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C H A P T E R
13
MODULE 2
E
Trigonometry
PL
How do we understand and apply the trigonometric ratios sine, cosine and tangent?
How do we solve right-angled triangles using trigonometric ratios?
How do we apply the definition of sine and cosine for any angle?
How do we solve non right-angled triangles using:
r the sine rule?
r the cosine rule?
How do we calculate the areas of triangles using the formulas:
13.1
M
Area = 1 bc sin A
2
Area = s(s − a)(s − b)(s − c) where s is the
a+b+c
semi-perimeter and s =
2
Defining sine, cosine and tangent
y
SA
(0, 1)
The unit circle is a circle of radius 1
with centre at the origin.
(0, 0)
(–1, 0)
(1, 0)
x
(0, –1)
y
θ°
(0, 0)
P(cos(θ°), sin(θ°))
x
Sine and cosine may be defined for any
angle through the unit circle.
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Chapter 13 – Trigonometry
393
For an angle of ◦ , a point P on the unit circle is defined as illustrated above.
The angle is measured in an anticlockwise direction from the positive direction of the x axis.
cos ( ◦ ) is defined as the x coordinate of the point P and sin ( ◦ ) is defined as the y coordinate
of P. Your calculator gives approximate values for these coordinates.
y
y
(–0.1736, 0.9848)
(–0.7071, 0.7071)
(0.8660, 0.5)
30°
y
135°
100°
x
In this chapter, angles greater than 180◦ or less than 0◦ will not be considered.
For a right-angled triangle OBC a similar triangle
OB C can be constructed that lies in the unit circle.
From the above, OC = cos ( ◦ ) and C B = sin ( ◦ ).
B'
The scale factor is the length OB.
1
◦
◦
Hence BC = OB sin ( ) and OC = OB cos ( ).
θ°
O
This implies
C'
and
B
C
OC
= cos ( ◦ ).
OB
M
BC
= sin ( ◦ )
OB
x
cos 100◦ = −0.1736
sin 100◦ = 0.9848
sin 135◦ ≈ 0.7071
cos 135◦ ≈ −0.7071
PL
sin 30◦ = 0.5 (an exact value)
cos 30◦ ≈ 0.8660
E
x
This gives the ratio definition of sine and cosine for a right-angled triangle. The naming of
sides with respect to an angle ◦ is as shown.
opposite
opp
hyp
hypotenuse
adjacent
adj
cos ( ◦ ) =
hyp
hypotenuse
◦
opp
opposite
sin
(
)
◦
tan ( ) =
=
cos ( ◦ )
adj adjacent
B
SA
sin ( ◦ ) =
From the unit circle note that
sin ( ◦ ) = sin (180 − )◦
e.g. sin 45◦ = sin 135◦ and
cos ( ◦ ) = −cos (180 − )◦
e.g. cos (45◦ ) = −cos (135◦ )
hypotenuse
opposite
O
θ°
C
adjacent
y
(cos(180 – θ)°, sin(180 – θ)°)
(180 – θ)°
(cos(θ°), sin(θ°))
θ°
O
x
This result will be used later in this chapter.
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394
Essential Further Mathematics – Module 2 Geometry and trigonometry
Example 1
The use of sine
C
Find the value of x correct to two decimal places.
80
A
cm
29.6°
x cm
B
Solution
E
x
= sin 29.6◦
80
∴ x = 80 sin 29.6◦
= 39.5153
PL
1 The hypotenuse length and the
angle A are given. The length of the
side opposite the angle is to be found.
2 Choose the appropriate trigonometric ratio.
OPP
.
In this case sine, as sin =
HYP
3 Solve the equation using a calculator to
evaluate sin 29.6.
∴ x = 39.52 correct to two
decimal places.
The use of cosine
M
Example 2
B
Find the length of the hypotenuse correct to two decimal places.
Solution
SA
1 The adjacent side length and the angle A are given.
The length of the hypotenuse is to be found.
2 Choose the appropriate trigonometric ratio.
ADJ
.
In this case cosine as cos =
HYP
3 Solve the equation using a calculator to
evaluate cos 15◦ .
4 Write down your answer.
A
15°
10 cm
AB is the hypotenuse
10
= cos 15◦
AB
∴ 10 = AB cos 15◦
10
∴ AB =
cos 15◦
The length of AB = 10.35 cm
correct to two decimal places.
Note: University
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Chapter 13 – Trigonometry
395
Example 3
The use of tan
A
Find the magnitude of ∠ ABC.
11 cm
B
x°
C
3 cm
Solution
tan x =
11
3
11
3
E
1 The adjacent side length and opposite side length are
given. The size of the angle at B is to be found.
2 Choose the appropriate trigonometric ratio. In this
OPP
.
case tan as tan(x) =
ADJ
3 Solve the equation
using a calculator to evaluate
11
tan−1
.
3
PL
∴ x = tan−1
∴ x = 74.74 correct to
two decimal places
4 Write down your answer.
M
Exercise 13A
1 Use sine to find x and cosine to
find y (correct to 4 decimal places).
2 Use tangent to find x (correct to 4 decimal
places).
