1. Trigonometry and Geometry
Maths for Further
Semester Two

2. Trigonometry and Geometry - Course

3. Right Angled Triangles
A square in the corner of the triangle means that the angle here is 90° - this is what makes it a right angled triangle
The internal angles within all triangles add up to 180°
This means that the other 2 angles add up to 90°

4. Labelling right angled triangles
We know that this triangle is right angled.
If a problem involves another angle in the triangle,
we can label the sides of the triangle based around the location of the angle.
We label the sides: * Hypotenuse
* Opposite
* Adjacent
Hypotenuse
Opposite 𝜃
Adjacent 𝜃
Hypotenuse
Adjacent
Opposite

5. Finding the length of an unknown side, given an angle
In problems involving right angled triangles, solving involves an angle and two side lengths.
First we analyse the triangle and label it’s sides according to the location of the angle.
For example:
1. Label the sides of the triangle.
2. Choose the appropriate rule to use.
3. Solve the unknown side
sin 𝜃 = 𝑂 𝐻
sin 50 = 𝑥 10
10 sin 50 =𝑥
𝑥=7.66 𝑚
50°
10 𝑚
Adjacent
Hypotenuse
Opposite 𝑥
SOH CAH TOA
sin 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
tan 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

6. SOH CAH TOA Choosing the rule Choosing a rule:
cos 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
tan 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Choosing a rule:
If we are asked to find the side adjacent to the 𝜃 and the ‘opposite’ side is known we use?
If we are asked to find the side opposite the 𝜃 and the ‘hypotenuse’ is known we use?
If we are asked to find the hypotenuse and the ‘opposite’ side is known we use?
If we are asked to find the hypotenuse and the ‘adjacent’ side is known we use?
If we are asked to find the side opposite the 𝜃 and the ‘adjacent’ side is known we use?

7. Finding the length of an unknown side, given an angle
1. Label the sides of the triangle.
2. Choose the appropriate rule to use.
3. Solve the unknown side
Example One
SOH CAH TOA
Example Two
Example Three
35°
8 𝑚 𝑥
60°
28°
10 𝑚 𝑥
5 𝑚
Adjacent
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite 𝑥
tan 𝜃 = 𝑂 𝐴
tan 60 = 𝑥 5
𝑥= 5 tan 60
𝑥= 8.66 𝑚
cos 𝜃 = 𝐴 𝐻
cos 35 = 𝑥 8
𝑥= 8 cos 35
𝑥=6.55 𝑚
sin 𝜃 = 𝑂 𝐻
sin 28 = 𝑥 10
𝑥= 10 sin 28
𝑥=4.69 𝑚

8. Finding the length of an unknown side, given an angle
1. Label the sides of the triangle.
2. Choose the appropriate rule to use.
3. Solve the unknown side
Example Four
SOH CAH TOA
Example Five
Example Six
55°
40°
4 𝑚 𝑥
30°
5.2 𝑚 𝑥
Adjacent 𝑥
Opposite
Hypotenuse
Adjacent
Hypotenuse
Opposite
8 𝑚
tan 𝜃 = 𝑂 𝐴
tan 55 = 8 𝑥
𝑥 tan 55 =8
𝑥= 8 tan =5.6 𝑚
sin 𝜃 = 𝑂 𝐻
sin 30 = 5.2 𝑥
𝑥 sin 30 =5.2
𝑥= 5.2 sin 30 = 10.4
cos 𝜃 = 𝐴 𝐻
cos 40 = 4 𝑥
𝑥 cos 40 =4
𝑥= 4 cos 40 =5.22 𝑚

9. Solving real problems using SOHCAHTOA
A 3 metre ladder is leant against a wall making a 70° angle to the ground.
How far up the wall does the ladder reach?
b) How far is the base of the ladder
from the wall?
sin 𝜃 = 𝑂 𝐻
sin 70 = 𝑥 3
𝑥=3 sin 70
𝑥=2.82 m
cos 𝜃 = 𝐴 𝐻
cos 70 = 𝑦 3
𝑦=3 cos 70
𝑥=1.03 m O H 𝑥
3 𝑚 H
70° 𝑦 A

