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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