TRIGONOMETRYRATIOS

LCOL and JCHL Revision

The diagram below shows two right-angled triangles, 𝐴𝐵𝐶 and 𝐴𝐶𝐷.They have right angles at 𝐵 and 𝐷, respectively.|𝐴𝐵| = 10, |𝐴𝐶| = 12, and |𝐴𝐷| = |𝐷𝐶| = 𝑥, as shown.The angle 𝐵𝐴𝐶 is marked 𝑌.

Use trigonometry to find the size of the angle 𝑌.Give your answer correct to one decimal place.

2017 JCHL Paper 2 – Question 8 (a) (i)

12

10

𝑌

cos =adjacent

hypotenuse

cos 𝑌 =10

12

𝑌 = cos−110

12

𝑌 ≈ 33.6°

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Hypotenuse

Adjacent

10 Marks

Find the value of 𝑥. Give your answer correct to two decimal places.

2017 JCHL Paper 2 – Question 8 (a) (ii)

𝑥

𝑥12

𝑐2 = 𝑎2 + 𝑏2

122 = 𝑥2 + 𝑥2

144 = 2𝑥2

72 = 𝑥2

𝑥 = 72

𝑥 = 8.49 units

Pythagoras

𝑐2 = 𝑎2 + 𝑏2

10 Marks

The diagram on the right shows a right-angled triangle with a hypotenuse of length 10 units.

Use trigonometry to find the length of the side marked 𝑥.Give your answer in surd form.

2017 JCHL Paper 2 – Question 12 (a)

sin =opposite

hypotenuse

sin 60° =𝑥

10𝑥 = 10 sin 60 °

𝑥 = 5 3 units

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Hypotenuse

Opposite

10 Marks

The diagram below shows a regular hexagon with sides of length 10 units.The hexagon is divided into 6 equilateral triangles.

Work out the area of this hexagon. Give your answer in the form 𝑎 3, where 𝑎 ∈ ℕ.

2017 JCHL Paper 2 – Question 12 (b)

5 3

10

Area of a Triangle

=1

2base perpendicular height

There are 6 equilateral triangles of base 𝟏𝟎 and height 𝟓 𝟑.

Area of Hexagon

= 6 ×1

210 5 3

= 150 3 units2

5 Marks

Use trigonometry to find the measure of the angle 𝐴𝐵𝐶.Give your answer in degrees, correct to two decimal places.

2016 JCHL Paper 2 – Question 4 (c)

𝑋6

5

tan =opposite

adjacent

tan 𝑋 =5

6

𝑋 = tan−15

6

𝑋 = 39.81°∠𝐴𝐵𝐶 = 39.81°

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Adjacent

Opposite

6

5

Let 𝑋=∠𝐴𝐵𝐶

𝑋

5 Marks

Write 2° 43′ 5“ in degrees in decimal form, correct to two decimal places.

2016 JCHL Paper 2 – Question 8 (a) (i)

Write 3·14° in DMS (i.e. degrees, minutes, and seconds).

(ii)

2° 43′5“ = 2.72°

Enter into the calculator using the degrees, minutes, seconds button.

Turn into decimal using the SD button.

3.14° = 3° 8′24“

5 Marks

The diagram shows a right-angled triangle, with the angle 𝐴 marked.Given that 𝐜𝐨𝐬 𝑨 = 𝐬𝐢𝐧 𝑨, show that this triangle must be isosceles.

2016 JCHL Paper 2 – Question 8 (b)

cos 𝐴 = sin 𝐴

adjacent

hypotenuse=

opposite

hypotenuse

∴ adjacent = opposite

The adjacent side is equal in length to the opposite side there fore the triangle is isosceles.

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Adjacent

Opposite

Hypotenuse

10 Marks

A right-angled triangle has sides of length 7 cm, 24 cm, and 25 cm.Find the size of the smallest angle in this triangle.Give your answer correct to one decimal place.

2016 JCHL Paper 2 – Question 8 (c)

24 25

7

X tan =opposite

adjacent

tan 𝑋 =7

24

𝑋 = tan−17

24

𝑋 = 16.3°

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Adjacent

Opposite

The smallest angle is opposite the smallest side.

10 Marks

A different triangular-based prism has the base shown in the diagram on the right.

Use trigonometry to find the length of the side marked 𝑥 cm.Give your answer correct to two decimal places.

2016 JCHL Paper 2 – Question 12 (b) (i)

70°

3.5

𝑥

cos =adjacent

hypotenuse

cos 70° =3.5

𝑥

𝑥 =3.5

cos 70 °𝑥 = 10.23 cm

3.5

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Adjacent

Hypotenuse

10 Marks

|𝑆𝑇| = 10 and |𝑅𝑆| = 30.

