### Sample pages from Linear Higher Student Book - Pearson Schools

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31
PythagorasвЂ™ theorem
and trigonometry (2)
CHAPTER 10
Linear equations
CHAPTER
In Chapter 19, PythagorasвЂ™ theorem and trigonometry were used to find the lengths of sides and the sizes
of angles in right-angled triangles. These methods will now be used with three-dimensional shapes.
31.1 Problems in three dimensions
In a cuboid all the edges are perpendicular to each other.
Problems with cuboids and other 3-D shapes involve identifying suitable right-angled
triangles and applying PythagorasвЂ™ theorem and trigonometry to them.
Example 1
H
ABCDEFGH is a cuboid with length 8 cm, breadth 6 cm and
height 9 cm.
a i Calculate the length of AC.
ii Calculate the length of AG.
b Calculate the size of angle GAC.
C
6 cm
A
E
8 cm
D
C
6 cm
8 cm
B
Look for a right-angled triangle where AC is one side and the lengths of the
other two sides are known.
ABC is a suitable triangle.
So draw triangle ABC marking the known lengths.
B
П­ 100
AC П­ 10 cm
ii
F
9 cm
A
Solution 1
a i
G
Use PythagorasвЂ™ theorem for this triangle.
G
Look for a right-angled triangle where AG is one side and the lengths of the
other two sides are known.
ACG is a suitable triangle.
G
So draw triangle ACG
marking the known lengths.
9 cm
C
A
10 cm
C
П­ 181
AG П­ Н™181
а·† П­ 13.4536 вЂ¦
AG П­ 13.5 cm (to 3 s.f.)
498
A
Use PythagorasвЂ™ theorem for this triangle.
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CHAPTER 31
31.1 Problems in three dimensions
b
G
9 cm
A
вњ“
10 cm
tan (angle GAC)П­ бЋЏ19бЋЏ0 П­ 0.9
angle GAC П­ 41.987 вЂ¦В°
Angle GAC П­ 42В°
(to the nearest degree)
For angle GAC.
9 cm is the opposite side.
10 cm is the adjacent side.
opp
tan П­ бЋЏбЋЏ
C
Exercise 31A
Where necessary give lengths correct to 3 significant figures and angles correct to one decimal place.
1 ABCDEFGH is a cuboid of length 8 cm, breadth 4 cm and
height 13 cm.
a Calculate the length of
ii GB
iii AC
iii FA
iv GA.
b Calculate the size of
ii angle GBC
iii angle GAC.
i angle FAB
H
G
E
F
13 cm
D
C
4 cm
A
B
8 cm
F
2 ABCDEF is a triangular prism.
In triangle ABC angle CAB П­ 90В°,
AB П­ 5 cm and AC П­ 12 cm.
In rectangle ABED the length
of BE П­ 15 cm.
a Calculate the length of CB.
b Calculate the length of
ii AF.
i CE
c Calculate the size of
i angle FED
C
D
E
12 cm
15 cm
A
5 cm
B
O
3 The diagram shows a square-based pyramid.
The lengths of sides of the square base, ABCD, are
10 cm and the base is on a horizontal plane.
The centre of the base is the point M and the vertex of
the pyramid is O, so that OM is vertical.
The point E is the midpoint of the side AB.
OA П­ OB П­ OC П­ OD П­ 15 cm.
D
a Calculate the length of i AC ii AM.
10 cm
b Calculate the length of OM.
c Calculate the size of angle OAM.
d Hence find the size of angle AOC.
e Calculate the length of OE.
f Calculate the size of angle OAB.
15 cm
C
M
B
E
A
B
Angle between a line and a plane
Imagine a light shining directly above AB onto the plane.
AN is the shadow of AB on the plane.
A line drawn from point B perpendicular to the plane will meet
the line AN and form a right angle with this line.
Angle BAN is the angle between the line AB and the plane.
N
A
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PythagorasвЂ™ theorem and trigonometry (2)
O
Example 2
The diagram shows a pyramid.
The base, ABCD, is a horizontal rectangle in which AB П­ 12 cm and AD П­ 9 cm.
The vertex, O, is vertically above the midpoint of the base and OB П­ 18 cm.
Calculate the size of the angle that OB makes with the horizontal plane.
Solution 2
18 cm
C
D
O
B
9 cm
A
18 cm
C
D
M
9 cm
B
12 cm
A
O
The base, ABCD, of the pyramid is horizontal so the angle that OB makes
with the horizontal plane is the angle that OB makes with the base ABCD.
Let M be the midpoint of the base which is directly below O.
Join O to M and M to B.
As OM is perpendicular to the base of the pyramid the angle OBM is the
angle between OB and the base and so is the required angle.
Draw triangle OBM marking OB П­ 18 cm.
To find the size of angle OBM find the
length of either MB or OM.
Calculate the length of MB which is 1бЋЏ2бЋЏ DB.
18 cm
D
Draw the right-angled
triangle ABD marking the
known lengths.
9 cm
A
M
12 cm
12 cm
B
B
DB2 П­ 225
DB П­ Н™225
а·† П­ 15
Use PythagorasвЂ™ theorem to calculate the length of DB.
MB П­ 1бЋЏ2бЋЏ DB П­ 7.5
O
For angle OBM, 18 cm is the hypotenuse, 7.5 cm is the adjacent side.
18 cm
M
7.5 cm
B
7.5
cos (angle OBM) П­ бЋЏбЋЏ
18
angle OBM П­ 65.37 вЂ¦В°
The angle between OB and the
horizontal plane is 65.4В° (to one d.p.)
500
cos П­ бЋЏбЋЏ
hyp
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31.1 Problems in three dimensions
вњ“
Exercise 31B
Where necessary give lengths correct to 3 significant figures and angles correct to one decimal place.
