1. Chapter 17: Trigonometry
Use the theorem of Pythagoras to solve problems (2D only)
Use the trigonometric ratios sin, cos and tan to solve problems
Define sin 𝜃 and cos 𝜃 for all values of 𝜃
Define tan 𝜃
Work with trigonometric ratios in surd form
Use trigonometry to calculate the area of a triangle
Solve problems using the Sine and Cosine Rules (2D only)
Solve problems involving the area of a sector of a circle and the length of an arc

2. Right-Angled Triangles and Pythagoras’ Theorem
Pythagoras’ theorem: In a right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
𝑐 2
𝑎 2
90°
𝑐 2 = 𝑎 2 + 𝑏 2
(F and T: P16)
𝑏 2
A vertical flagpole is 15 m high. It is held firm by a wire of length 17 m fixed to its top and to a point on the ground. How far is it from the foot of the flagpole to the point on the ground where the wire is secured?
Let 𝑥 be the distance from the foot of the flagpole to the point where the wire is secured.
17 2 = 𝑥 2
289=225+ 𝑥 2
289−225= 𝑥 2
64= 𝑥 2
𝑥=8 m

3. Right-Angled Triangles and the Trigonometric Ratios
In a right-angled triangle, we have the following special ratios:
90°
opposite A
adjacent
sin 𝐴= opposite hypotenuse
cos 𝐴= adjacent hypotenuse
tan 𝐴= opposite adjacent
(F and T: P16)
In the following right-angled triangle, write down the value of each of the following ratios: sin A, cos A and tan A; also sin B, cos B and tan B.
90° B A 8 6 10
sin 𝐴= 8 10 = 4 5
sin 𝐵= 6 10 = 3 5
cos 𝐴= 6 10 = 3 5
cos 𝐵= 8 10 = 4 5
tan 𝐴= 8 6 = 4 3
tan 𝐵= 6 8 = 3 4

4. Finding the Length of a Side in a Right-Angled Triangle
Consider the triangle ABC shown below. If |AB| = 8 cm and |∠BAC| = 35°,
then find |BC|, correct to one decimal place. 8
35° A B C
Let |BC|=𝑥
8 sin 35° =𝑥 𝒙
sin 35° =
𝑥 8 =𝑥
∴|BC|=4.6 cm
Find the values for 𝒙 and 𝒚 in the diagram below, correct to two decimal places. 𝑥 8
40°
60° A C D B 𝑦
sin 40° =
𝑥 8
cos 60° =
𝑥 𝑦
8 sin 40° =𝑥
cos 60° =
5.14 𝑦
𝑥= …
𝑥= 5.14
𝑦= (calculator)

5. Using Trigonometry to Solve Practical Problems
Compass Directions
Angles of Elevation and Depression
The angle of elevation is the angle above the horizontal. N S
N Ɵ° W
N Ɵ° E Ɵ Ɵ E W Ɵ Ɵ
The angle of depression is the angle below the horizontal.
S Ɵ° W
S Ɵ° E
John is standing on a cliff top and observes a boat drifting towards the base of the cliff. He decides to call the emergency services and give them the position of the boat. He measures the angle of depression of the boat from the cliff top to be 30°, and he knows the cliff top is 200 m above sea level. How far is the boat from the base of the cliff?
200 m
tan 30°= 200 𝑥
⇒ = 200 𝑥
⇒𝑥=200 3
30°
𝑥= m
x m

6. Special Angles 30°, 45° and 60° sin 60°= 3 2 sin 30°= 1 2 60° 30° 30°
90°
60° 2 1
sin 60°=
sin 30°= 1 2
30° 3
cos 30°=
cos 60°= 1 2
90°
tan 30°= 1 3
tan 60°=
90°
45° 1
sin 45°= 1 2 2
cos 45°= 1 2
tan 45°= 1 1
These ratios appear in
(F and T: P13)
Use these ratios when asked to give an answer in surd form.

7. Area of a Triangle Area of a triangle = 1 2 𝑎𝑏 sin 𝐶
(F and T: P16) c b a C
NOTE: To use this formula we need the lengths of two sides
and the angle between them.
Find the area of Δ ABC. A B C
12 cm
8 cm
30°
In the given triangle the area is 13.6 cm2. Find the measure of the angle A to the nearest degree. A 8 7
sin 𝐴 =13.6
Area = 1 2 𝑎𝑏 sin 𝐶
Area = 1 2 (12)(8) sin 30°
28 sin 𝐴 =13.6
sin 𝐴 = ⇒ 𝐴=sin −
Area = 1 2 (12)(8) 1 2
Area =24 cm 2
𝐴=29° OR 𝐴=151°
From the diagram, A is clearly acute. ⇒𝐴=29°

