TRIGONOMETRYSINE AND COSINE RULES

& AREA OF TRIANGLE

Leaving Cert Revision

Find the distance 𝑥 in the diagram below (not to scale).Give your answer correct to 2 decimal places.

2017 LCOL Paper 2 – Question 6 (a)

First fill in the missing angle in the triangle. 𝟏𝟖𝟎 − 𝟔𝟑 + 𝟔𝟓 = 𝟓𝟐

52°

Sine Rule𝑎

sin 𝐴=

𝑏

sin 𝐵

𝑎

sin 𝐴=

𝑏

sin 𝐵𝑥

sin 52=

10

sin 63

𝑥 sin 63 = 10 sin 52

𝑥 =10 sin 52

sin 63

𝑥 = 8.84 cm

10 Marks

Find the distance 𝑦 in the diagram below (not to scale).Give your answer correct to 2 decimal places.

2017 LCOL Paper 2 – Question 6 (b)

Cosine Rule

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

𝑦2 = 10.22 + 8.52 − 2 10.2 8.5 cos 53.8 °

𝑦2 = 73.88

𝑦 = 73.88

𝑦 = 8.6 cm

25 Marks

Find the area of the given triangle.

2016 LCOL Paper 2 – Question 2 (a)

=1

28 12 sin 30

= 24 cm2

Area of a Triangle

=1

2𝑎𝑏 sin 𝐶

5 Marks

7

3

5𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴72 = 32 + 52 − 2 3 5 cos 𝑋49 = 9 + 25 − 30 cos 𝑋30 cos 𝑋 = 9 + 25 − 49

cos 𝑋 = −15

30

cos 𝑋 = −1

2𝑋 = 120°

𝑋°

Cosine Rule

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

.

A triangle has sides of length 3 cm, 5 cm, and 7 cm.Find the size of the largest angle in the triangle.

2016 LCOL Paper 2 – Question 2 (b)

20 Marks

Joe wants to draw a diagram of his farm. He uses axes and co-ordinates to plot his farmhouse at the point 𝐹 on the diagram below.

Write down the co-ordinates of the point F.

2016 LCOL Paper 2 – Question 9 (a) (i)

4,1

4,6𝐵

A barn is 5 units directly North of the farmhouse. Plot the point representing the position of the barn on the diagram. Label this point 𝐵.

(ii)

𝐹 = 4,1

5 units

Combination of Co-ordinateGeometry and Trigonometry

5 Marks

5 Marks

Joe's quad bike is marked with the point 𝑄 on the diagram.Find the distance from the barn (𝐵) to the quad (𝑄).Give your answer correct to 2 decimal places.

2016 LCOL Paper 2 – Question 9 (b)

−2,7

4,1

4,6𝐵

𝑄𝐵 = 4 − −22

+ 6 − 7 2

𝑄𝐵 = 6 2 + −1 2

𝑄𝐵 = 36 + 1

𝑄𝐵 = 37𝑄𝐵 = 6.08 units

Distance

= 𝑥2 − 𝑥12 + 𝑦2 − 𝑦1

2𝑄 −2,7𝐵 4,6

5 Marks

Joe's tractor is at the point 𝑇, where 𝐹𝐵𝑄𝑇 is a parallelogram.Plot 𝑇 on the diagram and write the co-ordinates of 𝑇 in the space below.

2016 LCOL Paper 2 – Question 9 (c)

−2,7

4,1

4,6𝐵

𝑇

𝐵𝑄4,6 → −2,7

We can find the co-ordinates of 𝑻 by finding the image

of 𝑭 under the translation 𝑩𝑸.

↓ 6, ↑ 1

4,1 → −2,2

5 Marks

Joe's tractor is at the point 𝑇, where 𝐹𝐵𝑄𝑇 is a parallelogram.Plot 𝑇 on the diagram and write the co-ordinates of 𝑇 in the space below.

2016 LCOL Paper 2 – Question 9 (d)

−2,7

4,1

4,6𝐵

𝑇

Area of a ParallelogramA = base × perpendicular height

Base= 5

Height = 6

Area = 5 × 6= 30 units2

5 Marks

Given that |∠𝑄𝐹𝐵| = 45°, use trigonometric methods to find |∠𝐵𝑄𝐹|.Give your answer in degrees correct to one decimal place.

2016 LCOL Paper 2 – Question 9 (e)

𝐵

45°

6.08

5

𝑋°𝑎

sin 𝐴=

𝑏

sin 𝐵

6.08

sin 45°=

5

sin 𝑋

sin 𝑋 =5 sin 45°

6.08

𝑋 = sin−15 sin 45°

6.08

𝑋 = 35.6°

Sine Rule𝑎

sin 𝐴=

𝑏

sin 𝐵

20 Marks

The diagram shows the triangles 𝐵𝐶𝐷 and 𝐴𝐵𝐷, with some measurements given.

Find |𝐵𝐶|, correct to two decimal places.

2015 LCOL Paper 2 – Question 5 (a) (i)

16

sin 110=

𝐵𝐶

sin 42

𝐵𝐶 =16 sin 42

sin 110

𝐵𝐶 = 11.39

11.39

Sine Rule𝑎

sin 𝐴=

𝑏

sin 𝐵

15 Marks

Find the area of the triangle 𝐵𝐶𝐷, correct to two decimal places.