Using this information, and trigonometry, find the size of ∠𝑋.Give your answer in degrees, correct to one decimal place.

2015 JCHL Paper 2 – Question 8 (b)

𝑋°

30

10

sin =opposite

hypotenuse

sin 𝑋 =10

30

𝑋 = sin−110

30

∠𝑋 = 19.5°

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Opposite

Hypotenuse

10 Marks

Source: www.watertowersofireland.com. Altered.

Miriam is trying to find the volume of the water tank shown in the photograph on the right.She takes some measurements and draws a diagram. Part of her diagram is shown below.

Using the diagram, find the value of 𝑥. Give your answer in metres, correct to two decimal places.

2015 JCHL Paper 2 – Question 13 (a)

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

tan =opposite

adjacent

tan 30° =𝑥

20𝑥 = 20 tan 30 °𝑥 = 11.55 m

Adjacent

Opposite 𝑥

30°

20

5 Marks

The angle of elevation to the bottom of the water tank is 30°, as shown in the diagram.The angle of elevation to the top of the water tank is 38°.Find the distance marked ℎ on the photograph. Give your answer correct to one decimal place.

2015 JCHL Paper 2 – Question 13 (b)

ℎ = 15.63 − 11.55ℎ = 4.08 m

11.55

38°

𝑥

tan =opposite

adjacent

tan 38 =𝑥

20𝑥 = 20 tan 38𝑥 = 15.63 m

Adjacent

Opposite 𝑥

38°

20

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse10 Marks

Construct a right angled triangle ABC, where:𝐴𝐵 = 6 cm

∠𝐴𝐵𝐶 = 90°𝐴𝐶 = 10 cm.

2014 JCHL Paper 2 – Question 6 (i)

10 Marks

𝑋 = 53°

cos 53° = 0.602

∠𝐶𝐴𝐵 = 53°

On your diagram, measure the angle ∠𝐶𝐴𝐵. Give your answer correct to the nearest degree.

2014 JCHL Paper 2 – Question 6 (ii)

Let 𝑋 be the whole number you wrote as your answer to (ii).Use a calculator to find cos 𝑋 . Give your answer correct to 3 decimal places.

(iii)

𝑋 = 53°

5+5 Marks

Jacinta says that cos(∠𝐶𝐴𝐵) is exactly 0.6, because

cos(∠𝐶𝐴𝐵) =𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒Explain why your answer in (iii) is not the same as Jacinta’s.

2014 JCHL Paper 2 – Question 6 (iv)

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

cos =adjacent

hypotenuse

cos(∠𝐶𝐴𝐵) =6

10cos(∠𝐶𝐴𝐵) = 0.6∠𝐶𝐴𝐵 = cos−1 0. 6∠𝐶𝐴𝐵 = 53.1301°

So if 𝑋 is a whole number then cos 𝑋 can never be exactly 0.6.

Adjacent

Hypotenuse

∠𝐶𝐴𝐵

5 Marks

Madison draws the scale diagram of the triangle 𝑂𝐴𝐵 shown on the right.She marks in the angle 𝑋.

Recall that [𝐴𝐵] is a metal bar, which is part of the frame of the swing.

Write down the value of tan 𝑋, and hence find the size of the angle 𝑋.Give the size of the angle 𝑋 correct to two decimal places.

2014 JCHL Paper 2 – Question 7 (v)

tan =opposite

adjacent

tan 𝑋 =5

4

𝑋 = tan−15

4𝑋 = 51.34°

Adjacent

Opposite

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

10 Marks

∠𝑋 = 51.34°

∠𝑋′ = 1.2 51.34°∠𝑋′ = 61.608°

In order to increase the height of the swing, it is decided to increase 𝑋 by 20%.The distance 𝐴𝐵 will be kept the same.Find the new height of the swing. Give your answer in metres, correct to one decimal place.

2014 JCHL Paper 2 – Question 7 (vi)

sin 𝑋′ =opposite

hypotenuse

sin 61.608° =ℎ

4141 sin 61.608° = ℎ

ℎ = 5.632≈ 5.6 m

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

AdjacentOpposite

10 Marks

(a)Do you think that (i) is correct? Give an example to justify your answer.

During a trigonometry lesson a group of students wrote down some statements about what they expected to happen when they looked at the values of trigonometric functions of some angles. Here are some of the things they wrote down.(i) The value from any of these trigonometric functions will always be less than 1.(ii) If the size of the angle is doubled then the value from the trigonometric functions will not double.(iii) The value from all of the trigonometric functions will increase if the size of the angle is increased.(iv) I do not need to use a calculator to find sin 60°. I can do it by drawing an equilateral triangle. The

answer will be in surd form.They then found the sin, cos and tan of some angles, correct to three decimal places, to test their ideas.