O
1 The diagram shows a pyramid.
The base, ABCD, is a horizontal rectangle in which
AB П­ 15 cm and AD П­ 8 cm.
The vertex, O, is vertically above the centre of the
base and OA П­ 24 cm.
Calculate the size of the angle that OA makes with
the horizontal plane.
24 cm
C
D
B
8 cm
15 cm
A
2 ABCDEFGH is a cuboid with a rectangular base in which
AB П­ 12 cm and BC П­ 5 cm.
The height, AE, of the cuboid is 15 cm.
Calculate the size of the angle
a between FA and ABCD
b between GA and ABCD
d Write down the size of the angle between HE and ABFE.
H
G
E
F
15 cm
D
C
5 cm
A
3 ABCDEF is a triangular prism.
In triangle ABC, angle CAB П­ 90В°, AB П­ 8 cm and AC П­ 10 cm.
In rectangle ABED, the length of BE П­ 5 cm.
C
Calculate the size of the angle between
a CB and ABED
b CD and ABED
c CE and ABED
10 cm
B
12 cm
F
D
E
5 cm
A
4 The diagram shows a square-based pyramid.
The lengths of sides of the square base, ABCD,
are 8 cm and the base is on a horizontal plane.
The centre of the base is the point M and the vertex
of the pyramid is O so that OM is vertical.
The point E is the midpoint of the side AB.
OA П­ OB П­ OC П­ OD П­ 20 cm
Calculate the size of the angle between OE
and the base ABCD.
B
8 cm
O
20 cm
C
D
M
8 cm
A
B
E
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CHAPTER 31
PythagorasвЂ™ theorem and trigonometry (2)
5 ABCD is a horizontal rectangular lawn in a garden and
TC is a vertical pole. Ropes run from the top of the pole,
T, to the corners, A, B and D, of the lawn.
a Calculate the length of the rope TA.
b Calculate the size of the angle made with the lawn by
ii the rope TD
iii the rope TA.
i the rope TB
T
6m
D
C
12 m
A
B
8m
6 The diagram shows a learnerвЂ™s ski slope, ABCD, of length, AB, 500 m. Triangles BAF and CDE
are congruent right-angled triangles and ABCD, AFED and BCEF are rectangles.
The rectangle BCEF is horizontal and the rectangle AFED is vertical.
The angle between AB and BCEF is 20В° and the angle between AC and BCEF is 10В°.
D
A
E
C
500 m
F
Calculate
a the length of FB
c the distance AC
B
b the height of A above F
d the width, BC, of the ski slope.
7 Diagram 1 shows a square-based pyramid OABCD.
Each side of the square is of length 60 cm and
OA П­ OB П­ OC П­ OD П­ 50 cm.
O
C
50 cm
D
B
60 cm
60 cm
A
Diagram 2 shows a cube, ABCDEFGH, in which
each edge is of length 60 cm.
A solid is made by placing the pyramid on top of the cube
so that the base, ABCD, of the pyramid is on the top,
ABCD, of the cube.
The solid is placed on a horizontal table with the face,
EFGH, on the table.
a Calculate the height of the vertex O above the table.
b Calculate the size of the angle between OE and the
horizontal.
502
Diagram 1
C
60 cm
60 cm
D
B
A
G
60 cm
H
F
E
Diagram 2
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31.2 Trigonometric ratios for any angle
31.2 Trigonometric ratios for any angle
The diagram shows a circle, centre the origin O and radius 1 unit. Imagine a line, OP, of length 1 unit
fixed at O, rotating in an anticlockwise direction about O, starting from the x-axis.
The diagram shows OP when it has rotated through 40В°.
y
1
0.8
P
0.6
1
0.4
0.2
40В°
ПЄ1.2 ПЄ1 ПЄ0.8 ПЄ0.6 ПЄ0.4 ПЄ0.2 O
0.2
Q
0.4
0.6
0.8
1
x
ПЄ0.2
ПЄ0.4
ПЄ0.6
ПЄ0.8
ПЄ1
ПЄ1.2
The right-angled triangle OPQ has hypotenuse OP П­ 1
Relative to angle POQ, side PQ is the opposite side and side OQ is the adjacent side.
This means that
OQ П­ cos 40В° and PQ П­ sin 40В°
For P, x П­ cos 40В° and y П­ sin 40В° so the coordinates of P are (cos 40В°, sin 40В°).
In general when OP rotates through any angle вђЄ В°, the position of P on the circle, radius П­ 1 is given
by x П­ cos вђЄ В°, y П­ sin вђЄ В°.
The coordinates of P are (cos вђЄ В°, sin вђЄ В°).
So when OP rotates through 400В° the coordinates of P are (cos 400В°, sin 400В°).
A rotation of 400В° is 1 complete revolution of 360В° plus a further rotation of 40В°.
The position of P is the same as in the previous diagram so (cos 400В°, sin 400В°) is the same point as
(cos 40В°, sin 40В°), therefore cos 400В° П­ cos 40В° and sin 400В° П­ sin 40В°.
If OP rotates through ПЄ40В° this means OP rotates through 40В° in a clockwise direction.
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PythagorasвЂ™ theorem and trigonometry (2)
For вђЄ П­ 136, вђЄ П­ 225, вђЄ П­ 304 and вђЄ П­ ПЄ40 the position of P is shown on the diagram.
y
1.2
1
0.8
P
0.6
0.4
136В°
0.2
ПЄ1.2 ПЄ1 ПЄ0.8 ПЄ0.6 ПЄ0.4 ПЄ0.2 O
0.2
ПЄ0.2
225В°
0.4 0.6
ПЄ40В°
0.8
x
1.2
ПЄ0.4
ПЄ0.6
P
304В°
P
1
ПЄ0.8
P
ПЄ1
ПЄ1.2
For P when вђЄ П­ 136, x П­ cos 136В° and y П­ sin 136В°.
