1. Units 3 and 4 Further Maths Exam Revision Lecture 2 presented by Mr Bohni

2. The Exam Structure Both end of year exams will be broken up into two sections. Section A is made up of the core topics and must be answered by all students Section B is made up of 6 modules but you only have to answer 3 of them

3. The Modules The 6 Modules you can choose from are Number Patterns Geometry and Trigonometry Graphs and Relations Business Related Mathematics Networks and Decision Mathematics Matrices

4. The Modules The 6 Modules you can choose from are Number Patterns Geometry and Trigonometry Graphs and Relations Business Related Mathematics Networks and Decision Mathematics Matrices You will Choose...

5. The Modules The 6 Modules you can choose from are Number Patterns Geometry and Trigonometry Graphs and Relations Business Related Mathematics Networks and Decision Mathematics Matrices You will Choose...

6. The Modules The 6 Modules you can choose from are Number Patterns Geometry and Trigonometry Graphs and Relations Business Related Mathematics Networks and Decision Mathematics Matrices You will Choose...

7. The Modules The 6 Modules you can choose from are Number Patterns Geometry and Trigonometry Graphs and Relations Business Related Mathematics Networks and Decision Mathematics Matrices You will Choose... And you shall ignore the rest!!!

8. What this lecture will cover Today we will quickly cover the important parts from the three modules, with particular focus on areas that students commonly make mistakes on...

9. Geometry and Trigonometry

10. Geometry and Trigonometry is generally either very well done by students or not well done by students. Common issues keep cropping up in the exams each year. These include such things as – Bearings – Surface Area – Scale Factor

11. Bearings Bearing questions are best solved by using detailed diagrams to represent the situation. !!DRAW BIG CLEAR DIAGRAMS!! There is nothing worse than going to the effort of drawing a picture only to struggle to be able to label it because there isn’t enough room and then getting confused because you can’t read your diagram correctly

12. Example 1 On a school camp a group of students need to navigate around a small island. The students start at Point A and walk for 5km on a bearing of 120 0 T to Point B. Then from Point B they walk 8km on a bearing of 260 0 T to Point C, before returning from Point C to Point A.

13. Example 1 On a school camp a group of students need to navigate around a small island. The students start at Point A and walk for 5km on a bearing of 120 0 T to Point B. Then from Point B they walk 8km on a bearing of 260 0 T to Point C, before returning from Point C to Point A. Begin by highlighting the important parts and drawing a picture

14. Example 1 On a school camp a group of students need to navigate around a small island. The students start at Point A and walk for 5km on a bearing of 120 0 T to Point B. Then from Point B they walk 8km on a bearing of 260 0 T to Point C, before returning from Point C to Point A. Begin by highlighting the important parts and drawing a picture

15. A B 5km N 120

16. A B 5km N 120 Remember to draw a compass over each point, it helps to determine the other angles

17. A B 5km N 120 N

18. On a school camp a group of students need to navigate around a small island. The students start at Point A and walk for 5km on a bearing of 120 0 T to Point B. Then from Point B they walk 8km on a bearing of 260 0 T to Point C, before returning from Point C to Point A. Example 1 Continue to highlight the important parts and add to the picture

19. A B 5km N 120 N C 8km 260

20. A B 5km N 120 N N C 8km 260

21. A B 5km N 120 N N C 8km 260

22. A B 5km N 120 N N C 8km 260 Before doing anything else, we can now add in the unknown angles to the picture in the following manner

23. A B 5km N 120 N N C 8km 260 60 If this is 120 degrees, then the remaining angle must be 60 degrees.

24. A B 5km N 120 N N C 8km 260 60 If this is 60 degrees, then so must this angle be 60 degrees 60

25. A B 5km N 120 N N C 8km 260 60 If that was 60 degrees, then this angle be must be 30 degrees 60 30

26. A B 5km N 120 N N C 8km 260 60 If this is 260 degrees then this angle here must be 10 degrees 60 30 10

27. A B 5km N 120 N N C 8km 260 60 This angle must be 10 degrees also 60 30 10

28. A B 5km N 120 N N C 8km 260 60 Cleaning up, the total of this inside angle is 30 + 10 = 40 degrees 60 40 10

29. Now we can begin to answer some questions How far away from Point A is Point C? What is the bearing of Point A from Point C? and... What is the Area of the triangle contained by the lines joining Points A, B and C?