10. Now Do
Exercise 4A
Questions 2, 3*, 4*, 5*, 6a, 7, 9, 13a

11. Rearranging to solve unknown variables in triangles
If our unknown value is in the numerator:
We want x to be by itself. We move the denominator to the other side of the equation by multiplying each side by the number in the denominator.
If our unknown value is in the denominator:
sin 𝜃 = 𝑂 𝐻
sin 40 = 8 𝒙
𝑥= 8 sin 40
We want x to be by itself.
We can swap the “ sin 𝜃 ” and the “ x ”.
sin 𝜃 = 𝑂 𝐻
sin 𝜃 = 2 5
θ= sin −1 ( )
If our unknown value is the angle:
We want 𝜃 to be by itself.
We can move the “ sin ” part only to the other side of the equation by making it “ sin-1 ”

12. Finding the an angle given the lengths of 2 sides
1. Label the sides of the triangle.
2. Choose the appropriate rule to use.
3. Solve the unknown side
Example One
SOH CAH TOA
Example Two
Example Three 𝜃
10 𝑚 8 H A O H 𝜃
12 𝑚
6 𝑚 A 𝜃
3 𝑚
9 𝑚 O
tan 𝜃 = 𝑂 𝐴
tan 𝜃 = 9 3
tan 𝜃=3
𝜃= tan −1 (3)
𝜃=71.57°
cos 𝜃 = 𝐴 𝐻
cos 𝜃 = 8 10
cos 𝜃 =0.8
𝜃= cos −
𝜃=36.87°
sin 𝜃 = 𝑂 𝐻
sin 𝜃 = 6 12
sin 𝜃 =0.5
θ= sin −1 (0.5)
θ=30°

13. Now Do
Exercise 4B
Questions 3, 4*, 5*, 6ac, 7, 8

14. ANGLES OF ELEVATION & DEPRESSION
Angles of Elevation and Depression are used when dealing with directions which require us to look up or down at something.
The angle of elevation is the angle between the horizontal
and an object which is higher than the observer.
(eg. The top of a mountain or flagpole)
The angle of depression is the angle between the horizontal
and an object which is lower than the observer.
(eg. A boat at sea when the observer is on a cliff)

15. ANGLES OF ELEVATION & DEPRESSION
In a given problem, Angles of Elevation and Depression are equal.
We say that they are Alternate ‘ Z ’ angles.

16. ANGLES OF ELEVATION & DEPRESSION

17. ANGLES OF ELEVATION & DEPRESSION
eg. Standing on a cliff 50 metres high, the angle of depression
of a boat at sea is 12°. How far is the boat from the base of the cliff?
1. Sketch the scenario.
2. Fill in any values that we know from the question:
We want to know how far the boat is from the base of the cliff, call this ‘ x ’
Angle of Depression = Angle of Elevation
So our triangle to solve is:
SOH – CAH – TOA
tan 𝜃 = 𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
tan 12 = 50 𝑥
𝑥= 50 tan⁡(12)
𝑥= 𝑚𝑒𝑡𝑟𝑒𝑠
𝟓𝟎 𝒎
12°
Opposite
12°
BOAT 𝒙
Adjacent

18. Now Do
Exercise 4C
Questions 1, 2, 3, 4, 5, 7, 13

19. BEARINGS Bearings measure the direction of one object from another.
For this course we will cover true bearings.
True Bearings – are measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
The bearing corresponding to the image is 040°T
as it is 40 degrees from the North position (0˚) 0°
90°
180°
270°
360°
40°

20. BEARINGS True Bearings – measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
Find the bearing pictured:
( ) = 210°T
360° 0°
270°
90°
30°
180°

21. BEARINGS True Bearings – measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
Find the bearing pictured:
(360 – 60) = 300°T
360° 0°
60°
270°
90°
180°

22. BEARINGS True Bearings – measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
Find the bearing pictured:
(180 – 70) = 110°T
360° 0°
270°
90°
70°
180°

23. BEARINGS True Bearings – measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
Find the bearing pictured:
( ) = 255°T
360° 0°
270°
90°
75°
180°

24. BEARINGS True Bearings – measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
Find the bearing pictured:
( ) = 065°T
360° 0°
25°
270°
90°
180°