From the diagram, cos 136В° ПЅ 0 and sin 136В° Пѕ 0
For P when вђЄ П­ 225, x П­ cos 225В° and y П­ sin 225В°.
From the diagram, cos 225В° ПЅ 0 and sin 225В° ПЅ 0
For P when вђЄ П­ 304, x П­ cos 304В° and y П­ sin 304В°.
From the diagram, cos 304В° Пѕ 0 and sin 304В° ПЅ 0
For P when вђЄ П­ ПЄ40, x П­ cos ПЄ40В° and y П­ sin ПЄ40В°.
From the diagram, cos ПЄ40В° Пѕ 0 and sin ПЄ40В° ПЅ 0
The diagram shows for each quadrant whether the sine and
cosine of angles in that quadrant are positive or negative.
cos ПЄ
2nd
1st
3rd
4th
sin ПЄ
sin ПЄ
cos ПЄ
The sine and cosine of any angle can be found using your calculator. The following table shows some
of these values corrected where necessary to 3 decimal places.
вђЄ
0
30
40
45
60
90
136
180
sin вђЄ В°
0
0.5
0.643
0.707
0.866
1
0.695
0
cos вђЄ В°
1
0.866
0.766
0.707
0.5
0
225
270
304
ПЄ0.707 ПЄ1 ПЄ0.829
ПЄ0.719 ПЄ1 ПЄ0.707
0
0.559
Using these values and others from a calculator the graphs of y П­ sin вђЄ В° and y П­ cos вђЄ В° can be
drawn. A graphical calculator would be useful here.
504
360
0
1
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31.2 Trigonometric ratios for any angle
Graph of y вЂ« ШЌвЂ¬sin вђЄ В°
y
1
0.5
ПЄ180
O
180
360
540 Оё
ПЄ0.5
Notice that the graph:
в—Џ cuts the вђЄ-axis at вЂ¦ , ПЄ180, 0, 180, 360, 540, вЂ¦
в—Џ repeats itself every 360В°, that is, it has a period
of 360В°
в—Џ has a maximum value of 1 at вђЄ П­ вЂ¦ , 90, 450, вЂ¦
в—Џ has a minimum value of ПЄ1 at
вђЄ П­ вЂ¦ , ПЄ90, 270, вЂ¦
ПЄ1
Graph of y вЂ« ШЌвЂ¬cos вђЄ В°
y
1
0.5
ПЄ180
O
180
360
540 Оё
ПЄ0.5
Notice that the graph:
в—Џ cuts the вђЄ-axis at вЂ¦ ПЄ90, 90, 270, 450, вЂ¦
в—Џ repeats itself every 360В°, that is it has a period
of 360В°
в—Џ has a maximum value of 1 at вђЄ П­ вЂ¦ , 0, 360, вЂ¦
в—Џ has a minimum value of ПЄ1 at
вђЄ П­ вЂ¦ , ПЄ180, 180, 540, вЂ¦
ПЄ1
Notice also that the graph of y П­ sin вђЄ В° and the graph of y П­ cos вђЄ В° are horizontal translations of
each other.
sin вђЄ В°
To find the value of the tangent of any angle, use tan вђЄ В° П­ бЋЏбЋЏ
cos вђЄ В°
From the graph of y П­ cos вђЄ В° it can be seen that cos вђЄ В° П­ 0 at вђЄ П­ 90, 270, 450, вЂ¦ for example.
As it is not possible to divide by 0 there are no values of tan вђЄ В° at вђЄ П­ 90, 270, 450, вЂ¦ that is, the
graph is discontinuous at these values of вђЄ.
Graph of y вЂ« ШЌвЂ¬tan вђЄ В°
y
8
6
4
2
ПЄ180
O
ПЄ2
ПЄ4
ПЄ6
ПЄ8
180
360
540 Оё
Notice that the graph:
в—Џ cuts the вђЄ-axis where tan вђЄ В° П­ 0, that is, at вЂ¦
ПЄ180, 0, 180, 360, 540 вЂ¦
в—Џ repeats itself every 180В°, that is it has a period
of 180В°
в—Џ does not have values at
вђЄ П­ П®90, П®270, П®450, вЂ¦
в—Џ does not have any maximum or minimum
points.
Notice also that tan вђЄ В° can take any value.
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PythagorasвЂ™ theorem and trigonometry (2)
Example 3
For values of вђЄ in the interval вЂ“ 180 to 360 solve the equation
ii sin вђЄ В° П­ 0.7
ii 5 cos вђЄ В° П­ 2
Give each answer correct to one decimal place.
Solution 3
i sin вђЄ В° П­ 0.7
Use a calculator to find one value of вђЄ .