30. A B 5km N 120 N N C 8km 260 60 40 10 In order to determine the length of the line AC, we need to use a special rule...

31. A B 5km N 120 N N C 8km 260 60 40 10 The COS Rule

32. b 2 = a 2 + c 2 – 2ac Cos B

33. A B 5km N 120 N N C 8km 260 60 40 10 a = 8 c = 5 B = 40

34. b 2 = 8 2 + 5 2 – 2(8)(5) Cos 40 b 2 = 64 + 25 – 80 x 0.77 b 2 = 89 – 61.3 b 2 = 27.7 b = √ 27.7 b = 5.26 km

35. A B 5km N 120 N N C 8km 260 60 40 10 5.26km

36. So now how do we find the bearing of Point A from Point C? Remember that the bearing is the angle taken from North in a clockwise direction. So lets take another look at our diagram...

37. A B 5km N 120 N N C 8km 260 60 40 10 5.26km The Bearing angle of A from C must be this angle in here.

38. A B 5km N 120 N N C 8km 260 60 40 10 5.26km If only we knew this angle in here. Then we could add it to the 10 degrees below and subtract them both from 90 degrees to find the bearing.... Oh wait, I just remembered... we can find out this angle by using....

39. THE SINE RULE

40. We know Angle B is 40 degrees We know that the length of side b is 5.23km We know that the length of side c is 5km We are trying to find Angle C...

42. A B 5km N 120 N N C 8km 260 60 40 10 5.23km So now we know this is 37.9 degrees 37.9

43. A B 5km N 120 N N C 8km 260 60 40 10 5.23km The bearing is 90 – (37.9 + 10) = 42.1 degrees True 37.9

44. And what about the area of the Triangle? We have a few rules that allow us to find the area of the non right-angled triangle. Heron’s Formula uses the side lengths to determine the area A= √s(s-a)(s-b)(s-c) where s=(a+b+c)/2 The other formula we can use is A= ½ absinC

45. The Answer

46. Surface Area A common issue that seems to arise in exams is that students forget to calculate the area of ALL the sides of a shape. For example, in the case of a half sphere... 2cm

47. Most students are more than capable of working out the surface area of the sphere part What they forget is the top part of the sphere

48. 2cm If we had 12.57cm 2 of paint, we would only be able to cover this much of the half sphere. In order to get the top part, we need to add on the area of the circle that has a radius of 2cm

50. Scale Factors Another common problem encountered by students is scale factors. You can be guaranteed of getting a question that requires you to calculate the length or area of volume of a similar shape using a scale factor

51. Always draw a picture! A picture is worth a thousand words While I am not sure of the mathematics behind that statement, what it implies is true. A picture helps to clarify exactly what is going on and helps you to figure out how to answer the question.

52. Remember The scale factor for measurements of length is called k The scale factor for measurements of area is called k 2 The scale factor for measurement of volume is called k 3 Thus if you know the scale factor for the Volume, you can work out the scale factor for the lengths or for the areas and vice-versa.

53. Graphs and Relations

54. Constraint Inequations Students often struggle to identify the correct constraints on a problem. Be careful and take your time when you read the problems Practice these types of problems Keep practicing them AND PRACTICE LOTS OF THEM!

55. What are the constraints of the following? Griff is a Volleyball coach. He needs to take a squad of players to a tournament. He has two types of players, setters and spikers. The squad must contain at least 4 setters and at least 3 spikers. The squad must contain at least 8 players. Let x be the number of setters and y be the number of spikers. What are the constraints for this problem?