2014 Sample JCHL Paper 2 – Question 15

(b)Do you think that (ii) is correct? Give an example to justify your answer.

(c)Do you think that (iii) is correct? Give an example to justify your answer.

No – it IS possible for the opposite side to be greater than the adjacent. e.g. tan 50 = 1.19

Yes – the value of the ratio does NOT double when the angle does. e.g.tan 20 = 0.36tan 40 = 0.83

No - not all of the ratios increase when the angle does. cos 20 = 0.94cos 40 = 0.77

During a trigonometry lesson a group of students wrote down some statements about what they expected to happen when they looked at the values of trigonometric functions of some angles. Here are some of the things they wrote down.(i) The value from any of these trigonometric functions will always be less than 1.(ii) If the size of the angle is doubled then the value from the trigonometric functions will not double.(iii) The value from all of the trigonometric functions will increase if the size of the angle is increased.(iv) I do not need to use a calculator to find sin 60°. I can do it by drawing an equilateral triangle. The

answer will be in surd form.They then found the sin, cos and tan of some angles, correct to three decimal places, to test their ideas.

2014 Sample JCHL Paper 2 – Question 15

(d)Show how an equilateral triangle of side 2 cm can be used to find sin 60°in surd form. 2 2

122 = 𝑥2 + 12

4 = 𝑥2 + 1𝑥2 = 4 − 1𝑥2 = 3

𝑥 = 3

𝑥

60sin 60° =

opposite

hypotenuse

sin 60° =3

2

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Pythagoras

𝑐2 = 𝑎2 + 𝑏2

27.37

2.47

𝑥°

55.863

3.9

𝑥°

tan =opposite

adjacent

tan 𝑥 =55.863

3.9

𝑥 = tan−155.863

3.9𝑥 = 86°4° lean

tan =opposite

adjacent

tan 𝑥 =27.37

2.47

𝑥 = tan−127.37

2.47𝑥 = 84.84°5.16° lean

The Leaning Tower of Pisa is 55.863 m tall and leans 3.9 m from the perpendicular, as shown below. The tower of the Suurhusen Church in north-western Germany is 27.37 m tall and leans 2.47 m from the perpendicular. By providing diagrams and suitable calculations and explanations, decide which tower should enter the Guinness Book of Records as the Most Tilted Tower in the World.

2014 Sample JCHL Paper 2 – Question 16

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Adjacent

Opposite

Adjacent

Opposite

𝐌𝐎𝐒𝐓 𝐓𝐈𝐋𝐓𝐄𝐃 𝐓𝐎𝐖𝐄𝐑

𝑥°

4𝑥°The angles of a triangle sum to 180°

90 + 𝑥 + 4𝑥 = 1805𝑥 = 90𝑥 = 18°

4𝑥 = 4 18= 72°

In the right-angled triangle shown in the diagram, one of the acute angles is four times as large as the other acute angle.Find the measures of the two acute angles in the triangle.

2014 Sample JCHL Paper 2 – Question 17 (i)

The triangle in part (i) is placed on a co-ordinate diagram. The base is parallel to the 𝑥-axis.Find the slope of the line 𝑙 that contains the hypotenuse of the triangle.Give your answer correct to three decimal places.

2014 Sample JCHL Paper 2 – Question 17 (ii)

𝑙

𝑏𝑎𝑠𝑒

slope =rise

run

slope =opposite

adjacent

tan 18° =opposite

adjacenttan 18° = 0.3249

𝑟𝑖𝑠𝑒

𝑟𝑢𝑛

18°

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

In the triangle 𝐴𝐵𝐶, |𝐴𝐵| = 2 and |𝐵𝐶| = 1.

Find |𝐴𝐶|, giving your answer in surd form.

2013 JCHL Paper 2 – Question 10 (a)

𝐴

𝐵

𝐶

22 = 12 + 𝑥2

4 = 1 + 𝑥2

4 − 1 = 𝑥2

3 = 𝑥2

3 = |𝐴𝐶|

Write cos∠𝐵𝐴𝐶 and hence find ∠𝐵𝐴𝐶 .

(b)

cos =adjacent

hypotenuse

cos∠𝐵𝐴𝐶 =3

2

∠BAC = cos−13

2

∠𝐵𝐴𝐶 = 30°

Pythagoras

𝑐2 = 𝑎2 + 𝑏2

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

2

1

|𝐴𝐶|

∠𝐵𝐴𝐶

adjacent

hypotenuse

5 Marks 5 Marks

Sketch a right angled isosceles triangle in which the equal sides are 1 unit each and use it to write cos 45° in surd form.

