1. This triangle will provide exact values for
Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values
60º 2
60º
30º 2 3 1
This triangle will provide exact values for
sin, cos and tan 30º and 60º

2. ½ ½ 1 ∞ Exact Values x 0º 30º 45º 60º 90º 3 Sin xº Cos xº Tan xº 30º2
30º 3
Exact Values x 0º
30º
45º
60º
90º
Sin xº
Cos xº
Tan xº 1 ∞ 3

3. Exact Values 45º 2 1 1 45º 1 1 Consider the square with sides 1 unit
We are now in a position to calculate
exact values for sin, cos and tan of 45o

4. 1
45º 2
Exact Values
Tan xº
Cos xº
Sin xº
90º
60º
45º
30º 0º x 1 1

5. Naming the sides of a TriangleB a c C b A
26-Apr-17

6. 2 sides and the angle in between (SAS)
Key feature
To find the area
you need to know
2 sides and the angle in between (SAS)
Area of ANY Triangle
Remember:
Sides use lower case letters of angles
The area of ANY triangle can be found
by the following formula. B a C
Another version c b
Another version A
If you know C, a and b
If you know A, b and c
If you know B, a and c
26-Apr-17

7. Example : Find the area of the triangle.
Area of ANY Triangle
Example : Find the area of the triangle. B B
The version we use is a
20cm c C C
30o
25cm A A b

8. Example : Find the area of the triangle.
Area of ANY Triangle
Example : Find the area of the triangle. E
The version we use is d
10cm
60o f
8cm F e D

9. What Goes In The Box ? Key feature Remember (SAS)
Calculate the areas of the triangles below:
(1)
23o
15cm
12.6cm
A =36.9cm2
(2)
71o
5.7m
6.2m
A =16.7m2

10. Finding the Angle given the area
17 cm
95 cm2
12 cm P Q R
Find the size of angle P p r
Area = ½qrsinP
95 = ½ 17 × 12 sinP
95 = 102 sinP q
sinP = 95 ÷ = 0∙931
P = sin-1 0∙931 = 68∙6°
26-Apr-17

11. The Sine Rule The Rule The side opposite angle A is labelled a
The side opposite angle C is labelled c
The side opposite angle B is labelled b A B C c a b
The Rule

12. Calculating Sides Using The Sine RuleP
34o
41o x
10m Q R
Example 1
Find the length of x in this triangle. 10
sin 34° x
sin 41°
Now cross multiply.

13. Find the length of x in this triangle.
10m
133o
37o x D E F
Example 2
Find the length of x in this triangle. 10
sin 37° x
sin 133°
Cross multiply
= 12.14m

14. b c a Cross multiply Problem
The balloon is anchored to the ground as shown in the diagram.
Calculate the distance between the anchor points. A x
sin 35°
sin 75° 25
sin 70°
35°
25 m b c
75°
70°
Cross multiply
x m B C a

15. Find the unknown side in each of the triangles below:
x = 6.7cm
y = 21.8mm
(1)
12cm
72o
32o x
(2)
93o y
47o
16mm
(3)
87o
89m
35o a
(4)
143o g
12o
17m
a = 51.12m
g = 49.21m

16. Calculating Angles Using The Sine Rule
Example 1. ao
45m
23o
38m Z Y X
Find the angle ao 45
sin aº 38
sin 23º
Cross multiply
= 0.463
Use sin-1

17. Find the size of the angle bo
Example 2.
143o
75m
38m bo
Find the size of the angle bo
= 0.305

18. Calculate the unknown angle in the following:
(2)
14.7cm bo
14o
12.9cm
(1)
14.5m
8.9m ao
100o
bo = 16o
ao = 37.2o
(3)
93o
64mm co
49mm
c = 49.9o

19. Sine Rule
Basic Examples

20. In the examples which follow use the sine rule to find the required side or angle.B D
3.2 E
80°
5.8 C
5.7 A
74°
34°
4.6 A
76° C
Find AB
Find angle F F
9.6
Find angle B B

21. In the examples which follow use the sine rule to find the required side or angle.8 B C
72°
Find AB M 12 N
86° 7 T
Find angle N
13.8
127°
36° P R Q
Find PQ

22. A 9 B
75°
32° C
Find BC R
60° S
78°
3.8 T
Find RT I
27°
130°
4.1 G H
Find GI 8
39° 10 Y X Z
Find angle Z B
70° C 9 13 A
Find angle A

23. Sine Rule (Bearings)
Trigonometry

24. In the examples which follow use the sine rule
to find the required side or angle. 1.
Consider two radar stations Alpha and Beta.
Alpha is 140 miles west of Beta.
The bearing of an aero plane from Alpha is 032°
and from Beta it is 316°.
How far is the aeroplane from each of the radar stations?

