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TRIGONOMETRY
GENERAL MATHS
UNIT TWO
RIGHT ANGLED TRIANGLES – BACKGROUND INFO
The internal angles within the triangle add up to 180°
The square in the corner of the triangle means that the angle here is 90° and
means that this is a right angled triangle
This means that the other 2 angles add up to 90°
RIGHT ANGLED TRIANGLES
So we know this triangle is right angled (square in the corner).
If we are given any two of: An angle (other than the 90° angle) and/or any 2
sides lengths, we can then find every other length and angle of the triangle.
θ
Opposite
Adjacent
We label the sides of the triangle based around the position of the angle in
given in a problem.
Opposite
θ
Adjacent
I have labelled the triangle based on the location of the angle given by θ
In problems we are usually told 2 measurements and are asked to an unknown.
We are asked to find θ. The ‘opposite’ length and ‘hypotenuse’ are known we use?
We are asked to find θ. The ‘adjacent’ length and ‘hypotenuse’ are known we use?
We are asked to find ‘opposite’ length. The θ and ‘hypotenuse’ are known we use?
RIGHT ANGLED TRIANGLES
Applying the rule:
eg. Find the length of 𝑥 correct to 2 decimal places.
Step 1: Identify and label the parts of the triangle that are given in the question.
Step 2: Choose which formula to use to solve the unknown 𝑥.
𝑂
𝐻
Step 3: Apply the formula
𝑥
sin 50 =
4
𝑥 = 4 sin 50
𝜃
𝑥 = 3.06m
RIGHT ANGLED TRIANGLES
Applying the rule:
eg2. Find the length of 𝑥 correct to 2 decimal places.
Step 1: Identify and label the parts of the triangle that are given in the question.
Step 2: Choose which formula to use to solve the unknown 𝑥.
𝜃
𝐴
𝐻
Step 3: Apply the formula
𝑥
c𝑜𝑠 60 =
10
𝑥 = 10cos 60
𝑥 = 5cm
RIGHT ANGLED TRIANGLES
Applying the rule:
eg3. Find the length of 𝑥 correct to 2 decimal places.
Step 1: Identify and label the parts of the triangle that are given in the question.
Step 2: Choose which formula to use to solve the unknown 𝑥.
𝜃
𝐴
𝑂
Step 3: Apply the formula
𝑥
tan 55 =
8
𝑥 = 8tan 55
𝑥 = 11.43cm
RIGHT ANGLED TRIANGLES
Applying the rule:
eg4. Find the angle given by 𝑥 (round to 2 decimal places)
Step 1: Identify and label the parts of the triangle that are given in the question.
Step 2: Choose which formula to use to solve the unknown 𝑥.
𝜃
Step 3: Apply the formula
𝐴
tan 𝑥
𝑂
6
=
8
6
−1
tan
8
𝑥=
𝑥 = 36.87°
NOW DO EXERCISE 7A
Questions
1, 2AFG, 3ABJ, 7, 10,
11, 13, 15, 16, 19
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
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)
ANGLES OF ELEVATION & DEPRESSION
ANGLES OF ELEVATION & DEPRESSION
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?
Sketch the scenario.
Then fill in anything else we can:
• Angle of Depression = Angle of Elevation
• We want to know how far the boat is from the base
So our triangle to solve is:
Opposite
𝜃
Adjacent
SOH – CAH – TOA
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 =
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
50
tan 12 =
𝑥
50
𝑥=
tan(12)
𝑥 = 235.23 𝑚𝑒𝑡𝑟𝑒𝑠
𝟏𝟐°
50m
𝟏𝟐°
𝒙
eg. Standing on a cliff 80 metres high, the angle of depression of a boat
at sea is 23°. Over time it moves to a new position with an angle of
depression of 28°. How far has the boat moved in this time?
23°
28°
Sketch the scenario.
Then fill in anything else we can:
• Angle of Depression = Angle of Elevation
• We want to know how far the boat has moved
So we have 2 triangles to solve:
SOH – CAH – TOA
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 =
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
80
tan 23 =
𝑎
80
𝑎=
tan(23)
𝑎 = 188.47 𝑚𝑒𝑡𝑟𝑒𝑠
80m
𝟐𝟑°
𝒙
𝟐𝟖°
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 =
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
80
tan 28 =
𝑏
80
𝑏=
tan(28)
𝑏 = 150.46 𝑚𝑒𝑡𝑟𝑒𝑠
∴ 𝑥 = 𝑎 − 𝑏 = 188.47 − 150.46 = 38.01 𝑚𝑒𝑡𝑟𝑒𝑠
NOW DO EXERCISE 7B
Questions
1-8
BEARINGS
360°
0°
Bearings measure the direction of one object to another.