2013 JCHL Paper 2 – Question 10 (c)

Show that cos 75° ≠ cos 45° + cos 30°.

(d)

45°

1

1

𝑥

𝑥2 = 12 + 12

𝑥2 = 1 + 1𝑥2 = 2

𝑥 = 2

cos =adjacent

hypotenuse

cos 45 =1

2

cos 75 ° ≠ cos 45 ° + cos 30 °0.2588 ≠ 0.7071 + 0.86600.2588 ≠ 1.5731

Pythagoras

𝑐2 = 𝑎2 + 𝑏2

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

adjacent

hypotenuse

5 Marks

5 Marks

A tree 32 m high casts a shadow 63 m long. Calculate 𝜃, the angle of elevation of the sun.Give your answer in degrees and minutes (correct to the nearest minute).

2013 JCHL Paper 2 – Question 13

32 m

tan 𝜃 =opposite

adjacent

tan 𝜃 =32

63

𝜃 = tan−132

63

𝜃 = 26.93°𝜃 = 26°56′

Adjacent

Opposite

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

10 Marks

A homeowner wishes to replace the three identical steps leading to her front door with a ramp. Each step is 10 cm high and 35 cm long.Find the length of the ramp. Give your answer correct to one decimal place.

2012 JCHL Paper 2 – Question 12

𝑥2 = 302 + 1052

𝑥2 = 900 + 11025𝑥2 = 11925𝑥 = 109.2 cm

𝑥30

105

Pythagoras

𝑐2 = 𝑎2 + 𝑏2

35 cm

10 cm

Height= 3 × 10= 30 cm

Width= 3 × 35= 105 cm

10 Marks

Two vertical poles 𝐴 and 𝐵, each of height ℎ, are standing on opposite sides of a level road.They are 24 m apart. The point 𝑃, on the road directly between the two poles, is a distance 𝑥 from pole 𝐴. The angle of elevation from 𝑃 to the top of pole A is 60°.

Write ℎ in terms of 𝑥.

2012 JCHL Paper 2 – Question 13 (a)

tan =opposite

adjacent

tan 60 =ℎ

𝑥ℎ = 𝑥 tan 60

ℎ = 𝑥 3

ℎ = 3𝑥

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Adjacent

Opposite

5 Marks

From 𝑃 the angle of elevation to the top of pole 𝐵 is 30°. Find ℎ, the height of the two poles.

2012 JCHL Paper 2 – Question 13 (b)

3𝑥 =24 − 𝑥 3

33 3𝑥 = 24 − 𝑥 33𝑥 = 24 − 𝑥3𝑥 + 𝑥 = 244𝑥 = 24𝑥 = 6

tan =opposite

adjacent

tan 30 =ℎ

24 − 𝑥ℎ = 24 − 𝑥 tan 30

ℎ = 24 − 𝑥3

3

ℎ =24 − 𝑥 3

3

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

Adjacent

Opposite

24 − 𝑥 Let the value of 𝒉 from (a) and (b) equal and solve for 𝒙.

2 Marks

ℎ =24 − 𝑥 3

3

ℎ =24 − 6 3

3

ℎ =18 3

3ℎ = 10.39 m

The circumference can be used to calculate the radius, which will give the full distance that Maria is from the centre of the base of the spire.

7.07

𝑟2𝜋𝑟 = 7.07

𝑟 =7.07

2𝜋𝑟 = 1.12523𝑟 ≈ 1.13 m

A group of students wish to calculate the height of the Millennium Spire in Dublin. The spire stands on flat level ground. Maria, who is 1.72 m tall, looks up at the top of the spire using a clinometer and records an angle of elevation of 60°. Her feet are 70 m from the base of the spire. Ultan measures the circumference of the base of the spire as 7.07 m.

Explain how Ultan’s measurement will be used in the calculation of the height of the Spire.

2011 JCHL Paper 2 – Question 15 (a)

Circumference of a Circle

= 2𝜋𝑟

5 Marks for both (a) AND (b)!

60°

1.72 m

ℎ

70 m

1.13

ℎ

71.13

60°

tan =opposite

adjacent

tan 60 ° =ℎ

71.13ℎ = 71.13 tan 60°

= 123.2

123.2 + 1.72= 124.92≈ 125 m

Maria

Draw a suitable diagram and calculate the height of the spire, to the nearest metre, using measurements obtained by the students.

2011 JCHL Paper 2 – Question 15 (b)

Ratios

tan =opposite

adjacent

sin =opposite

hypotenuse

cos =adjacent

hypotenuse

opposite

adjacent