1. Trigonometry Learning Objectives:
Grade B
08/11/2018
Learning Objectives:
Able to label a right angled triangle
correctly
Able to find the length of a side using trigonometry
Able to find an angle using trigonometry

2. Located between the right angle and the included angle
Trigonometry
HYPOTENUSE
OPPOSITE
Located opposite the included angle
Greek letter ‘Pheta’ – represents the angle Ѳ
ADJACENT
Located between the right angle and the included angle

3. Which side? A B D C F E

4. SOH CAH TOA Trig Ratios Sine Ratio: SinѲ = Cosine Ratio: CosѲ =
Tangent Ratio:
TanѲ =
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
Adjacent
Hypotenuse
Hypotenuse
Adjacent
Opposite
Adjacent
Opposite
SOH
CAH
TOA

5. Which Ratio?
SOH CAH TOA x
35o 7

6. Which ratio?
SOH CAH TOA x 10
40o

7. Which ratio?
SOH CAH TOA
35o x 20

8. Which ratio?
SOH CAH TOA 12
87o 4

9. Which ratio?
SOH CAH TOA 13 9
65o

10. Trigonometry SOH CAH TOA Opposite Hypotenuse sinѲ = sin63 = x
10cm
sinѲ =
sin63 =
x = 10sin63
x = 8.91cm (2dp) x x 10
63o

11. Trigonometry SOH CAH TOA Opposite Hypotenuse sinѲ = sin49 = 8
ysin49 = 8
y =
y = 10.60m (2dp) 8m 8 y
49o 8
sin49

12. Trigonometry SOH CAH TOA Opposite Hypotenuse sinѲ = 7 Ѳ = sin-1 12 Ѳ
Ѳ = 35.69o (2dp) 7m 7 12 Ѳ 7 12

13. Trigonometry Ѳ 𝜃 15cm 21cm 9cm 27o 71 𝑜 c cm b cm 17cm a cm 38o 11m
d cm
11cm 𝜃
14cm
84 𝑜

14. Starter
2cm
11cm a 𝜃
27 𝑜
14cm 𝑥
79 𝑜
31cm

15. Angles of Elevation and Depression
08/11/2018
Grade B
Learning Objectives:
Able to find the length of a side using trigonometry
Able to find an angle using trigonometry
Able to apply this to worded problems including angles of elevation and
depression

17. Angles of Elevation and Depression
Angles of Elevation: the angle between the horizontal and an object above the observer’s line of sight Angle of Depression: the angle between the horizontal and an object below the observer’s line of sight

18. Angles of Elevation How can I work out the height of the London Eye?
Opposite
Adjacent
How can I work out the height of the London Eye?
tanѲ =
tan50 =
h = 113.3tan50
h = m (2dp) h
113.3
50o
113.3m

19. Angles of Elevation
If I stand 140m from Horseguard’s Parade, the building is 60m tall, what is my angle of elevation?

20. Angles of Depression
I want to work out how far away the beach is from where I’m standing. I know I’m currently 53m above sea level.
I need to know my angle of depression.
Opposite
Adjacent
tanѲ =
tan21 =
𝑥 = 53 𝑡𝑎𝑛21
𝑥 = 138.1m (2dp) 53 𝑥 𝑥
21o
53m

21. Angles of Elevation and Depression
EXAMPLE 1: A boat is at sea and sees the light of a lighthouse in the distance. The captain knows the lighthouse is 86m tall. Given his angle of elevation when looking at the light is 13o, how far away is the boat from the lighthouse? EXAMPLE 2: While cleaning a hall, a cleaner spots a mouse on the floor 12m away. Given the woman is 1.6m tall, what is her angle of depression?

22. Some practice Complete the worksheet in your workbooks
You may find it useful to sketch the triangles
Show all working out!