25. Two ports Dundee and Stonehaven are 54 miles apart with Dundee approximately due south of Stonehaven.
The bearing of a ship at sea from Stonehaven is 098°
and from Dundee it is 048°.
How far is the ship from Dundee?
A ship is sailing north. At noon its bearing from a lighthouse is 240°. Five hours later the ship is 84km further north and its new bearing from the lighthouse is 290°.
How far is the ship from the lighthouse now?

26. Two ports P and Q are 35 miles apart with P due east of Q
Two ports P and Q are 35 miles apart with P due east of Q. The bearings of a ship from P and Q are 190° and 126° respectively.
a) What is the bearing of port Q from the ship?
b) How far is the ship from port P
The bearing of an aeroplane from Aberdeen is 068° and from Dundee it is 030°. Dundee is 69 miles south of Aberdeen.
How far is the aero plane from Aberdeen?

27. Two oil rigs are 80 miles apart with rig
B being 80 miles east of rig A.
The bearing of a ship from rig B is 194° and from A the bearing is 140°.
Find the distance of the ship from rig A.
At noon a ship lies on a bearing 070° from Dundee.
The ship sails a distance of 45 miles south and at 3pm its
new bearing from Dundee is 130°.
How far is the ship from Dundee now?
If the ship maintains the same speed,
how long will it take to return to Dundee?

28. An aeroplane leaves Leuchers airport and flies 80 miles north.
It then changes direction and flies 120 miles east.
The plane now turns onto a bearing of 126°
and flies a further 164 miles.
Calculate the bearing and distance of
the plane from Leuchers.

29. Question Solution 1 122.4 miles ; 103.8 miles 2 69.8 miles 3
95.0 kilometers 4
Bearing 306° ; 22.9 miles 5
56.0 miles 6
114.9 miles 7
48.8 miles ; 3 hours 15 minutes 8
236.2 miles ; bearing 269°
Note: the last question requires
the cosine rule as well as the sine rule.

30. The Cosine Rule can be used with ANY triangle
as long as we have been given enough information. B a c C b A
Given angle C
Given angle B
Given angle A
26-Apr-17

31. Identify sides a, b, c and angle Ao Write down the Cosine Rule for a
Using The Cosine Rule
Example 1 : Find the unknown side in the triangle below: x 5m
12m
43o A B C
Identify sides a, b, c and angle Ao
a = x
b = 5
c = 12
Ao =
43o
Write down the Cosine Rule for a
a2 = b2 + c2 - 2bc cos A
x2 = 52 +
122
- 2 x 5 x 12 cos 43o
Substitute values
x2 =
81.28
Square root to find “x”.
x = 9.02m

32. Using The Cosine Rule
137o
17.5 m
12.2 m y P Q R
Example 2 :
Find the length of side QR
Identify the sides and angle.
p = y
r = 12.2
q = 17.5
P = 137o
Write down Cosine Rule for p
p2 = q2 + r2 – 2pq cos P
Substitute
y2 = – 2 x 12.2 x 17.5 x cos 137o
y2 =
y = 27.7m

33. a2 = b2 + c2 – 2bc cosA p2 = 432 + 312 – 2 × 43 × 31 × cos78°
Find the length of the unknown side in the triangles: b
(1)
78o
43cm
31cm p A C a c B
a2 = b2 + c2 – 2bc cosA
p2 = – 2 × 43 × 31 × cos78°
p2 = 2255∙7
p = 47∙5 cm

34. Find the length of the unknown side in the triangles:
(2) 8m
5∙2m
38o m
m = 5∙05m
112º
17 mm
28 mm k
k = 37∙8 mm
(3)

35. Basic Examples 1. Finding a side.
Cosine Rule
Basic Examples Finding a side.
5.6cm
39° C B
4.7cm A
Find a C
Find c
6cm B
32° A 1. 2. A
Find b C
Find a
2.7cm B 3. C 4.
2.4cm
7.4cm
3.8cm
36°
104° B A

36. 5. 7. 6. 8. 10. 9. Find a. Find a 4.5cm 22° A C 1.7cm B 6cm A B 6.6cm
44° C B
8.7cm A
Find c
67° 7. 6.
8cm P 4 R Q
134°
Find q C M 8.
71°
3cm
5cm
10. N
40°
11.4cm T U
8.8cm
Find v V
Find m P 9.