вђЄ П­ 44.4
y
1
ПЄ180
y П­ 0.7
O
180
360 Оё
To find the other solutions draw a sketch
of y П­ sin вђЄ В° for вђЄ from ПЄ180 to 360
The sketch shows that there are two
values of вђЄ in the interval ПЄ180 to
360 for which sin вђЄ В° П­ 0.7
One solution is вђЄ П­ 44.4 and by symmetry
the other solution is вђЄ П­ 180 ПЄ 44.4
ПЄ1
вђЄ П­ 44.4, 180 ПЄ 44.4
вђЄ П­ 44.4, 135.6
ii 5 cos вђЄ В° П­ 2
Divide each side of the equation by 5
cos вђЄ В° П­ 2бЋЏ5бЋЏ П­ 0.4
вђЄ П­ 66.4
Use a calculator to find one value of вђЄ .
y
To find the other solutions draw a sketch of
y П­ cos вђЄ В° for вђЄ from ПЄ180 to 360
1
y П­ 0.4
ПЄ180
O
180
The sketch shows that there are three values of вђЄ in
the interval ПЄ180 to 360 for which cos вђЄ В° П­ 0.4
360 Оё
ПЄ1
вђЄ П­ 66.4, ПЄ66.4, 360 П© ПЄ66.4
вђЄ П­ 66.4, ПЄ66.4, 293.6
One solution is вђЄ П­ 66.4 and by symmetry
another solution is вђЄ П­ ПЄ66.4
Using the period of the graph the other solution
is вђЄ П­ 360 П© ПЄ66.4
Exercise 31C
1 For ПЄ360 СЂ вђЄ СЂ 360 sketch the graph of
a y П­ sin вђЄ В°
b y П­ cos вђЄ В°
2 Find all values of вђЄ in the interval 0 to 360 for which
a sin вђЄ В° П­ 0.5
b cos вђЄ В° П­ 0.1
c y П­ tan вђЄ В°.
c tan вђЄ В° П­ 1
3 a Show that one solution of the equation 3 sin вђЄ В° П­ 1 is 19.5, correct to 1 decimal place.
b Hence solve the equation 3 sin вђЄ В° П­ 1 for values of вђЄ in the interval 0 to 720
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31.3 Area of a triangle
4 a Show that one solution of the equation 10 cos вђЄ В° П­ ПЄ3 is 107.5 correct to 1 decimal place.
b Hence find all values of вђЄ in the interval ПЄ360 to 360 for which 10 cos вђЄ В° П­ ПЄ3
5 Solve 4 tan вђЄ В° П­ 3 for values of вђЄ in the interval ПЄ180 to 360
31.3 Area of a triangle
Labelling sides and angles
The vertices of a triangle are labelled with capital letters.
The triangle shown is triangle ABC.
B
c
a
C
A
b
The sides opposite the angles are labelled so that a is the length of the side opposite angle A, b is the
length of the side opposite angle B and c is the length of the side opposite angle C.
Area of a triangle П­ бЋЏ12бЋЏ base П« height
B
Area of triangle ABC П­ 1бЋЏ2бЋЏ bh
In the right-angled triangle BCN
a
h П­ a sin C
So area of triangle ABC П­ 1бЋЏ2бЋЏ b П« a sin C that is
C
c
h
A
N b
area of triangle ABC вЂ« ШЌвЂ¬ab sin C
1
бЋЏбЋЏ
2
The angle C is the angle between the sides of length a and b and is called the included angle.
The formula for the area of a triangle means that
Area of a triangle П­ 1бЋЏ2бЋЏ product of two sides П« sine of the included angle.
For triangle ABC there are other formulae for the area.
Area of triangle ABC П­ 1бЋЏ2бЋЏ ab sin C П­ 1бЋЏ2бЋЏ bc sin A П­ 1бЋЏ2бЋЏ ac sin B.
These formulae give the area of a triangle whether the included angle is acute or obtuse.
Example 4
Find the area of each of the triangles correct to 3 significant figures.
a
b
B
7.3 cm
C
16.2 m
37В°
5.8 cm
A
118В°
7.4 m
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PythagorasвЂ™ theorem and trigonometry (2)
Solution 4
a Area П­ 1бЋЏ2бЋЏ П« 7.3 П« 5.8 П« sin 37В°
Substitute a П­ 7.3 cm, b П­ 5.8 cm, C П­ 37В° into area П­ 1бЋЏ2бЋЏ ab sin C
Area П­ 12.74 вЂ¦
Area П­ 12.7 cm2
Give the area correct to 3 significant figures and state the units.
b Area П­ 1бЋЏ2бЋЏ П« 7.4 П« 16.2 П« sin 118В°
Area П­ 52.92 вЂ¦
Substitute into
area of a triangle П­ бЋЏ12бЋЏ product of two sides П« sine of the included angle.
Area П­ 52.9 m2
Example 5
xВ°
2
The area of this triangle is 20 cm .
Find the size of the acute angle xВ°.
Give your angle correct to one decimal place.
Solution 5
1
бЋЏбЋЏ П« 8.1 П« 6.4 П« sin xВ° П­ 20
2
8.1 cm
6.4 cm
Use
area of a triangle П­ 1бЋЏ2бЋЏ product of two sides П« sine of the included angle.
2 П« 20
sin xВ° П­ бЋЏбЋЏ П­ 0.7716
8.1 П« 6.4
Find the value of sin xВ°.
xВ° П­ 50.49 вЂ¦В°
xВ° П­ 50.5В°
Give the angle correct to one decimal place.
Exercise 31D
Give lengths and areas correct to 3 significant figures and angles correct to one decimal place.
1 Work out the area of each of these triangles.
i
ii
iii
28В°
9.3 cm
10.6 cm
13.5 cm
9.2 cm
34.7В°
43В°
6.9 cm
9.1 cm
iv
v
vi
8.6 cm
148.6В°
13.4 cm
76.3В°
4.6 cm
4.6 cm
9.6 cm
137В°
4.7 cm
Work out the area of the quadrilateral.
D
57В°
9.4 cm
C
12.6 cm
8.6 cm
80В°
A
508
B
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CHAPTER 31
31.3 Area of a triangle
3
The area of triangle ABC is 15 cm2
Angle A is acute.
Work out the size of angle A.
C
6.5 cm
A
B
8.4 cm
C
4 The area of triangle ABC is 60.7 m2
Work out the length of BC.
A
35В°
B
12.6 m
5 a Triangle ABC is such that a П­ 6 cm, b П­ 9 cm and angle C П­ 25В°.