23. Starter
Finish the questions from yesterday’s lesson

24. Trigonometry in 3D Learning Objectives:
Grade A
08/11/2018
Learning Objectives:
Able to calculate a side or angle in a right-angled triangle using trigonometry
Able to solve worded problems involving trigonometry
Able to calculate an angle or side in a 3D shape

25. Trigonometry in 3D
The diagram shows a pyramid with a rectangular base. Calculate the angle that OA makes with the horizontal plane. O O
Draw the triangle we need to use.
24cm
24cm
We need to find AM first. B A M C A
10cm M
13cm D

26. Trigonometry in 3D O Since AM is half of AC, let’s calculate that.
AC = 16.4cm (1dp) C
24cm B
13cm A M C
10cm
13cm D D A
10cm
Therefore AM = 8.2cm

27. Trigonometry in 3D O O 𝑐𝑜𝑠𝜃= 8.2 24 𝜃= 𝑐𝑜𝑠 −1 8.2 24 𝜃= 70.0 𝑜 (1𝑑𝑝)
𝑐𝑜𝑠𝜃=
𝜃= 𝑐𝑜𝑠 −
𝜃= 70.0 𝑜 (1𝑑𝑝)
24cm
24cm B A 𝜃 M C A
10cm M
8.2cm
13cm D

28. Trigonometry in 3D
The image below is of a cuboid. Calculate the size of the angle between:
(a) CH and CDGH (b) CE and EFGH B C C D
𝑡𝑎𝑛𝜃= 11 3
𝜃= 𝑡𝑎𝑛 −
𝜃= 74.7 𝑜 (1𝑑𝑝) A
11cm
11cm F G 𝜃
3cm E H H G
4cm
3cm

29. Trigonometry in 3D
The image below is of a cuboid. Calculate the size of the angle between:
(a) CH and CDGH (b) CE and EFGH B C C D
𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒
𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝐸𝐺 𝑓𝑖𝑟𝑠𝑡. A
11cm
11cm F G 𝜃
3cm E H E G
4cm

30. Trigonometry in 3D
The image below is of a cuboid. Calculate the size of the angle between:
(a) CH and CDGH (b) CE and EFGH G B C C
3cm D A E H
4cm
EG2 =
EG2 = 25
EG = √25
EG= 5cm
11cm
11cm F G 𝜃
3cm E H E G
4cm

31. Trigonometry in 3D
The image below is of a cuboid. Calculate the size of the angle between:
(a) CH and CDGH (b) CE and EFGH B C C D
𝑡𝑎𝑛𝜃= 11 5
𝜃= 𝑡𝑎𝑛 −
𝜃= 65.6 𝑜 (1𝑑𝑝) A
11cm
11cm F G 𝜃
3cm E H E G
4cm
5cm

32. Trigonometry in 3D – Your turn…
The triangular prism below is made up from equilateral triangles and rectangles. F
Calculate the size of the angle between:
EC and ABCD
AE and ADE E B C
41cm A D
25cm

33. Trigonometry in 3D – Your turn…
Pages 555 – 556
Exercise 29B
Q1 to 3 Q4
Q5 and 6

34. Area of a Triangle Learning Objectives:
Grade A
08/11/2018
Learning Objectives:
Able to calculate missing angles and sides in
right-angled triangles
Able to calculate the area of a triangle using trigonometry
Able to calculate a side or angle given the area

35. Area of a triangle = 𝟏 𝟐 𝒂𝒃𝒔𝒊𝒏𝑪B
𝑠𝑖𝑛𝐶= ℎ 𝑎
ℎ=𝑎𝑠𝑖𝑛𝐶
Therefore:
Area = 1 2 𝑏×𝑎𝑠𝑖𝑛𝐶
Area of a triangle = 𝟏 𝟐 𝒂𝒃𝒔𝒊𝒏𝑪 c a h A C b

36. Area of a triangle Area of a triangle = 𝟏 𝟐 𝒂𝒃𝒔𝒊𝒏𝑪
C is the “included” angle – the angle between the two given sides
Area of a triangle = 𝟏 𝟐 𝒂𝒃𝒔𝒊𝒏𝑪 = 𝟏 𝟐 𝒃𝒄𝒔𝒊𝒏𝑨= 𝟏 𝟐 𝒂𝒄𝒔𝒊𝒏𝑩

37. Area of a triangle Calculate the area of the triangle, correct to 1dp.
Area = 121.7cm2 A
21cm
43 𝑜 C B
17cm

38. Area of a triangle
Given the area of the triangle below is 97cm2, calculate the value of 𝜃.
Area = 1 2 𝑎𝑏𝑠𝑖𝑛𝐶
97 = 1 2 ×14×15×𝑠𝑖𝑛𝜃
97 = 105sin𝜃
= sin𝜃
𝜃= 𝑠𝑖𝑛 −
𝜃= 67.5 𝑜 (1𝑑𝑝)
15cm 𝜃
14cm