25. BEARINGS Eg 1. A person walks 11 m on a bearing of 200°.
a) How far West is point A from the starting point?
b) How far South is point A from the starting point? a)
Step 1: Draw a diagram of the situation, labelling compass points and the given information
Step 2: Find the triangle that we are interested in and label H, O, A, θ
Step 3: Decide which rule to apply & solve the problem
Θ = 70°
Hypotenuse = 11 m
Adjacent = ? H A θ
cos 𝜃 = 𝑎𝑑𝑗 ℎ𝑦𝑝
cos 70 = 𝑥 11
𝑥=11 cos 70
𝑥=3.76 𝑚

26. BEARINGS Eg 1. A person walks 11 m on a bearing of 200°.
a) How far West is point A from the starting point?
b) How far South is point A from the starting point? b)
Step 1: Draw a diagram of the situation, labelling compass points and the given information
Step 2: Find the triangle that we are interested in and label H, O, A, θ
Step 3: Decide which rule to apply & solve the problem
Θ = 20°
Hypotenuse = 11 m
Adjacent = ?
cos 𝜃 = 𝑎𝑑𝑗 ℎ𝑦𝑝
cos 20 = 𝑥 11
𝑥=11 cos 20
𝑥=10.34 𝑚 H A θ

27. BEARINGS Eg 2. A ship sails 20km on a bearing of 300°.
How far West from the starting point has the ship sailed?
Step 1: Draw a diagram of the situation, labelling compass points and the given information
Step 2: Find the triangle that we are interested in and label H, O, A, θ
Step 3: Decide which rule to apply & solve the problem
Θ = 60°
Hypotenuse = 20km
Opposite = ? H O θ
sin 𝜃 = 𝑜𝑝𝑝 ℎ𝑦𝑝
sin 60 = 𝑥 20
𝑥=20 sin 60
𝑥=17.32𝑘𝑚

28. BEARINGS Eg 3. A boat race circuit consists of four legs:
Leg 1: 4km West. Leg 2: 6km South. Leg 3: 2km East.
Leg 4: Back to the starting point.
How long is the final leg (Leg 4)?
b) On what bearing must the final leg be sailed?
a) Step 1: Draw a diagram using the given information
Step 2: Find the triangle that we are interested in and label relevant points
Step 3: Decide which rule to apply & solve the problem
𝑐 2 = 𝑎 2 + 𝑏 2 𝑥=
𝑥= 40
∴𝑥≅6.32𝑘𝑚

29. So the bearing to sail at is 18.43°𝑇
BEARINGS
Eg 2. A boat race circuit consists of four legs:
Leg 1: 4km West. Leg 2: 6km South. Leg 3: 2km East.
Leg 4: Back to the starting point.
b) On what bearing must the final leg be sailed?
b) Step 1: Use the diagram from part a to choose how to solve
This is our starting point which we are considering
Step 2: Find the triangle that we are interested in
and label relevant points
Step 3: Decide which rule to apply &
solve the problem
tan 𝜃 = 𝑜𝑝𝑝 𝑎𝑑𝑗
𝜃= 𝑡𝑎𝑛 −1 ( 2 6 )
𝜃=18.43°
So the bearing to sail at is 18.43°𝑇

30. Now Do
Exercise 4D
Questions 1, 2, 4, 5, 6, 7, 8, 10

31. Triangles can be classified as:
GEOMETRY - INTRODUCTION
Triangles can be classified as:
All 3 side lengths are different
All angles within the triangle are different
2 of the side lengths are the equal
2 angles within the triangle are equal
All 3 side lengths are equal
All angles within the triangle are equal

32. Quadrilaterals can be classified as:
• Parallelogram –
Quadrilaterals with two pairs of parallel sides.
• Rectangle –
Parallelograms with all angles 90°.
• Square –
Rectangles with equal length sides.

33. Quadrilaterals can be classified as:
• Rhombus –
Parallelograms with equal length sides.
• Kite –
Quadrilaterals with two pairs of equal adjacent sides.
• Trapezium –
Quadrilaterals with one pair of parallel sides.

34. Where n is the number of sides of the polygon.
Polygons are closed shapes with straight edges and any number of sides.
The sum of the internal angles of Irregular Polygons can be calculated using:
S = 180 (n – 2)
Where n is the number of sides of the polygon.