40°
There are two systems used for describing bearings:
 True Bearings – measured in a clockwise direction starting
270°
90°
at North (0° T) all the way around to 360°T
040°T
 Compass (or Conventional) Bearings – are measured:
First relative to North and South, then relative to East or West
N40°E
180°
BEARINGS
360°
0°
Bearings measure the direction of one object to another.
There are two systems used for describing bearings:
 True Bearings – measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
(180 + 30) = 210°T
 Compass (or Conventional) Bearings – are measured:
First relative to North and South, then relative to East or West
S30°W
270°
90°
30°
180°
BEARINGS
360°
0°
Eg 3. Find the True and Compass Bearings of the following:
 True Bearings – measured in a clockwise direction starting
60°
270°
90°
at North (0° T) all the way around to 360°T
(360 – 60) = 300°T
 Compass (or Conventional) Bearings – are measured:
First relative to North and South, then relative to East or West
N60°W
E
180°
BEARINGS
360°
0°
Eg 4. Find the True and Compass Bearings of the following:
 True Bearings – measured in a clockwise direction starting
at North (0° T) all the way around to 360°T
270°
90°
70°
(180 – 70) = 110°T
 Compass (or Conventional) Bearings – are measured:
First relative to North and South, then relative to East or West
S70°E
180°
BEARINGS
360°
0°
Eg 5. Find the True and Compass Bearings of the following:
 True Bearings – measured in a clockwise direction starting
270°
90°
75°
at North (0° T) all the way around to 360°T
(180 + 75) = 255°T
 Compass (or Conventional) Bearings – are measured:
First relative to North and South, then relative to East or West
S75°W
180°
BEARINGS
360°
0°
Eg 6. Find the True and Compass Bearings of the following:
 True Bearings – measured in a clockwise direction starting
25°
270°
at North (0° T) all the way around to 360°T
(90 - 25) = 065°T
 Compass (or Conventional) Bearings – are measured:
First relative to North and South, then relative to East or West
N65°E
180°
90°
BEARINGS - APPLICATIONS
Eg 1. A ship sails 20km in a direction of N60°W.
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
O
Step 3: Decide which rule to apply & solve the problem
Θ = 60°
Hypotenuse = 20km
Opposite = ?
sin 𝜃 =
𝑜𝑝𝑝
ℎ𝑦𝑝
sin 60 =
𝑥
20
𝑥 = 20 sin 60
𝑥 = 17.32𝑘𝑚
H
θ
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.
a) 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 = 𝑎2 + 𝑏 2
𝑥=
22 + 62
𝑥 = 40
∴ 𝑥 ≅ 6.32𝑘𝑚
Eg 2. A boat race circuit consists of four legs:
N
Leg 1: 4km West. Leg 2: 6km South. Leg 3: 2km East.
Leg 4: Back to the starting point.
𝜽
b) On what bearing must the final leg be sailed?
Step 1: Use the diagram from part a to choose how to solve
This is our starting point which we are considering
W
E
Step 2: Find the triangle that we are interested in
S
Opp
and label relevant points
𝑜𝑝𝑝
tan 𝜃 =
𝑎𝑑𝑗
Step 3: Decide which rule to apply &
solve the problem
Adj
𝜃=
𝑡𝑎𝑛−1 (
2
)
6
𝜃 = 18.43°
So the bearing to sail at is 18.43°𝑇 𝑜𝑟 𝑁18.43°𝐸
NOW DO EXERCISE 7B
Questions
10 – 18; 20
NON RIGHT ANGLED TRIANGLES - THE SINE RULE
So far we have been solving right angled triangles (where one corner is 90°).
Now we look at solving every other triangle. The internal angles still add up to 180°.
One method is the sine rule
THE SINE RULE
Two angles and a side length are known.
Find the unknown length 𝑥 correct to 2 decimal points.
Step 1: Label the triangle.
A
Step 2: Identify variables & choose which rule to use.
𝑩 = 𝟏𝟎𝟐°
𝑨 = 𝟓𝟒°
𝒃=𝟖
𝒂 =?
𝑎
sin 𝐴
=
B
c
b
𝑏
sin 𝐵
Step 3: Substitute in values and solve the unknown variable.