39. Area of a triangle
The image below shows the sector of a circle, centre O. Calculate the area of the shaded section.
Area of shaded section = Area of sector – Area of triangle
Area of sector = 𝜋 ×𝑟 2 × 𝜃 360
Area of sector = 𝜋× 8 2 ×
Area of sector = 𝒄𝒎 𝟐 (3dp)
Area of triangle = 1 2 𝑎𝑏𝑠𝑖𝑛𝐶
Area of triangle = 1 2 ×8×8×𝑠𝑖𝑛68
Area of triangle = 𝒄𝒎 𝟐 (3dp)
8cm
68 𝑜
8cm
Area of shaded region = –
Area of shaded region = 8.3 𝒄𝒎 𝟐 (1dp)

40. Area of a triangle – your turn…
Pages 562 – 563
Exercise 29D
Q1 – 6

41. Starter B C D A G 𝑥 F 2cm 4cm 7cm 29 𝑜 E H 3cm
Calculate the value of 𝑥.
Calculate the value of the angle formed between BH and EFGH. J
The area of triangle MNO is 36 𝑐𝑚 2 . Calculate the value of 𝑥. M
18cm
7cm
41 𝑜
52 𝑜 K
6cm L O 𝑥 N
Calculate the area of 𝐽𝐾𝐿.

42. The Sine Rule Learning Objectives:
08/11/2018
Grade A
Learning Objectives:
Able to calculate the area of a triangle
Able to calculate a missing side using the
Sine rule
Able to calculate a missing angle using the

43. The Sine Rule Think back to the formulae for the area of a triangle:
1 2 𝑎𝑏𝑠𝑖𝑛𝐶= 1 2 𝑎𝑐𝑠𝑖𝑛𝐵
We can cancel the 1 2 s and the a’s to leave us with:
𝑏𝑠𝑖𝑛𝐶=𝑐𝑠𝑖𝑛𝐵
Re-arranging, gives us the Sine Rule:
𝑏 𝑠𝑖𝑛𝐵 = 𝑐 𝑠𝑖𝑛𝐶

44. The Sine Rule 𝑎 𝑠𝑖𝑛𝐴 = 𝑏 𝑠𝑖𝑛𝐵 = 𝑐 𝑠𝑖𝑛𝐶
The Sine Rule allows us to calculate a missing side or angle for ANY triangle
We need two sides and an angle (SSA) or two angles and a side (ASA)

45. The Sine Rule Calculate the length 𝑥: 𝑎 𝑠𝑖𝑛𝐴 = 𝑏 𝑠𝑖𝑛𝐵
𝑥 𝑠𝑖𝑛86 = 12 𝑠𝑖𝑛56
𝑥= 12𝑠𝑖𝑛86 𝑠𝑖𝑛56
𝑥=14.4𝑐𝑚 (1𝑑𝑝) 𝑥
12cm
86 𝑜
56 𝑜

46. The Sine Rule Calculate the length 𝑦: 𝑎 𝑠𝑖𝑛𝐴 = 𝑏 𝑠𝑖𝑛𝐵 6.3cm 42 𝑜
𝑦 𝑠𝑖𝑛42 = 6.3 𝑠𝑖𝑛43
𝑦= 6.3𝑠𝑖𝑛42 𝑠𝑖𝑛43
𝑦=6.2𝑐𝑚 (1𝑑𝑝)
6.3cm
42 𝑜 𝑦
43 𝑜

47. The Sine Rule
When calculating an angle, we can use the reciprocal of the formula:
𝑠𝑖𝑛𝐴 𝑎 = 𝑠𝑖𝑛𝐵 𝑏 = 𝑠𝑖𝑛𝐶 𝑐

48. The Sine Rule Calculate the missing angle below: 27 𝑜 15.4cm 𝑚 𝑜 7cm
𝑠𝑖𝑛𝐴 𝑎 = 𝑠𝑖𝑛𝐵 𝑏
𝑠𝑖𝑛𝑚 15.4 = 𝑠𝑖𝑛27 7
𝑠𝑖𝑛𝑚= 15.4𝑠𝑖𝑛27 7
𝑚= 𝑠𝑖𝑛 − 𝑠𝑖𝑛27 7
𝑚=87.2 𝑜 (1𝑑𝑝)
27 𝑜
15.4cm
𝑚 𝑜
7cm