Work out the area of triangle ABC.
b Triangle PQR is such that p П­ 6 cm, q П­ 9 cm and angle R П­ 155В°.
Work out the area of triangle PQR.
6 The diagram shows a regular octagon with centre O.
a Work out the size of angle AOB.
OA П­ OB П­ 6 cm.
b Work out the area of triangle AOB.
c Hence work out the area of the octagon.
A
B
O
7 Work out the area of the parallelogram.
5.7 cm
63В°
12.8 cm
8 a An equilateral triangle has sides of length 12 cm.
Calculate the area of the equilateral triangle.
b A regular hexagon has sides of length 12 cm.
Calculate the area of the regular hexagon.
9 The diagram shows a sector, AOB, of a circle, centre O.
The radius of the circle is 8 cm and the size of angle AOB is 50В°.
a Work out the area of triangle AOB.
b Work out the area of the sector AOB.
c Hence work out the area of the segment shown shaded in
the diagram.
A
8 cm
50В°
O
B
8 cm
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PythagorasвЂ™ theorem and trigonometry (2)
31.4 The sine rule
B
a
c
C
A
b
The last section showed that
1
1
1
Area of triangle ABC П­ 2 ab sin C П­ 2 bc sin A П­ 2 ca sin B
1
бЋЏбЋЏ
2
ab sin C вЂ« ШЌвЂ¬1бЋЏ2бЋЏ bc sin A
and
cancelling 1бЋЏ2бЋЏ and b from both sides
a sin C вЂ« ШЌвЂ¬c sin A
1
бЋЏбЋЏ
2
bc sin A вЂ« ШЌвЂ¬1бЋЏ2бЋЏ ca sin B
cancelling 1бЋЏ2бЋЏ and c from both sides
and
or
b sin A вЂ« ШЌвЂ¬a sin B
or
a
c
бЋЏбЋЏ вЂ« ШЌвЂ¬бЋЏбЋЏ
sin A sin C
and
b
a
бЋЏбЋЏ вЂ« ШЌвЂ¬бЋЏбЋЏ
sin B sin A
Combining these results
a
b
c
бЋЏбЋЏ вЂ« ШЌвЂ¬бЋЏбЋЏ вЂ« ШЌвЂ¬бЋЏбЋЏ
sin A sin B sin C
This result is known as the sine rule and can be used in any triangle.
Using the sine rule to calculate a length
Example 6
Find the length of the side marked a in the triangle.
74В°
a
37В°
Solution 6
a
8.4
бЋЏбЋЏ П­ бЋЏбЋЏ
sin
74В°
sin 37В°
8.4 П« sin 37В°
a П­ бЋЏбЋЏ
sin 74В°
a П­ 5.258 вЂ¦
a П­ 5.26 cm
510
8.4 cm
a
b
Substitute A П­ 37В°, b П­ 8.4, B П­ 74В° into бЋЏбЋЏ П­ бЋЏбЋЏ.
sin A sin B
Multiply both sides by sin 37В°.
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CHAPTER 31
31.4 The sine rule
Example 7
Find the length of the side marked x in the triangle.
18В°
x
Solution 7
Missing angle П­ 180 ПЄ (18 П© 124)
П­ 38В°
9.7 cm
124В°
18В°
The angle opposite 9.7 cm must be found before the sine rule can be used.
Use the angle sum of a triangle.
x
9.7 cm
124В°
38В°
x
9.7
бЋЏбЋЏ П­ бЋЏбЋЏ
sin 124В° sin 38В°
9.7 П« sin 124В°
x П­ бЋЏбЋЏ
sin 38В°
Write down the sine rule with x opposite 124В° and 9.7 opposite 38В°.
Multiply both sides by sin 124В°.
x П­ 13.06 вЂ¦
x П­ 13.1 cm
Using the sine rule to calculate an angle
When the sine rule is used to calculate an angle it is a good idea to turn each fraction upside down
(the reciprocal). This gives
sin A sin B sin C
бЋЏбЋЏ П­ бЋЏбЋЏ П­ бЋЏбЋЏ
a
b
c
Example 8
Find the size of the acute angle x in the triangle.
74В°
7.9 cm
x
Solution 8
sin x sin 74В°
бЋЏбЋЏ П­ бЋЏбЋЏ
8.4
7.9
8.4 cm
Write down the sine rule with x opposite 7.9 and 74В° opposite 8.4
7.9 П« sin 74В°
sin x П­ бЋЏбЋЏ
8.4
Multiply both sides by 7.9
sin x П­ 0.904 вЂ¦
Find the value of sin x.
x П­ 64.69 вЂ¦В°
x П­ 64.7В°
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CHAPTER 31
PythagorasвЂ™ theorem and trigonometry (2)
Exercise 31E
Give lengths and areas correct to 3 significant figures and angles correct to 1 decimal place.
1 Find the lengths of the sides marked with letters in these triangles.
a
b
c
a
79В°
b
27В°
46В°
58В°
4.2 cm
51В°
62В°
11 cm
6.1 cm
d
e
64В°
d
c
f
17В°
13.6 cm
113В°
f
e
g
22В°
76В°
134В°
14.9 cm
6.1 cm
2 Calculate the size of each of the acute angles marked with a letter.
a
b
c
6 cm
A
8 cm
32В°
17 cm
21В°
6.8 cm
d
C
18.4 cm
73В°
9.1 cm
B
E
104В°
7.6 cm
3 The diagram shows quadrilateral ABCD
and its diagonal AC.
a In triangle ABC, work out the length of AC.
b In triangle ACD, work out the size of angle DAC.
c Work out the size of angle BCD.