56. What are the constraints of the following? Griff is a Volleyball coach. He needs to take a squad of players to a tournament. He has two types of players, setters and spikers. The squad must contain at least 4 setters and at least 3 spikers. The squad must contain at least 8 players. Let x be the number of setters and y be the number of spikers. What are the constraints for this problem? x > 4

57. What are the constraints of the following? x > 4 y > 3 Griff is a Volleyball coach. He needs to take a squad of players to a tournament. He has two types of players, setters and spikers. The squad must contain at least 4 setters and at least 3 spikers. The squad must contain at least 8 players. Let x be the number of setters and y be the number of spikers. What are the constraints for this problem?

58. What are the constraints of the following? Griff is a Volleyball coach. He needs to take a squad of players to a tournament. He has two types of players, setters and spikers. The squad must contain at least 4 setters and at least 3 spikers. The squad must contain at least 8 players. Let x be the number of setters and y be the number of spikers. What are the constraints for this problem? x > 4 y > 3 x + y < 8

59. Graphing these inequations is important too So often, students do the right thing in obtaining the correct constraints but then when graphing them, highlight the incorrect region. Remember, draw the line and then choose a point that is clearly on one side of the line. Check to see if this point satisfies the constraint. – If it does, then shade everything on the same side of the line as the dot – If it doesn’t, shade everything on the other side of the line to the dot

60. 3x-5y < 15

61. Choose a point that is clearly on one side of the line

62. 3x-5y < 15 Choose a point that is clearly on one side of the line

63. 3x-5y < 15 It doesn’t matter which point I choose, but there is one that makes things easier for me

64. 3x-5y < 15 This one!

65. 3x-5y < 15 This one! When we use (0,0) it makes our substitution into the inequation so simple

66. 3x-5y < 15 Watch....

67. 3x-5y < 15 3x – 5y < 15

68. 3x-5y < 15 3x – 5y < 15 Sub (0,0) in and we get

69. 3x-5y < 15 3x – 5y < 15 Sub (0,0) in and we get 3 (0) – 5 (0) < 15

70. 3x-5y < 15 3x – 5y < 15 Sub (0,0) in and we get 3 (0) – 5 (0) < 15 0 – 0 < 15 0 < 15 Which is true, so we shade everything on the same side of the line as (0,0)

71. 3x-5y < 15 3x – 5y < 15 Sub (0,0) in and we get 3 (0) – 5 (0) < 15 0 – 0 < 15 0 < 15 Which is true, so we shade everything on the same side of the line as (0,0)

72. Business Related Mathematics

73. Finance Solver Your CAS Calculator is your greatest tool when it comes to Business Related Mathematics. The finance solver can be used for any problem that involves compound interest (which is most things). USE IT!

74. What do all the parts mean? N – is the number of years that the interest is being calculated over I(%) – is the interest rate per year PV – is the Primary Value (or Principle amount) Pmt - is the Payment amount FV - is the value of the loan/investment at the end of the period of time PpY - is the number of payments per year

75. When is a number positive or negative? When using the program, it is important to remember that any money that is given away (eg, investment, repayment, etc) is negative. Any money that is given to the person with the investment or loan is positive. So if I borrow $1000 from the bank, that $1000 is positive because I have it. All my repayments however, will be negative because I am giving that money back.

77. Use me!

78. Read the Problems Carefully When looking through the examiners reports of previous exams it becomes evident that so many students lose easy marks simply because they don’t read the question properly If they ask you to shade the solution region, don’t shade the areas that you don’t want and leave the solution region blank... If you do, you will get the question wrong!

79. SHOW ALL YOUR WORKING! You will lose a mark if you do the following a 2 = 25 a = 5 From the examiners point of view you have simply guessed the answer is a=5. In order to get full marks you need to write a 2 = 25 a = √25 a = 5

80. And most importantly When it comes to the multiple choice answers remember the following...

81. Multiple Choice The most common multiple choice answer for Further Maths is D, making up 25% of all answers. This was closely followed by C and B (each making up 21% of the answers). The least common answer was A. That being said... If you have absolutely no idea what the answer to a question is from the core, choose C If you have absolutely no idea what the answer is in Geometry and Trigonometry, choose D

82. Multiple Choice If you have absolutely no idea what the answer to a question is from the Graphs and Relations, choose C If you have absolutely no idea what the answer is in Business Related Mathematics, choose B

83. The End