49. The Sine Rule Calculate the missing angle below: 39 𝑜 8.4cm 𝑥 𝑜 5.6cm
𝑠𝑖𝑛𝐴 𝑎 = 𝑠𝑖𝑛𝐵 𝑏
𝑠𝑖𝑛𝑥 8.4 = 𝑠𝑖𝑛39 5.6
𝑠𝑖𝑛𝑥= 8.4𝑠𝑖𝑛39 5.6
𝑥= 𝑠𝑖𝑛 − 𝑠𝑖𝑛39 5.6
𝑥=70.7 𝑜 (1𝑑𝑝)
39 𝑜
8.4cm
𝑥 𝑜
5.6cm

50. The Sine Rule – Your Turn
Page 565, Exercise 29E
Question 1
Page 566, Exercise 29F
Page , Exercise 29F
Questions 2 – 5

51. Starter Calculate the missing side or angle below. 81 𝑜 4cm 9cm
Area = 8 𝑐𝑚 2
23 𝑜 𝜃 𝑥
7cm
14cm
63 𝑜
13cm
𝑦 𝑜

52. The Cosine Rule Learning Objectives:
08/11/2018
Grade A
Learning Objectives:
Able to calculate missing sides or angles
using the Sine rule
Able to calculate a missing side using the Cosine rule
Able to calculate a missing angle using the Cosine rule

53. The Cosine Rule 𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐cosA
We can use the Cosine rule with ANY triangle to calculate a side or angle
We need to be given two sides and an included angle (SAS)

54. The Cosine Rule Calculate the value of 𝑥. 𝑥 𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐cosA
𝑥 2 = − 2×12×10×𝑐𝑜𝑠44
𝑥 2 = −240𝑐𝑜𝑠44
𝑥 2 =71.358…
𝑥=8.5𝑐𝑚 (1𝑑𝑝)
10cm
12cm
44 𝑜

55. The Cosine Rule Calculate the value of 𝑦. 𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐cosA 4cm
𝑦 2 = − 2×4×6×𝑐𝑜𝑠156
𝑦 2 =16+36−48𝑐𝑜𝑠156
𝑦 2 =95.850…
𝑦=9.8𝑐𝑚 (1𝑑𝑝)
4cm
156 𝑜 𝑦
6cm

56. The Cosine Rule Calculate the size of angle 𝑚. 9cm 11cm 𝑚 𝑜 7cm
𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐cosA
11 2 = − 2×7×9×𝑐𝑜𝑠𝑚
121=81+49−126𝑐𝑜𝑠𝑚
121=130−126𝑐𝑜𝑠𝑚
−9=−126𝑐𝑜𝑠𝑚
9 126 =𝑐𝑜𝑠𝑚
𝑚= 𝑐𝑜𝑠 −
𝑚= 85.9 𝑜 (1𝑑𝑝)
9cm
11cm
𝑚 𝑜
7cm

57. The Cosine Rule Calculate the size of angle 𝑟. 7.2cm 3.7cm 𝑟 𝑜 5.2cm
𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐cosA
7.2 2 = − 2×5.2×3.7×𝑐𝑜𝑠𝑟
51.84= −38.48𝑐𝑜𝑠𝑟
51.84=40.73−38.48𝑐𝑜𝑠𝑟
11.11=−38.48𝑐𝑜𝑠𝑟
11.11 −38.48 =𝑐𝑜𝑠𝑟
𝑟= 𝑐𝑜𝑠 − −38.48
𝑟= 𝑜 (1𝑑𝑝)
7.2cm
3.7cm
𝑟 𝑜
5.2cm

58. The Cosine Rule – your turn…
Pages 568 – 569, Exercise 29G
Question 1
Page 570, Exercise 29H
Pages 570 – 571, Exercise 29H
Question 2, 3
Page 571, Exercise 29H
Questions 4 – 6

59. Starter
Should you use the Sine or Cosine rule to find the missing side/angle in the following questions? Why?
67 𝑜
𝑥 𝑜
13cm
4cm
4.4cm
9cm
7cm
81 𝑜
74 𝑜
11cm
𝑥 𝑜
51 𝑜
8cm
55 𝑜 𝑥
10cm
5.2cm 𝑥
84 𝑜 𝑥

60. Trigonometry Learning Objectives: Able to determine which rule to use
08/11/2018
Grade A/A*
Learning Objectives:
Able to determine which rule to use
Able to find the missing angle or side for
any triangle
Able to apply what you’ve learnt on
Trigonometry to difficult worded problems

61. Trigonometry Work through the worksheet on mixed trigonometry problems
In pairs, complete the Trigonometry Tarsia exercise