D
D
12.7 cm
8.6 cm
C
102В°
5.7 cm
46В°
18В°
B
A
4 In triangle ABC, BC П­ 8.6 cm, angle BAC П­ 52В° and angle ABC П­ 63В°.
a Calculate the length of AC.
b Calculate the length of AB.
c Calculate the area of triangle ABC.
5 In triangle PQR all the angles are acute. PR П­ 7.8 cm and PQ П­ 8.4 cm.
Angle PQR П­ 58В°.
a Work out the size of angle PRQ.
b Work out the length of QR.
6 The diagram shows the position of a port (P),
a lighthouse (L) and a buoy (B).
The lighthouse is due east of the buoy.
B
The lighthouse is on a bearing of 035В° from the port
and the buoy is on a bearing of 312В° from the port.
ii angle PLB.
a Work out the size of
i angle PBL
The lighthouse is 8 km from the port.
b Work out the distance PB.
c Work out the distance BL.
d Work out the shortest distance from the port (P) to the line BL.
512
N
L
P
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CHAPTER 31
31.5 The cosine rule
31.5 The cosine rule
The diagram shows triangle ABC.
The line BN is perpendicular to AC and meets the line AC at N so that AN П­ x and NC П­ (b ПЄ x).
The length of BN is h.
In triangle ANB
PythagorasвЂ™ theorem
gives
c2 П­ x2 П© h2 1
In triangle BNC
PythagorasвЂ™ theorem
gives
a2 П­ (b ПЄ x)2 П© h2
Using 1 substitute c2 for x2 П© h2
a2 П­ b2 ПЄ 2bx П© c2 2
B
c
a
h
x
(b ПЄ x)
N
A
b
C
In the right-angled triangle ANB, x П­ c cos A
Substituting this into 2
a2 вЂ« ШЌвЂ¬b2 Ш‰ c2 ШЉ 2bc cos A
This result is known as the cosine rule and can be used in any triangle.
b2 вЂ« ШЌвЂ¬a2 Ш‰ c2 ШЉ 2ac cos B
c2 вЂ« ШЌвЂ¬a2 Ш‰ b2 ШЉ 2ab cos C
Similarly
and
Using the cosine rule to calculate a length
Example 9
Find the length of the side marked with a letter in each triangle.
a
b
B
a
x
8 cm
7.3 cm
117В°
24В°
C
12 cm
Solution 9
a a2 П­ 122 П© 82 ПЄ 2 П« 12 П« 8 П« cos 24В°
a П­ 144 П© 64 ПЄ 175.4007 вЂ¦
2
5.8 cm
A
Substitute b П­ 12, c П­ 8, A П­ 24В° into a2 П­ b2 П© c2 ПЄ 2bc cos A.
Evaluate each term separately.
a2 П­ 32.599 27 вЂ¦
a П­ Н™32.599
а·†а·†
27 вЂ¦
a П­ 5.709 577 вЂ¦
a П­ 5.71 cm
b x2 П­ 7.32 П© 5.82 ПЄ 2 П« 7.3 П« 5.8 П« cos 117В°
x2 П­ 53.29 П© 33.64 ПЄ 84.68 П« (ПЄ0.4539 вЂ¦)
x2 П­ 86.93 П© 38.44 вЂ¦
x2 П­ 125.37 вЂ¦
x П­ Н™125.37
а·†а·†
вЂ¦
x П­ 11.19 вЂ¦
x П­ 11.2 cm
Take the square root.
Substitute the two given lengths and the included
angle into the cosine rule.
cos 117В° ПЅ 0
Take the square root.
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CHAPTER 31
PythagorasвЂ™ theorem and trigonometry (2)
Using the cosine rule to calculate an angle
To find an angle using the cosine rule, when the lengths of all three sides of a triangle are known,
rearrange a2 П­ b2 П© c2 ПЄ 2bc cos A.
2bc cos A П­ b2 П© c2 ПЄ a2
b2 Ш‰ c2 ШЉ a2
cos A вЂ« ШЌвЂ¬бЋЏбЋЏ
2bc
Similarly
a2 Ш‰ c2 ШЉ b2
cos B вЂ« ШЌвЂ¬бЋЏбЋЏ
2ac
and
a2 Ш‰ b2 ШЉ c2
cos C вЂ« ШЌвЂ¬бЋЏбЋЏ
2ab
Example 10
Find the size of
a angle BAC
b angle X.
a
b
B
12.7 cm
16 cm
13 cm
8.6 cm
X
A
11 cm
C
Solution 10
a cos A П­ бЋЏбЋЏ
2 П« 11 П« 16
208
cos A П­ бЋЏбЋЏ
352
cos A П­ 0.590 909 вЂ¦
A П­ 53.77 вЂ¦В°
A П­ 53.8В°
b cos X П­ бЋЏбЋЏ
2 П« 8.6 П« 6.9
ПЄ39.72
cos X П­ бЋЏбЋЏ
118.68
cos X П­ ПЄ0.334 68 вЂ¦
X П­ 109.55 вЂ¦В°
X П­ 109.6В°
514
6.9 cm
Substitute b П­ 11, c П­ 16, a П­ 13 into cos A П­ бЋЏбЋЏ.
2bc
Substitute the three lengths into the cosine rule noting that 12.7 cm is
opposite the angle to be found.
The value of cos X is negative so X is an obtuse angle.
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CHAPTER 31
31.5 The cosine rule
Exercise 31F
Where necessary give lengths and areas correct to 3 significant figures and angles correct to
1 decimal place.