35. Polygons are closed shapes with straight edges and any number of sides.
The sum of the internal angles of any Polygon can be calculated using:
S = 180 (n – 2)
Where n is the number of sides of the polygon.

36. Regular Polygons are also closed shapes with straight edges and any number of sides but all sides are equal and all internal angles are equal.
A single angle within a Regular Polygon can be calculated using:
𝑆 = 180(𝑛−2) 𝑛
Where n is the number of sides of the polygon.

37. All interior angles add up to 180°
Angles in triangles
All interior angles add up to 180°
The exterior angles ‘ d ’ and the interior angle ‘ c ’ are Supplementary Angles, which means together they form a 180° angle.
The exterior angle theorem states that the exterior angle is equal to the sum of the two opposite interior angles.

38. Parallel and Transversal Lines

39. Equal Angles formed by Parallel and Transversal Lines Vertical Angles

40. Equal Angles formed by Parallel and Transversal Lines
Corresponding Angles
Alternate Interior Angles
Alternate Exterior Angles

41. Equal Angles formed by Parallel and Transversal Lines
Put simpler by the example…..

42. Example 1 Use the exterior angle theorem to solve 𝑥

43. Irregular Polygon – use sum of angles inside shape to solve
Example 3 Find the value of x
Irregular Polygon – use sum of angles inside shape to solve
Sum of angles ‘S’ = 180˚(n – 2)
S = 180 (5 - 2)
S = 180 x 3
S = 540 ˚
540 = x
540 = x
540 – 420 = 2x
120 = 2x
x = = 60°

44. Example 4 Find the value of x
Regular Polygon – angles within the shape are equal Use single angle formula to calculate each angle.
𝑆 = 180(𝑛−2) 𝑛
𝑆 = 180(5−2) 5
𝑆 = 180(3) 5
𝑆 = 540 5
S = 108 ˚

45. Example 4 Find the value of x
Regular Polygon – angles within the shape are equal Use single angle formula to calculate each angle.
𝑆 = 180(𝑛−2) 𝑛
𝑆 = 180(5−2) 5
𝑆 = 180(3) 5
𝑆 = 540 5
S = 108 ˚

46. Now Do
Exercise 2A
Questions 3, 4ab, 5*, 6acf, 7b

47. Congruent Triangles SSS
If two objects are exactly the same size and shape then we say they are congruent.
When comparing 2 triangles, they are congruent if corresponding:
* Side lengths are equal
* Angles are equal
Any two triangles can be tested for congruence by using the following rules:
SSS

48. Congruent Triangles SAS AAS
Any two triangles can be tested for congruence by using the following rules:
SAS
AAS

49. Congruent Triangles RHS
Any two triangles can be tested for congruence by using the following rules:
RHS

50. Congruent Triangles Yes they are congruent
Example 1 Are the following triangles congruent?
List the pairs of corresponding sides and angles.
Each include a right angle
Each have a hypotenuse of 15
Each have a length of 6 adjacent to right angle
Yes they are congruent
Satisfies the RHS rule (Right Angle Hypotenuse Another Side)
Congruent Triangles

51. Congruent Triangles Yes they are congruent
Example 2 Are the following triangles congruent?
List the pairs of corresponding sides and angles.
Each have a side length of 4
Each have a side length of 6
Each have a side length of 7
Yes they are congruent
Satisfies the SSS rule (Side Side Side)
Corresponding sides are equal
Congruent Triangles

52. Example 3 Prove that the following triangles are congruent.
List the pairs of corresponding sides and angles.
BA = ED (side length)
AC = DF (side length)
BAC = EDF
Satisfies the SAS rule (Side Angle Side)
So ABC = DEF
Congruent Triangles

53. Example 4 Prove that following triangles are congruent
List the pairs of corresponding sides and angles.
ABC = DEF (Angle)
BCA = EFD (Angle)
AB = DE (side length)
Satisfies the AAS rule (Angle Angle Side)
So ABC = DEF
Congruent Triangles

