications I

2.  trigonometry – definition and practical applications of the sin, cos and tan ratios  simple practical applications of the sine and cosine rules

3. Model Calculate the value of the pronumeral in each of the following: cos x = 12 19 x = cos -1 12 19 = 50  50’ X° 19 cm 12 cm

4. Page 33 Exercise 2.1

5. Find the perimeter of the figure below.

6.  Perimeter = 30+43+30+51+67 = 221m

7. Page 36 Exercise 2.2 No. 1 – 6 Page 42 Exercise 2.4 No. 1, 3, 5 - 8 Worked solution Ex 2.4 No. 6

8. The Sine Rule A a b B c C

9. Model Find x in the following 85  c = x 65  A B C b = 15

10. Example 11, page 46 (Modelling and Problem Solving) Jermaine sees the top of a sand dune at an angle of 16°. On walking 40m closer, she finds the angle of elevation increases to 22°. What is the height of the sand dune? 16°22° 40m h A C B D * If we knew the length of line DB, we could easily work out the height.

11. 16°22° 40m h A C B D Consider triangle ABD 16° 6°6° 158° 40m d b a

12. 16°22°A B D C h 105.5m

13. Page 46 Exercise 2.5 N0. 1, 2(b,c,d,e), 3, 4, 7-9

14. The Cosine Rule A a b B c C Similarly

15. Model Find the length of the unknown side in the triangle below: 12 17 32  A B C c = b= a

16. Model Find the size of the largest angle in the triangle below: 12 17 A B C c = b= a=8 N.B. Largest angle is always opposite the largest side

17. Page 50 Exercise 2.6

18. Examples 15 & 16, page 51 & 52 Page 53 Exercise 2.7 No. 1(a,b,c,f), 2(a,b,c,f), 4, 7, 11-15 Worked solution Ex 2.7 No. 13

19. 3D Applications Example 17: page 56 - From point A, a mountain point due north is at an elevation of 20°. From point B, 2Km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above A and B.

20. Consider triangle ABM Angle BAM is a right angle

21. - From point A, a mountain point due north is at an elevation of 20°. From point B, 2Km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above A and B. Consider triangle ABM A M B 50° 2 Km Side AM, which is common to both triangles, can now be calculated

22. - From point A, a mountain point due north is at an elevation of 20°. From point B, 2Km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above A and B. Now consider triangle AMP A P M 20° 2 tan50

23. Example 18: page 57 John observes that the top of a transmission tower at bearing 038° is at an elevation of 12°. Mary is 575m due east of John and can see the tower on a bearing of 295°. What is the height of the tower?

24. Example 18: page 57 John observes that the top of a transmission tower at bearing 038° is at an elevation of 12°. Mary is 575m due east of John and can see the tower on a bearing of 295°. What is the height of the tower? 52° 25° 103° FH W 575m We need side WH (f) f w =

25. Example 18: page 57 John observes that the top of a transmission tower at bearing 038° is at an elevation of 12°. Mary is 575m due east of John and can see the tower on a bearing of 295°. What is the height of the tower? 12° T HW 249.4m We can now calculate tower height

26. Page 59 Exercise 2.8 Worked solution Ex 2.7 No. 13