1 Calculate the length of the sides marked with letters in these triangles.
a
b
c
a
8 cm
11.3 cm
b
18В°
16.2 cm
62В°
15.5 cm
75В°
9 cm
9.2 cm
c
d
e
d
f
10.2 cm
9.6 cm
8.4 cm
e
147В°
134В°
52В°
9.6 cm
f
8.4 cm
6.3 cm
2 Calculate the size of each of the angles marked with a letter in these triangles.
a
b
7 cm
9 cm
9.4 cm
15.3 cm
A
B
11 cm
13.6 cm
c
d
8.6 cm
C
D
8.7 cm
8.7 cm
7.2 cm
14.4 cm
6.8 cm
3 The diagram shows the quadrilateral ABCD.
a Work out the length of DB.
b Work out the size of angle DAB.
c Work out the area of quadrilateral ABCD.
C
26.4 cm
D
56В°
8.4 cm
9.8 cm
A
16.3 cm
B
4 Work out the perimeter of triangle PQR.
R
8.6 cm
Q
27В°
P
10.9 cm
5 In triangle ABC, AB П­ 10.1 cm, AC П­ 9.4 cm and BC П­ 8.7 cm.
Calculate the size of angle BAC.
6 In triangle XYZ, XY П­ 20.3 cm, XZ П­ 14.5 cm and angle YXZ П­ 38В°.
Calculate the length of YZ.
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CHAPTER 31
PythagorasвЂ™ theorem and trigonometry (2)
7 AB is a chord of a circle with centre O.
The radius of the circle is 7 cm and the length of the chord is 11 cm.
Calculate the size of angle AOB.
O
7 cm
B
11 cm
A
8 The region ABC is marked on a school field.
The point B is 70 m from A on a bearing of 064В°.
The point C is 90 m from A on a bearing of 132В°.
a Work out the size of angle BAC.
b Work out the length of BC.
N
B
70 m
A
90 m
C
9 Chris ran 4 km on a bearing of 036В° from P to Q. He then ran in a straight line from Q to R
where R is 7 km due East of P. Chris then ran in a straight line from R to P.
Calculate the total distance run by Chris.
10 The diagram shows a parallelogram.
Work out the length of each diagonal of the parallelogram.
6 cm
65В°
8 cm
31.6 Solving problems using the sine rule, the cosine rule and
1
бЋЏбЋЏ ab sin C
2
Example 11
The area of triangle ABC is 12 cm2
AB П­ 3.8 cm and angle ABC П­ 70В°.
ii AC.
a Find the length of
i BC
b Find the size of angle BAC.
A
3.8 cm
70В°
B
Solution 11
a i 1бЋЏ2бЋЏ П« 3.8 П« BC sin 70В° П­ 12
C
Substitute c П­ 3.8, B П­ 70В° into area П­ бЋЏ12бЋЏ ac sin B.
2 П« 12
BC П­ бЋЏбЋЏ
3.8 sin 70В°
BC П­ 6.721 вЂ¦
BC П­ 6.72 cm
ii b2 П­ 6.721 вЂ¦2 П© 3.82 ПЄ 2 П« 6.721вЂ¦ П« 3.8 cos 70В°
b2 П­ 59.613 вЂ¦ ПЄ 17.470 вЂ¦
b2 П­ 42.142 вЂ¦
b П­ 6.491 вЂ¦
AC П­ 6.49 cm
516
Substitute a П­ 6.721 вЂ¦, c П­ 3.8 and B П­ 70В°
into b2 П­ a2 П© c2 ПЄ 2ac cos B.
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1
31.6 Solving problems using the sine rule, the cosine rule and бЋЏ2бЋЏab sin C
sin A
sin 70В°
b бЋЏбЋЏ П­ бЋЏ бЋЏ
6.721 вЂ¦ 6.491 вЂ¦
6.721 вЂ¦ П« sin 70В°
sin A П­ бЋЏбЋЏ
6.491 вЂ¦
CHAPTER 31
sin A sin B
Substitute a П­ 6.721 вЂ¦, b П­ 6.491 вЂ¦ and B П­ 70В° into бЋЏбЋЏ П­ бЋЏбЋЏ.
a
b
sin A П­ 0.9728 вЂ¦
A П­ 76.62 вЂ¦В°
Angle BAC П­ 76.6В°
Exercise 31G
Where necessary give lengths and areas correct to 3 significant figures and angles correct to
1 decimal place, unless the question states otherwise.
1 A triangle has sides of lengths 9 cm, 10 cm and 11 cm.
a Calculate the size of each angle of the triangle.
b Calculate the area of the triangle.
2 In the diagram ABC is a straight line.
a Calculate the length of BD.
b Calculate the size of angle DAB.
c Calculate the length of AC.
D
12.9 cm
A 5.4cm B
12 cm
65В°
47В°
C
2
3 The area of triangle ABC is 15 cm .
AB П­ 4.6 cm and angle BAC П­ 63Лљ.
a Work out the length of AC.
b Work out the length of BC.
c Work out the size of angle ABC.
4 ABCD is a kite with diagonal DB.
a Calculate the length of DB.
b Calculate the size of angle BDC.
c Calculate the value of x.
d Calculate the length of AC.
D
x cm
8 cm
A 50В°
C
30В°
8 cm
x cm
B
5 Kultar walked 9 km due South from point A to point B.
He then changed direction and walked 5 km to point C.
Kultar was then 6 km from his starting point A.
a Work out the bearing of point C from point B. Give your answer correct to the nearest degree.
b Work out the bearing of point C from point A. Give your answer correct to the nearest degree.
6 The diagram shows a pyramid. The base of the pyramid, ABCD, is
a rectangle in which AB П­ 15 cm and AD П­ 8 cm.
The vertex of the pyramid is O where OA П­ OB П­ OC П­ OD П­ 20 cm.
Work out the size of angle DOB correct to the nearest degree.