2015 LCOL Paper 2 – Question 5 (a) (ii)

180 − 42 + 110= 28°

11.39

28°

Area of a Triangle

=1

2𝑎𝑏 sin 𝐶

=1

216 11.39 sin 28°

= 42.78 m2

First fill in the missing angle in the triangle, 𝚫𝑫𝑪𝑩.

5 Marks

Find |𝐴𝐵|, correct to two decimal places.

2015 LCOL Paper 2 – Question 5 (b)

180 − 63 + 42= 75

75

16.53

First fill in the missing angle in the triangle, 𝚫𝑫𝑨𝑩.

Cosine Rule

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

𝐴𝐵 2 = 102 + 162 − 2 10 16 cos 75𝐴𝐵 2 = 273.18𝐴𝐵 = 16.53 m

5 Marks

20 1822

25

A stand is being used to prop up a portable solar panel. It consists of a support that is hinged to the panel near the top, and an adjustable strap joining the panel to the support near the bottom.

By adjusting the length of the strap, the angle between the panel and the ground can be changed.

The dimensions are as follows:𝐴𝐵 = 30 cm𝐴𝐷 = 𝐶𝐵 = 5 cm𝐶𝐹 = 22 cm𝐸𝐹 = 4 cm.

2014 LCOL Sample Paper 2 – Question 8

25 2018

22

Two diagrams are given below – one showing triangle 𝐶𝐴𝐹 and the other showing triangle 𝐶𝐷𝐸. Use the measurements given above to record on the two diagrams below the lengths of two of the sides in each triangle.

2014 LCOL Sample Paper 2 – Question 8 (a)

Taking α = 60°, as shown, use the triangle 𝐶𝐴𝐹 to find ∠𝐶𝐹𝐴 , correct to one decimal place.

2014 LCOL Sample Paper 2 – Question 8 (b)

25

25

sin 𝑥=

22

sin 60

sin 𝑥 =25 sin 60

22𝑥 = 79.78°

180 − 79.78 − 60 = 40.22°

Hence find ∠𝐴𝐶𝐹 , correct to one decimal place.

(c)

Sine Rule𝑎

sin 𝐴=

𝑏

sin 𝐵

20 1822

40.22°

60° 𝑥°

Use triangle 𝐶𝐷𝐸 to find 𝐷𝐸 , the length of the strap, correct to one decimal place.

2014 LCOL Sample Paper 2 – Question 8 (d)

60°

25

79.78°

𝐷𝐸 2 = 202 + 182 − 2 20 18 cos 40.22°𝐷𝐸 2 = 174.23𝐷𝐸 = 13.2

Cosine Rule

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

20 1822

40.22°

A triangle in which the three sides have different lengths.

The planned supports for the roof of a building form scalene triangles of different sizes.Explain what is meant by a scalene triangle.

2012 Paper 2 – Question 7 (a)

5 Marks

The triangle 𝐸𝐹𝐺 is the image of the triangle 𝐶𝐷𝐸 under an enlargement and the triangle 𝐶𝐷𝐸 is the image of the triangle 𝐴𝐵𝐶 under the same enlargement.The proposed dimensions for the structure are 𝐴𝐵 = 7.2 m, 𝐵𝐶 = 8 m, |𝐶𝐷| = 9 m and |∠𝐷𝐶𝐵| = 60° .

Find the length of [𝐹𝐺].

2012 Paper 2 – Question 7 (b)

𝑘 =9

7.2𝑘 = 1.25

𝐹𝐺 = 8 × 1.25 × 1.25= 12.5 m

𝐒𝐜𝐚𝐥𝐞 𝐅𝐚𝐜𝐭𝐨𝐫

𝑘 =Image Length

Object Length

15 Marks

Find the length of [𝐵𝐷], correct to three decimal places.

2012 Paper 2 – Question 7 (c)

𝐵𝐷 2 = 82 + 92 − 2 8 9 cos 60°𝐵𝐷 2 = 73

𝐵𝐷 = 73𝐵𝐷 = 8.544 m

Cosine Rule

𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴

15 Marks

The centre of the enlargement is 𝑂. Find the distance from 𝑂 to the point 𝐵.

2012 Paper 2 – Question 7 (d)

𝑂𝐷

𝑂𝐵= 1.25

𝑥 + 8.544

𝑥= 1.25

𝑥 + 8.544 = 1.25𝑥0.25𝑥 = 8.544𝑥 = 34.176 m

𝑂

𝑥

8.544

Scale Factor = 1.25

L𝐞𝐭 𝒙 be the section from O to D, 𝑶𝑫

5 Marks

A condition of the planning is that the height of the point 𝐺 above the horizontal line 𝐵𝐹 cannot exceed 11.6 m. Does the plan meet this condition? Justify your answer by calculation.

2012 Paper 2 – Question 7 (e)

ℎ12.5

𝛼𝛼

9

sin 𝛼=

8.544

sin 60

sin 𝛼 =9 sin 60

8.544

ℎ

sin 𝛼=

12.5

sin 90

sin 𝛼 =ℎ sin 90

12.5

ℎ sin 90

12.5=

9 sin 60

8.544ℎ 1

12.5=

9 sin 60

8.5448.544ℎ = 12.5 9 sin 60

ℎ =12.5 9 sin 60

8.544ℎ = 11.4

11.4 < 11.6

Yes, the plan meets the condition. Sine Rule

𝑎

sin 𝐴=

𝑏

sin 𝐵

Sine Rule𝑎

sin 𝐴=

𝑏

sin 𝐵

10 Marks