8. The Unit Circle
The unit circle has its centre at (0,0) and has a radius length of 1 unit.
(0,1)
cos 𝜃= 𝑥 1 ⇒𝑥=cos 𝜃
(x,y) x y 1 Ɵ 1
sin 𝜃= 𝑦 1 ⇒𝑦=sin 𝜃
(−1,0)
(1,0) Ɵ
(0,0)
tan 𝜃= 𝑦 𝑥 ⇒tan 𝜃= sin 𝜃 cos 𝜃
(0,−1)
cos 90° = 0, sin 90° = 1 1
(0,1)
(1,0)
(0,−1)
(−1,0)
(0,0) Ɵ
(cos Ɵ,sin Ɵ)
cos 0° = 1, sin 0° = 0
cos 180° = −1, sin 180° = 0
cos 360° = 1, sin 360° = 0
cos 270° = 0, sin 270° = −1

9. Evaluating the Trigonometric Ratios of All Angles Between 0° and 360°
In the 1st quadrant, all three ratios are positive.
In the 2nd quadrant, sin is positive; cos and tan are negative. S
(sin +)
A (all +)
In the 3rd quadrant, tan is positive; sin and cos are negative.
In the 4th quadrant, cos is positive; sin and tan are negative. T
(tan +) C
(cos +)
The diagram summarises this.
CAST
Reference Angles
|∠AOB| = 140°
|∠AOB| = 250°
|∠AOB| = 330°
Reference angle
Reference angle
Reference angle
= 180° − 140°
= 250° − 180°
= 360° − 330°
= 40°
= 70°
= 30° O B A
180° B O A
180° B O A
30°
330°
40°
140°
250°
70°

10. Evaluating the Trigonometric Ratios of All Angles Between 0° and 360°
Write in surd form:
(i) cos 225°
(ii) tan 330°
(iii) sin 135°
(i) cos 225°
(ii) tan 330°
(iii) sin 135°
Step 1
Draw an angle of 225°.
Step 1
Draw an angle of 330°.
Step 1
Draw an angle of 135°.
180°
135°
180°
180°
45°
225°
45°
30°
330°
135°
225°
330°
Step 2
Step 2
Step 2
cos is negative (CAST).
3rd quadrant
tan is negative (CAST).
4th quadrant
sin is positive (CAST).
2nd quadrant
Step 3
Step 3
Step 3
Reference angle = 225° – 180°
Reference angle = 360° – 330°
Reference angle = 180° – 135°
= 45°
= 30°
= 45°
Step 4
Step 4
Step 4
cos 45° =
1 2
tan 30° =
1 3
sin 45° =
1 2
∴ cos 225° =
− 1 2
∴ tan 330° =
− 1 3
∴ sin 135° =
1 2

11. The Sine Rule 𝑥 Sine Rule: 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶
(F and T: P16)
To use the Sine Rule you need: two angles and one side (opposite one of the angles) OR two sides and one angle (opposite one of the sides) 12
50°
30° 𝑥 10 A
30° 15
Find the distance 𝒙. Give your answer to two decimal places.
Find the value of A. Give your answer to the nearest degree.
𝑥 sin 50° = 12 sin 30°
15 sin 𝐴 = 10 sin 30°
10 sin 𝐴=15 sin 30°
𝑥 = 12 sin 50° sin 30°
sin 𝐴= 15 sin 30° 10
𝑥=18.39
sin 𝐴=0.75
𝐴= sin −1 0.75
𝐴=49°

12. The Cosine Rule Cosine Rule: 𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐 cos 𝐴
(F and T: P16)
NOTE: If the lengths of two sides and the angle between these sides are known, use Cosine Rule.
NOTE: If the lengths of all three sides are known, use Cosine Rule.
In the triangle given, find |AB|. A B C 6 8 x
40°
Calculate the measure of the angle A. A 8 5 7
𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐 cos 𝐴
𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐 cos 𝐴
7 2 = −2(5)(8) cos 𝐴
𝑥 2 = −2 6 (8) cos 40°
49=25+64−80 cos 𝐴
𝑥 2 = …
80 cos 𝐴 =89−49
𝑥=5.1439…
cos 𝐴 = 1 2
𝑥=5.14
∴𝐴=60°

13. Length of an Arc and Area of a Sector𝜃 𝑙 𝐴 𝐵
Length of an arc: 𝑙=(2π𝑟) 𝜃 360
(F and T: P9)
Area of a sector: 𝐴=(𝜋 𝑟 2 ) 𝜃 360
(F and T: P9)
Find: (i) The area of the sector AOB (ii) The length of the minor arc AB. Give your answers in terms of 𝝅.
48° 𝐴 𝐵
12 cm 𝑂
NOTE: The minor arc is the shorter arc joining two points on the circumference of the circle.
(i) Area of the sector AOB
(ii) Length of the arc AB
𝑙=(2π𝑟) 𝜃 360
Area =(𝜋 𝑟 2 ) 𝜃 360
=𝜋 (12)
𝑙=2π(12)
= 96𝜋 5 cm 2
= 16𝜋 5 cm