O
20 cm
C
D
8 cm
A
15 cm
B
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CHAPTER 31
PythagorasвЂ™ theorem and trigonometry (2)
7 The diagram shows a vertical pole, PQ, standing on a hill.
The hill is at an angle of 8В° to the horizontal.
The point R is 20 m downhill from Q and the line PR is at 12В° to the hill.
a Calculate the size of angle RPQ.
b Calculate the length, PQ, of the pole.
12В°
R
P
Q
20 m
8В°
8 A, B and C are points on horizontal ground so that
AB П­ 30 m, BC П­ 24 m and angle CAB П­ 50В°.
AP and BQ are vertical posts, where AP П­ BQ П­ 10 m.
a Work out the size of angle ACB.
b Work out the length of AC.
c Work out the size of angle PCQ.
d Work out the size of the angle between
QC and the ground.
P
Q
10 m
10 m
30 m
A
B
50В°
24 m
C
Chapter summary
You should now be able to:
в�…
use PythagorasвЂ™ theorem to solve problems in 3 dimensions
в�…
use trigonometry to solve problems in 3 dimensions
в�…
work out the size of the angle between a line and a plane
в�…
draw sketches of the graphs of y П­ sin x В°, y П­ cos x В°, y П­ tan x В° and use these graphs to
solve simple trigonometric equations
в�…
use the formula area П­ бЋЏ12бЋЏ ab sin C to calculate the area of any triangle
в�…
a
b
c
use the sine rule бЋЏбЋЏ П­ бЋЏбЋЏ П­ бЋЏбЋЏ and the cosine rule a2 П­ b2 П© c2 ПЄ 2bc cos A in
sin A sin B sin C
triangles and in solving problems.
Chapter 31 review questions
1 In the diagram, XY represents a vertical tower
on level ground.
A and B are points due West of Y.
The distance AB is 30 metres.
The angle of elevation of X from A is 30Вє.
The angle of elevation of X from B is 50Вє.
Calculate the height, in metres, of the tower XY.
X Diagram NOT
accurately drawn
50В°
30В°
30 m
B
Y
(1384 June 1996)
2 The diagram shows triangle ABC.
AC П­ 7.2 cm
BC П­ 8.35 cm
Angle ACB П­ 74В°.
a Calculate the area of triangle ABC.
b Calculate the length of AB.
518
C
Diagram NOT
accurately drawn
74В°
7.2 cm
A
8.35 cm
B
(1385 June 2002)
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CHAPTER 31
Chapter 31 review questions
3 In triangle ABC
AC П­ 8 cm
CB П­ 15 cm
Angle ACB П­ 70В°.
a Calculate the area of triangle ABC.
3 significant figures.
X
8 cm
70В°
C
4 The diagram shows a cuboid.
A, B, C, D and E are five vertices of the cuboid.
AB П­ 5 cm
BC П­ 8 cm
CE П­ 3 cm.
B
15 cm
X is the point on AB such that angle CXB П­ 90В°.
b Calculate the length of CX.
(1387 June 2003)
Diagram NOT
accurately drawn
E
3 cm
D
C
Calculate the size of the angle the
diagonal AE makes with the plane
A
ABCD.
5 In triangle ABC
AC П­ 8 cm
BC П­ 15 cm
Angle ACB П­ 70В°.
a Calculate the length of AB.
3 significant figures.
b Calculate the size of angle BAC.
place.
Diagram NOT
accurately drawn
A
8 cm
5 cm
B
Diagram NOT
accurately drawn
A
8 cm
70В°
B
6 This is a sketch of the graph of y П­ cos xВ°
for values of x between 0 and 360.
Write down the coordinates of the point
ii A
ii B.
C
15 cm
(1387 June 2003)
y
A
360 x
O
B
7 Angle ACB П­ 150В°
BC П­ 60 m.
The area of triangle ABC is 450 m2
Calculate the perimeter of triangle ABC.
C
Diagram NOT
accurately drawn
150В°
60 m
A
B
(1385 November 2000)
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CHAPTER 31
PythagorasвЂ™ theorem and trigonometry (2)
8 The diagram shows a quadrilateral ABCD.
AB П­ 4.1 cm
BC П­ 7.6 cm
Angle ABC П­ 117В°
a Calculate the length of AC.
b Calculate the area of triangle ABC.
c Calculate the area of the quadrilateral ABCD.
Diagram NOT
accurately drawn
A
5.4 cm
4.1 cm
62В°
B 117В°
D
7.6 cm
C
(1385 June 2000)
9 This is a graph of the curve y П­ sin xВ° for 0 СЂ x СЂ 180
y
1
0.5
O
45
90
135
x
180
ПЄ0.5
ПЄ1
a Using the graph or otherwise, find estimates of the solutions in the interval 0 СЂ x СЂ 360 of
the equation
ii sin xВ° П­ ПЄ0.6.
i sin xВ° П­ 0.2
cos xВ° П­ sin (x П© 90)В° for all values of x.
b Write down two solutions of the equation cos xВ° П­ 0.2
(1385 November 2002)
10 In the diagram, ABCD, ABFE and EFCD are rectangles.
The plane EFCD is horizontal and the plane ABFE is vertical.
EA П­ 10 cm
DC П­ 20 cm
ED П­ 20 cm.
B
A
10 cm
20 cm
Calculate the size of the angle that the line AC makes with
the plane EFCD.
11 In triangle ABC
AB П­ 10 cm
AC П­ 14 cm
BC П­ 16 cm.
a Calculate the size of the smallest angle in the triangle.
b Calculate the area of triangle ABC.
520
F
E
D
C
Diagram NOT
accurately drawn
A
14 cm
10 cm
B
20 cm
16 cm
C
```