54. Now Do
Exercise 2B
Questions 1*, 2*, 3, 4

55. Similar Figures
Similar Figures are two figures that have the same shape but a different size.
* Angles within the shape are the same
* All side lengths are increased by the same proportion (or ratio)
The size ratio between two similar objects is called the scale factor and can be found by:
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑎𝑟𝑔𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑎𝑙𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ

56. Proof – lets use the other side and check
Similar Figures
The size ratio between two similar objects is called the scale factor and can be found by:
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑎𝑟𝑔𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑎𝑙𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
Calculate the scale factor of the given the rectangles.
Choose equivalent side lengths:
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑟𝑔 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ = 8 4 =2
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑟𝑔 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ = 2 1 =2
We can use either side to calculate the scale factor as all sides increase by the same factor.
Proof – lets use the other side and check

57. Example 1 Find the scale factor for these similar figures
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑟𝑔 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
= 9 3 =3
Example 2 Find the scale factor for these similar figures
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑟𝑔 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
= =1.5

58. Similar Figures
Example 3 Find the scale factor and use it to find the value of x for these similar figures
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑟𝑔 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
= =1.5
Use this value to find x.
The big figure is 1.5 times larger than the small one.
x = 6 × scale factor
= 6 × 1.5
= 9

59. Example 4 These two shapes are similar.
a) Write a similarity statement for the two shapes
b) Complete the following: 𝐸𝐻 = 𝐹𝐺
c) Find the scale factor
d) Find the value of x
x = 3 cm × scale factor
x = 3 cm × 2
x = 6 cm
ABCB ||| EFGH or ABCD ~ EFGH
𝐸𝐻 𝐴𝐷 = 𝐹𝐺 𝐵𝐶
𝑠𝑐𝑎𝑙𝑒 𝑓𝑎𝑐𝑡𝑜𝑟= 𝑙𝑟𝑔 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑠𝑚𝑙 𝑜𝑏𝑗𝑒𝑐𝑡 𝑙𝑒𝑛𝑔𝑡ℎ
= =2

60. Now Do
Exercise 2D
Questions 1, 2*, 4, 6*, 8, 9*

61. Proving Similar Triangles
As we have just seen in the previous exercise, similar figures are 2 of the same shape, but one is enlarged.
Like the tests for congruence, there are 4 tests for proving whether 2 triangles are similar.
SSS All pairs of sides have the same ratio
SAS Two pairs of sides are given and have the same ratio AND they have an equal angle

62. Proving Similar Triangles
AAA All pairs of angles within the triangles are equal
RHS The triangles are right angled (90˚ angle) and another pair of corresponding sides have the same ratio

63. Proving Similar Triangles
Example 1 Prove that the following pair of triangles are similar
SAS ( Side Angle Side )
Check ratio for each side
Side 1: 𝐸𝐷 𝐵𝐴 = =2
Side 2: 𝐹𝐷 𝐶𝐴 = 4 2 =2
Angles: BAC = EDF = 20˚
This proves that Δ BAC ||| Δ EDF

64. Proving Similar Triangles
Example 2 Prove that the following pair of triangles are similar
RHS ( Right Hypotenuse Side)
Check ratio for each side:
Hypotenuse: 𝐷𝐹 𝐴𝐶 = 12 4 =3
Side: 𝐷𝐸 𝐵𝐶 = 9 3 =3
Angles: ABC = DEF = 90˚
This proves that Δ ABC ||| Δ DEF

65. Proving Similar Triangles
Example 3 Prove that the following pair of triangles are similar
AAA ( Angle Angle Angle)
Check each pair of angles:
Angle 1: BAC = EDF = 80˚
Angle 2: ABC = DEF = 70˚
Angle 3: BCA = EFD = 180 – 70 – 80 = 30˚
This proves that Δ ABC ||| Δ DEF

66. Proving Similar Triangles
Example 4 Prove that the following pair of triangles are similar
SSS ( Side Side Side)
Check ratio for each pair of sides:
Side 1: 𝐴𝐶 𝐷𝐹 = 20 5 =4
Side 2: 𝐵𝐶 𝐸𝐹 = 8 2 =4
Side 3: 𝐵𝐴 𝐸𝐷 = 16 4 =4
This proves that Δ ABC ||| Δ DEF

67. Now Do
Exercise 2E
Questions 4, 5abcf, 7b, 8b, 11