GCSE: Non-Right Angled Triangles

Dr J Frost (jfrost@tiffin.kingston.sch.uk)www.drfrostmaths.com

Last modified: 31st August 2015

RECAP: Right-Angled TrianglesWe’ve previously been able to deal with right-angled triangles, to find the area, or missing sides and angles.

54

3

6

5

Area = 15

5

3

30.96°

?

? ?

Using Pythagoras: Using : Using trigonometry:

Labelling Sides of Non-Right Angle Triangles

Right-Angled Triangles:

h𝑜

𝑎

Non-Right-Angled Triangles:

𝑎𝑏

𝑐

𝐶

𝐴𝐵?

?

?

We label the sides and their corresponding OPPOSITE angles

OVERVIEW: Finding missing sides and angles

You have You want Use#1: Two angle-side opposite pairs

Missing angle or side in one pair

Sine rule

#2 Two sides known and a missing side opposite a known angle

Remaining side Cosine rule

#3 All three sides An angle Cosine rule

#4 Two sides known and a missing side not opposite known angle

Remaining side Sine rule twice

The Sine Rule

65°

85°

30°

105.02

9.10

For this triangle, try calculating each side divided by the sin of its opposite angle. What do you notice in all three cases?

! Sine Rule:

c

C

b

B

a

A

?

You have You want Use#1: Two angle-side opposite pairs

Missing angle or side in one pair

Sine rule

Examples

45°

8

11.27

85°

?

Q1

You have You want Use#1: Two angle-side opposite pairs

Missing angle or side in one pair

Sine rule

100°

8

15.7630° ?

Q2

50°

sin𝜃5 =

sin 856

Examples

85°

6

5

56.11°?

Q3 8

When you have a missing angle, it’s better to ‘flip’ your formula to get

i.e. in general put the missing value in the numerator.

40.33°

10

126°?

Q4

sin𝜃8 =

sin 126 °10

Test Your Understanding

𝑃𝑄

𝑅

85 °20 °

5𝑐𝑚

Determine the length .

82 °𝜃 10𝑚

12𝑚

Determine the angle .

? ?

Exercise 1Find the missing angle or side. Please copy the diagram first! Give answers to 3sf.

Q1

85 °

𝑥

15

40 °

𝑥=23.2?

Q2

𝑥 30 °

𝑥=53.1 °?

1610Q3

30 °

𝑦=56.4 °?

𝑦12

40 °

𝑥

10

𝑥=6.84?

Q4

20

Q5

𝛼=16.7 °?

𝛼20

1035 °Q6

𝑥=5.32?

70 °𝑥

5

Cosine RuleThe sine rule could be used whenever we had two pairs of sides and opposite angles involved.However, sometimes there may only be one angle involved. We then use something called the cosine rule.

15

12

115°𝑥

Cosine Rule:

𝑎𝐴

𝑏

𝑐

The only angle in formula is , so label angle in diagram , label opposite side , and so on ( and can go either way).

How are sides labelled ?

Calculation?

Sin or Cosine Rule?If you were given these exam questions, which would you use?

Sine Cosine Sine Cosine

Sine Cosine

10

15

𝑥70 °

10

15

𝑥

70 °

10

15

7𝛼

Sine Cosine

10

70 °12

𝛼

Test Your Understanding

𝑥

7 8

e.g. 1

𝑥=6.05

47 °

e.g. 2

7

4106.4 °

𝑥

𝑥=8.99? ?

You have You want UseTwo sides known and a missing side opposite a known angle

Remaining side Cosine rule

Exercise 2

Use the cosine rule to determine the missing angle/side. Quickly copy out the diagram first.

5 𝑥

7

60 °

𝑥=6.24?

100 °5 8

𝑦𝑦=10.14?

135 °58

70𝑥

𝑥=50.22?

6𝑥

643 °

𝑥=4.398?

Q1 Q2 Q3

Q4 Q5

10

3

8

𝑥

65 °

𝑥=9.513? 𝑥 3

54 75 °

𝑥=6.2966?

Q6

Dealing with Missing Angles

7

49

𝛼

𝛼=25.2 °? 𝟒𝟐=𝟕𝟐+𝟗𝟐− (𝟐×𝟕×𝟗×𝐜𝐨𝐬𝜶 )

Label sides then substitute into formula.