37. Finding Angles Using The Cosine Rule
The Cosine Rule formula can be rearranged to allow us to find the size of an angle
This formula is cyclic, depending on the angle to be found

38. Finding Angles Using The Cosine Rule
Example 1 : Calculate the
unknown angle, xo . F d e
Label and identify angles and sides E D f
D = xo
d = 11
e = 9
f = 16
e 2 +
f 2 -
d 2
cos D =
Write the formula for cos D
2ef 92 +
162 -
112
cos x =
Substitute values into the formula.
2 x 9 x 16
cos x =
0.75
Use cos-1 0∙75 to find x
x = 41∙4o 38

39. Label and identify the sides and angle.
Example 2: Find the unknown angle in the triangle: B
Label and identify the sides and angle.
B = yo
b = 26
a = 13
c = 15 C A
a 2 +
c 2 -
b 2
cos B =
Write down the formula for cos B
2ac
132 +
152 -
262
cos y =
Substitute values
2 x 13 x 15
cos y =
- 0∙723
The negative tells you the angle is obtuse.
Use cos-1 -0∙723 to find y
y = 136.3o

40. Calculate the unknown angles in the triangles below:
(1)
12.7cm
7.9cm
8.3cm C A ao
ao = 37.3o B B
(2)
10m 7m 5m bo
bo =111.8o C D

41. Basic Examples 2. Finding an angle.
4.7cm C A A
3.6cm B
6cm
4.1 cm 1. 2.
5.6 cm C 3.
5cm
5.2cm
3.2cm
Find C
Find A
6cm
Find B B C A A
10.5cm B C
1.9cm C
4cm
2.7 cm 4. C
3.6cm 5. 6.
5.8 cm
7.4 cm B
2.4cm
4.7cm A A
Find B B
Find A
Find C

42. 6.9cm Y L
6cm X B A
4.5cm 8. 9.
8.7cm
2.8 cm 7. M
6.6cm
7cm
8cm
Find X
4.5cm
Find the
largest angle. Z K C
Find L M
3cm
11.9cm V T
5cm N
11.
12. Q
8.8 cm
4.5cm 4 4
9.6 cm
Find the
smallest angle Q
10. U R P
Find the
largest angle
7.6cm
Find all the angles

43. Cosine Rule or Sine Rule
How to determine which rule to use
Two questions
1. Do you know the length of ALL the sides? OR
SAS
2. Do you know 2 sides and the angle in between?
If YES to either of the questions then Cosine Rule
Otherwise use the Sine Rule

44. Calculate the size of x in each of these diagrams

45. The Sine Rule
Application Problems A D
The angle of elevation of the top of a building measured from point A is 25o. At point D which is 15m closer to the building, the angle of elevation is 35o Calculate the height of the building. T B
35o
25o
10o
36.5
145o
15 m
Angle TDA =
180 – 35 = 145o
Angle DTA =
180 – 170 = 10o

46. The Sine Rule
The angle of elevation of the top of a column measured from point A, is 20o. The angle of elevation of the top of the statue is 25o. Find the height of the statue when the measurements are taken 50 m from its base
180 – 110 = 70o
180 – 70 = 110o
Angle ATC =
180 – 115 = 65o
Angle BCA =
Angle ACT = B T C A
25o
65o
110o
20o
70o
=5.1 m
53.21 m 5o
50 m

47. The Cosine Rule Application Problems L N H B Bearing = 90 – 20 = 070°
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
Bearing = 90 – 20 = 070° N
20°

48. Not to Scale The Cosine Rule a2 = b2 + c2 – 2bcCosA P Q W
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
P = 48.7° (49°)
Bearing = = 229°

49. 1. Find the length of the third side of triangle ABC when
COSINE RULE
1. Find the length of the third side of triangle ABC when
i) b = 2, c = 5,  A = 60°
ii) a = 2, b = 5,  A = 65°
iii) a = 2, c = 5,  B = 115°
iv) b = 6, c = 8,  A = 50°
Find QR in ∆PQR in which PR = 4, PQ = 3,  P = 18°
Find XY in ∆ XYZ in which YZ = 25, XZ = 30,  Z = 162°
In ∆ ABC, a = 7, b = 4,  C = 53°. Calculate c.

50. 5. In ∆ ABC, b = 4·2, c = 6·5,  A = 24°. Calculate a.
In ∆ ABC, a = 1·64, c = 1·64,  B = 110°. Calculate b.
In ∆ ABC, a = 18·5, b = 22·6,  C = 72·3°. Calculate c.
8. In ∆ABC, a = 100, b = 120,  C = 15°. Calculate c.
9. In ∆ABC, b = 80, c = 100,  A = 123°. Calculate a.

51. 10. A town B is 20 km due north of town A and a town C
10. A town B is 20 km due north of town A and a town C is 15 km north-west of A. Calculate the distance between B and C.
11. Two ships leave port together. One sails on a course of 045° at 9 km/h and the other on a course of 090° at 12 km/h.
After 2h 30 min, how far apart will they be?
12. From a point O, the point P is 3 km distant on a bearing of 040° and the point Q is 5 km distant on a bearing of 123°.
What is the distance between P and Q ?