𝑎
8
=
sin(54)
sin(102)
a = sin 54 ×
8
sin 102
𝒙a
= 6.62
C
THE SINE RULE
Two angles and a side length are known.
a) Find the unknown length 𝑥 correct to 2 decimal points.
Step 1: Label the triangle.
Step 2: Identify variables and choose equation to use.
𝒂 =?
𝑨 = 𝟒𝟔°
𝒃 = 𝟏𝟐 𝑩 = 𝟖𝟕°
𝑎
sin 𝐴
=
𝑏
sin 𝐵
B
c
b
12
sin 87
= 8.64
b) Then solve for the remaining unknowns on the triangle.
Internal angles = 180. So, A + B + C =180.
C = 180 – 46 – 87 = 47°
𝑏
𝑐
12
𝑐
=
→
=
→
𝑐 = sin 47 ×
sin 𝐵
sin 𝐶
sin 87
sin 47
47°
A
Step 3: Substitute in values and solve.
𝑎
12
=
→ a = sin 46
sin 46
sin 87
a
𝒙 8.64
8.79
12
sin 87
= 8.79
C
THE SINE RULE
Two angles and a side length are known.
a) Find the unknown length 𝑥 correct to 2 decimal points.
Step 1: Label the triangle.
Step 2: Identify variables and choose which equation to use.
𝒃 = 𝟑𝟎 𝑩 = 𝟕𝟓°
𝑪 = 𝟏𝟖 𝒄 = ?
𝑏
sin 𝐵
=
𝑐
sin 𝐶
9.60
sin 𝐵
sin 87
sin 75
a
87°
A
C
b
30
sin 75
= 9.60
b) Then solve for the remaining unknowns on the triangle.
Internal angles = 180. So, A + B + C =180°.
A = 180 – 75 – 18 = 87°
𝑎
𝑏
𝑎
30
=
→
=
→
𝑐 = sin 87 ×
sin 𝐴
31.06
c𝒙
Step 3: Substitute values into equation and solve.
30
𝑐
=
→ c = sin 18
sin 75
sin 18
B
30
sin 75
= 31.06
THE SINE RULE
One angle and 2 side lengths are known.
a) Find the unknown angle 𝑥 correct to 2 decimal points.
Step 1: Label the triangle.
Step 2: Identify variables & choose equation to use.
𝑨 =?
𝒃 = 𝟐𝟒
𝒂 = 𝟏𝟕
𝑩 = 𝟗𝟒°
𝑏
sin 𝐵
=
41.04°
A
𝒙 44.96°
=
17
sin 𝐴
→
sin 94
24
=
sin 𝐴
17
a
15.80c
𝑎
sin 𝐴
B
Step 3: Substitute values into equation and solve.
24
sin 94
C
b
→ sin 𝐴 =
17 sin 94
24
→
𝐴=
−1 17 sin 94
sin (
)
24
∴ 𝐴 = 44.96°
b) Now solve for the remaining unknowns on the triangle.
Internal angles = 180. So, A + B + C =180°. ∴ C = 180 – 94 – 44.96 = 41.04°
𝑐
sin 𝐶
=
𝑏
sin 𝐵
→
𝑐
sin 41.04
=
24
sin 94
→
𝑐 = sin 41.04 ×
24
sin 94
= 15.80
THE SINE RULE
Two side lengths and an angle are known.
a) Find the unknown angle 𝑥 correct to 2 decimal points.
c
Step 1: Label the triangle.
Step 2: Identify variables and choose the appropriate equation.A 67.2°
𝒃 = 𝟐𝟎
𝑩 = 𝟖𝟎°
𝒄 = 𝟏𝟏
𝑪 =?
B
18.72
a
32.8° 𝒙
b
𝑏
𝑐
=
sin 𝐵
sin 𝐶
Step 3: Substitute given values into the equation. Solve the unknown variable θ.
20
sin 80
=
11
sin 𝐶
→
sin 80
20
=
sin 𝐶
11
→
11sin 80
20
= sin 𝐶 → 𝐶 =
−1 11sin 80
sin (
20
b) Then solve for the remaining unknowns on the triangle.
Internal angles = 180. So, A + B + C =180°.
A = 180 – 32.8 – 80 = 67.2°
𝑎
𝑏
𝑎
20
=
→
=
→
𝑎 = sin 67.2 ×
sin 𝐴
sin 𝐵
sin 67.2
sin 80
20
sin 80
) = 32.8°
= 18.72
C
NOW DO EXERCISE 7C
Questions
1 - 3; 6 - 8;
12, 14, 18, 21
NON RIGHT ANGLED TRIANGLES - THE COSINE RULE
THE COSINE RULE
Eg1. Use the cosine rule to find the angle 𝑥
(to 2 decimal places)
b
C Step 1. Label the triangle A, B, C, a, b, c.