Simplify each bit of formula.

Rearrange (I use ‘swapsie’ trick to swap thing you’re subtracting and result)

? ? ?

?

𝒂𝟐=𝒃𝟐+𝒄𝟐−𝟐𝒃𝒄𝐜𝐨𝐬 𝑨

You have You want UseAll three sides An angle Cosine rule

Test Your Understanding

𝜃5

82=72+52− (2×7×5× cos𝜃 )?

8

7𝜃7𝑐𝑚

4𝑐𝑚

9𝑐𝑚

42=72+92− (2×7×9× cos𝜃 )?

𝜃

76

6

𝜃=71.4 °?

12

513.2

𝛽

𝛽=92.5 °?

5.211

8

𝜃

𝜃=111.1 °?

Exercise 3

1 2 3

Using sine rule twiceYou have You want Use#4 Two sides known and a missing side not opposite known angle

Remaining side Sine rule twice

4

𝑥

3 32 °

Given there is just one angle involved, you might attempt to use the cosine rule:

This is a quadratic equation!It’s possible to solve this using the quadratic formula (using ). However, this is a bit fiddly and not the primary method expected in the exam…

?

Using sine rule twiceYou have You want Use#4 Two sides known and a missing side not opposite known angle

Remaining side Sine rule twice

4

𝑥

3 32 °

𝟏𝟖𝟎−𝟑𝟐−𝟒𝟒 .𝟗𝟓𝟓𝟔=𝟏𝟎𝟑 .𝟎𝟒𝟒𝟒

1: We could use the sine rule to find this angle.

2: Which means we would then know this angle.

3: Using the sine rule a second time allows us to find

𝑥sin 103.0444=

3sin 32

!

?

?

?

Test Your Understanding

61 °

53 °

10

9

𝑦

34

𝑦

𝑦=6.97

𝑦=5.01

?

?

Area of Non Right-Angled Triangles

Area = Where C is the angle wedged between two sides a and b.

59°

3cm

7cm

Area = 0.5 x 3 x 7 x sin(59) = 9.00cm2

!

?

Test Your Understanding

61 ° 10

9

6.97

5 5

5

𝐴=12×6.97×10×𝑠𝑖𝑛61

𝐴=12×5×5×sin 60

?

?

Harder Examples

Q1 (Edexcel June 2014)

Finding angle :

Area of

67

8Using cosine rule to find angle opposite 8:

? ?

Q2

Exercise 4

64 ° 49 °

8 .7𝑐𝑚Q5

5100°

Q1

Area = 7.39? 8

3Q2

𝐴𝑟𝑒𝑎=√34

=0.433?

1 1

1 5.2

3.63.8

Q3

75°

Area = 9.04?

Q4

Area = 8.03

5

70°

? Q6

𝐴𝑟𝑒𝑎=29.25𝑐𝑚2 is the midpoint of and the midpoint of . is a sector of a circle. Find the shaded area.

( 12×62×sin 60)− 16 𝜋 (32 )=10.9𝑐𝑚2

?

?

Q7

3cm

2cm

110°

Area = ? Q8

3m

4.2m

5.3m

Area = ?

Segment Area

𝑂

𝐴

𝐵

70 °

10𝑐𝑚 is a sector of a circle, centred at .Determine the area of the shaded segment.

? ?

?

𝐴=3𝜋 −9?

Test Your Understanding

𝐴=119𝑚2?

Exercise 5 - Mixed ExercisesQ1

8 0°

𝑥

27

40 °

b) ? ?

8 𝑦

10

70 °

𝑦=10.45?

Q2

?

Q3

𝛼=17.79 °?

𝛼18

1130 °𝑧

? ?

𝑄𝑅=12.6𝑐𝑚?

130 °90𝑚

60𝑚

𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟?

Q4

Q5

Q6

4.615

12

𝜃

𝜃=122.8 °?

6𝑐𝑚52 °

𝐴𝑟𝑒𝑎=2.15𝑐𝑚2?

Q7

61 °57

𝑥

𝑥=7.89

Q8

? ?