52. COSINE RULE (1) Solutions
1. Find the length of the third side of triangle ABC when
i) a2 =  225cos60° =  200·5 = 19
 a = 4·36
ii) c2 =  225cos65° =  200·4226
= 20·54763  c = 4·533
iii) b2 =  225cos115° =  225(0· )
= · = 37·45237
 b = 6·12
iv) a2 =  268cos50° =  960·64279
= 100  661·7076 = 38·29239
 a = 6·188

53. QR2 =  243cos18° =  240· = 2·17464
 QR = 1·4747
XY2 =  22530cos162° =  1500(0· )
= ·5848
= 2951·5848
 XY = 54·33
4. c2 =  274cos53°
= 31·2984
 c = 5·5945

54. 5. a2 = 4·22 + 6·52  24·26·5cos24°
= 17· ·25  554·60· = 10·0104
 a = 3·164
6. b2 = 1· ·642  21·641·64cos110°
= 22·6896  22·6896(0· ) = 7·2190
 b = 2·6868
7. c2 = 18· ·62  218·522·6cos72·3° = 598·778
 c = 24·50
8. c2 =  2100120cos15° = 1217·780
 c = 34·90

55. 9. a2 =  280100cos123° = ·225
 a = 158·47
North
10. BC2 =  21520cos45°
=  6000·7071
= 200·7359
BC = 14·17 km B
20 km C
11. QR2 = ·52
 23022·5cos45°
= ·25  954·59
= 451·6558
QR = 21·25 km
15 km
45°
West A
North Q
22·5 km
12. PQ2 =  235cos83°
=  300·12187
= 30·344
PQ = 5·51 km
45° R P
30 km

56. Cosine Rule
Bearings problems

57. Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships.
1. Ship1 [ 74km, 053° ] ; Ship2 [ 104km, 112° ] 
Port
Ship1
Ship2
North
112°
53°
59°

58. Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships.
2. Ship1 [ 56km,021° ] ; Ship2 [ 66km,090° ] 
Port
Ship1
Ship2
North

59. Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships.
3. Ship1 [ 83km,060° ] ; Ship2 [ 80km,159° ] 
Port
Ship1
Ship2
North

60. Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships.
4. Ship1 [ 50km,090° ] ; Ship2 [ 62km,142° ] 
Port
Ship1
Ship2
North

61. Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships.
5. Ship1 [ 80km,146° ] ; Ship2 [ 70km,190° ] 
Port
Ship1
Ship2
North

62. Cosine Rule- Bearings:-Two Ships.
In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships.
6. Ship1 [ 47km,180° ] ; Ship2 [ 54km,235° ] 
Port
Ship1
Ship2
North

63. Bearings Problems
Dundee is 84 km due south of Aberdeen.
A ship at sea is on a bearing of 078° from Aberdeen and 048° from Dundee.
How far is the ship from Dundee and from Aberdeen ?
An aeroplane is 240 km from an airport on a bearing of 100° while a helicopter is
165 km from the airport on a bearing of 212°. How far apart are the aircraft ?

64. Bearings Problems
A ship sails 80 km on a bearing of 060° from its home port.
It then sails 93 km on a bearing of 134°.
How far is it now from its home port ?
Glasgow airport is 73 km from Edinburgh airport and lies
to the west of Edinburgh airport.
The bearing of an aeroplane from Glasgow airport is 040° while
its bearing from Edinburgh airport is 300°.
How far is the aeroplane from each airport ?

65. Bearings Problems
A ship sails 74 km south from its port to a lighthouse.
It then sails 85 km on a bearing of 160°.
How far is the ship from the port ?
From a port P, ship A is 144 km distant on a bearing of 036° and
ship B is 97 km distant on a bearing of 114°.
What is the distance between the two ships ?

66. Bearings Problems
A ship sails 93 km on a bearing 054° and then
another 108 km on a bearing of 110°.
How far is it now from its starting point ?
8. Two radar stations Alpha and Beta pick up signals from an incoming aircraft.
Alpha is 40 km east of Beta and picks up the signals on a bearing of 300°.
Beta picks up the signals on a bearing of 070°.
How far is the aircraft from Alpha and from Beta ?

67. aeroplane Q is 170 km from the same airport on a bearing of 094°.
Bearings Problems
9. Aeroplane P is 200 km from an airport on a bearing of 208° while
aeroplane Q is 170 km from the same airport on a bearing of 094°.
How far apart are the two aeroplanes ?
10. The bearings and distances from Aberdeen of three oil platforms are :
a) 028° km b) 081° km c) 138° 97 km
A supply boat leaves Aberdeen, visits each platform and then
returns to Aberdeen.
Find the total length of the journey.