A
Step 2. Identify variables and choose the equation to use.
c
a
𝒙
B
𝒂 = 𝟏𝟎
𝒄 = 𝟏𝟐
𝒃=𝟖
𝑩 =?
Step 3. Substitute in given values and solve.
102 + 122 − 82
cos 𝐵 = (
)
2 × 10 × 12
𝐵=
102 +122 −82
−1
cos (
)
2 ×10×12
=
100+144−64
−1
cos (
)
240
= 41.41°
THE COSINE RULE
Eg2. Use the cosine rule to find the angle 𝑥
(to 2 decimal places)
B Step 1. Label the triangle A, B, C, a, b, c.
Step 2. Identify variables and choose the equation to use.
c
a
𝒂 = 𝟐𝟑
𝒄 = 𝟐𝟏
Step 3. Solve
A 𝒙
b
C
𝒃 = 𝟏𝟔
𝑨 =?
cos 𝐴 =
162 +212 −232
(
)
2 ×16×21
2 + 212 − 232
16
256 + 441 − 529
−1
−1
A = cos (
) = cos (
)
2 × 16 × 21
672
∴𝐴 =
168
−1
cos ( )
672
= 75.52°
THE COSINE RULE
B
Eg3. Use the cosine rule to find the length 𝑥 (to 2 decimal places)
Step 1. Label the triangle A, B, C, a, b, c.
𝒙a
c
Step 2. Identify variables and choose the equation to use.
C
A
b
𝒂 =?
𝒄 = 𝟏𝟒
𝒃 = 𝟏𝟏
𝑨 = 𝟔𝟒°
Step 3. Substitute in values and solve.
𝑎2 = 112 + 142 − 2 11 14 cos(64)
𝑎2 = 317 − 135.01 = 181.98
∴ 𝑎 = 181.98 = 13.49
Eg4. a) Use the cosine rule to find the length 𝑥 correct to 2 decimal places.
b) Use your answer to then solve for all other unknown values on the triangle.
a) Step 1. Label the triangle A, B, C, a, b, c.
B
Step 2. Identify variables and choose which equation to use.
a
c
C
5.57
𝒙
𝒂=𝟔 𝒃=𝟕
𝒄 =? 𝑪 = 𝟓𝟎°
Step 3. Substitute in given values and solve.
𝑐 2 = 62 + 72 − 2 6 7 cos 50 → 𝑐 2 = 36 + 49 − 83 cos 50
𝑐 = 36 + 49 − 53.99 → ∴ 𝑐 = 31.01 = 5.57
b
A
b)
cos 𝐴 =
72 + 5.572 −(62 )
2×7×5.57
→𝐴=
49+31.01−36
−1
cos (
)
77.96
= cos −1 0.56 = 55.63°
Sum of Internal Angles = 180° = A+B+C → B =180° - 50°- 55.63° = 74.37°
NOW DO EXERCISE 7D
Questions
1 – 9 & 13.
FINDING THE AREA OF THE TRIANGLE
B
If two length’s and the angle between them are
known, we find the area using:
𝟏
𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢𝐧(𝑨)
𝟐
Which can also be represented as:
𝑨𝒓𝒆𝒂 =
or
𝑨𝒓𝒆𝒂 =
𝟏
𝒂𝒄 𝒔𝒊𝒏(𝑩)
𝟐
𝟏
𝒂𝒃 𝒔𝒊𝒏(𝑪)
𝟐
c
A
a
b
C
FINDING THE AREA OF THE TRIANGLE
B
Find the area for the triangle pictured.
Step 1 : Label A, B, C, a, b, c.
c6m
a
A
𝟏
𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢 𝐧 𝑨
𝟐
𝟏
𝑨𝒓𝒆𝒂 = 𝒂𝒄 𝒔𝒊 𝒏 𝑩
𝟐
𝟏
𝑨𝒓𝒆𝒂 = 𝒂𝒃 𝒔𝒊𝒏(𝑪)
𝟐
72°
𝟏
𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢𝐧(𝑨)
𝟐
𝐴 = 72°
𝑐 = 6𝑚
𝑏 = 9𝑚
9m
b
Step 2 : Identify what we know on the triangle and decide
which rule to use.
C
Step 3 : Substitute values into equation and solve.
𝑨𝒓𝒆𝒂 =
𝟏
𝒃𝒄
𝟐
𝐬𝐢𝐧(𝑨)
𝟏
(
𝟐
=
× 𝟗 × 𝟔) sin 𝟕𝟐
= 27 sin 72 = 25.68𝑚2
FINDING THE AREA OF THE TRIANGLE – HERON’S FORMULA
B
If all 3 length’s are known, we can use
𝑨𝒓𝒆𝒂 =
where
c
𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
𝒔=
𝟏
(𝒂
𝟐
a
+ 𝒃 + 𝒄)
A
b
C
FINDING THE AREA OF THE TRIANGLE
HERON’S FORMULA
𝑨𝒓𝒆𝒂 =
𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
Where,
𝒔 = (𝒂 + 𝒃 + 𝒄)
𝟏
𝟐
Find the area for the triangle pictured.
B
a
c
Step 1 : Label A, B, C, a, b, c.
Step 2 : Identify what we know on the triangle and decide
𝑎 = 6𝑚
which rule to use.
𝑏 = 8𝑚
𝑐 = 12𝑚
C Step 3 : Substitute values into equation and solve.
A
b
𝟏
𝟐
First find 𝐬;
𝒔=
Then solve Area;
𝑨𝒓𝒆𝒂 =
=
𝒂+𝒃+𝒄 =
𝟏
𝟐
𝟔 + 𝟖 + 𝟏𝟐
=
𝟏
𝟐
× 𝟐𝟔 = 𝟏𝟑
𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
𝟏𝟑(𝟏𝟑 − 𝟔)(𝟏𝟑 − 𝟖)(𝟏𝟑 − 𝟏𝟐)
= 𝟏𝟑 × 𝟕 × 𝟓 × 𝟏
= 21.33𝑐𝑚2
FINDING THE AREA OF THE TRIANGLE
HERON’S FORMULA
𝑨𝒓𝒆𝒂 =
𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
Where,
𝒔 = (𝒂 + 𝒃 + 𝒄)
𝟏
𝟐
Find the area for the triangle pictured, correct to 2 decimals.
B
a
c
Step 1 : Label A, B, C, a, b, c.
Step 2 : Identify what we know on the triangle and decide
𝑎 = 5𝑚
which rule to use.
𝑏 = 4𝑚
𝑐 = 7𝑚
C Step 3 : Substitute values into equation and solve.
A
b
𝟏
𝟐
First find 𝐬;
𝒔=
Then solve Area;
𝑨𝒓𝒆𝒂 =
=
𝒂+𝒃+𝒄 =
𝟏
𝟐
𝟓+𝟒+𝟕
=
𝟏
𝟐
× 𝟏𝟔 = 𝟖
𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
𝟖(𝟖 − 𝟓)(𝟖 − 𝟒)(𝟖 − 𝟕)
= 𝟖×𝟑×𝟒×𝟏
= 9.80𝑚2
= 𝟗𝟔
B
𝑨𝒓𝒆𝒂 =
𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
Where,
𝒔 = (𝒂 + 𝒃 + 𝒄)
𝟏
𝟐
Find the area for the triangle pictured.
𝟏
𝑨𝒓𝒆𝒂 = 𝒃𝒄 𝐬𝐢 𝐧 𝑨
𝟐
𝟏
𝑨𝒓𝒆𝒂 = 𝒂𝒄 𝒔𝒊 𝒏 𝑩
𝟐
𝟏
𝑨𝒓𝒆𝒂 = 𝒂𝒃 𝒔𝒊𝒏(𝑪)
𝟐
Step 1 : Label A, B, C, a, b, c.
c
a
𝟏
𝑨𝒓𝒆𝒂 = 𝒂𝒃 𝐬𝐢𝐧(𝑪)
𝟐
C = 62°
𝑎 = 6𝑚
𝑏 = 7𝑚
A
b
Step 2 : Identify what we know on the triangle and decide
which rule to use.
62°
C
Step 3 : Substitute values into equation and solve.
𝑨𝒓𝒆𝒂 =
𝟏
𝒂𝒃
𝟐
𝐬𝐢𝐧(𝑨)
𝟏
(
𝟐
=
× 𝟔 × 𝟕) sin 𝟔𝟐
= 21 sin 62 = 18.54𝑚2
NOW DO EXERCISE 7E
Questions
1, 3, 8, 10, 12, 15,
16, 18 – 20