B
B
10 cm
SA
x cm
A
y cm
31°
x cm
35°
10 cm
A
C
3 Find the value of x in each of the following:
a
c
b
10 cm
m
5c
x cm
x cm
5°
20.16°
35°
x cm
d
C
8 cm
e
f
x cm
10 cm
10 cm
30.25°
7 cm
x°
15 cm
40°
x cm
4 An equilateral triangle has altitudes of length 20 cm. Find the length of one side.
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396
Essential Further Mathematics – Module 2 Geometry and trigonometry
5 The base of an isosceles triangle is 12 cm long and the equal sides are 15 cm long. Find the
magnitude of each of the three angles of the triangle.
6 A pole casts a shadow 20 m long when the altitude of the
sun is 49◦ . Calculate the height of the pole.
pole
49°
20 m
7 This figure represents a ramp.
E
a Find the magnitude of angle ACB.
b Find the distance BC.
A
6m
1m
C
P
PL
8 This figure shows a vertical mast PQ, which
stands on horizontal ground. A straight wire
20 m long runs from P at the top of the mast
to a point R on the ground, which is 10 m from
the foot of the mast.
B
a Calculate the angle of inclination, ◦ , of the wire to
the ground.
b Calculate the height of the mast.
20
m
θ°
R
Q
10 m
M
9 A ladder leaning against a vertical wall makes an angle of 26◦ with the wall. If the foot of
the ladder is 3 m from the wall calculate:
a the length of the ladder b the height it reaches above the ground
10 An engineer is designing a straight concrete entry ramp, 60 m long, for a car park 13 m
above street level. Calculate the angle of the ramp to the horizontal.
SA
11 A vertical mast is secured from its top by straight cables 200 m long fixed at the ground.
The cables make angles of 66◦ with the ground. What is the height of the mast?
12 A mountain railway rises for 400 m at a uniform slope of 16◦ with the horizontal. What is
the distance travelled by a train for this rise?
13 The diagonals of a rhombus bisect each
other at right angles.
If BD = AC = 10 cm, find:
B
O
a the length of the sides of the rhombus
b the magnitude of angle ABC
14 A pendulum swings from the vertical through
an angle of 15◦ on each side of the vertical. If
the pendulum is 90 cm long, what is the distance
x cm between its highest and lowest points?
C
A
D
15°
90 cm
90 cm
x cm
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Chapter 13 – Trigonometry
397
15 A picture is hung symmetrically by means of a string
passing over a nail with its ends attached to two rings
on the upper edge of the picture. The distance
between the rings is 30 cm and the angle between
the two portions is 105◦ . Find the length of the string.
16 The distance AB = 50m. If the line of sight of a
person standing at A to the tree makes an angle
of 32◦ with the bank, how wide is the river?
105°
E
tree
32°
50 m
B
A
PL
17 A ladder 4.7 m long is to be placed against a wall. The foot of the ladder must not be placed
in a flower bed, which extends a distance of 1.7 m from the foot of the wall. How high up
the wall can the ladder reach?
18 A river is known to be 50 m wide. A swimmer
sets off at A to cross the river, and the path of
the swimmer is as shown. How far does the swimmer
swim?
B
50 m
60°
A
M
The sine rule
In section 13.1, methods for finding unknown lengths and angles for right-angled triangles are
discussed. In this section and the next, methods for finding unknown quantities in non
right-angled triangles are discussed.
The sine rule is used to find unknown quantities in a triangle when one of the following
situations arises:
one side and two angles are given
two sides and a non-included angle are given
In the first of the two cases a unique triangle is defined, but for the second it is possible for
two triangles to exist.
SA
13.2
Labelling convention
The following convention is followed in the remainder of this module. Interior angles are
denoted by upper-case letters and the length of the side opposite an angle is denoted by the
corresponding lower-case letter.
c
A
B
b
a
C
The magnitude of angle BAC is denoted by A.
The length of side BC is denoted by a.
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398
Essential Further Mathematics – Module 2 Geometry and trigonometry
The sine rule
The sine rule states that for a triangle ABC:
a
b
c
=
=
sin A
sin B
sin C
c
A
B
a
C
b
Example 4
PL
E
A proof will only be given for the acute-angled triangle case. The proof for obtuse-angled
triangles is similar.
h
∴ h = b sin A
In triangle ACD, sin A =
C
b
h
In triangle BCD, sin B =
∴ h = a sin B
b
a
a
h
∴ a sin B = b sin A
b
a
A
B
=
i.e.
D
sin A
sin B
Similarly, starting with a perpendicular from A to BC would give
c
b
=
sin B
sin C
Sine rule given two angles and a side
B
M
Find the length of AB.
c
A
70°
a
31°
10 cm
C
Solution
SA
1 Observe that the information given is two angles
and a side. Therefore the sine rule is used.
10
b
.
=
The known ratio is
sin(B)
sin 70◦
10
c
=
◦
sin 31
sin 70◦
10 × sin 31◦
∴c =
sin 70◦
2 The side to find is AB which has length c and
the opposite angle is C which has size 31◦ .
Example 5
The length of AB is 5.48 cm,
correct to two decimal places.
Sine rule given two sides and a non-included angle: ambiguous case
Find possible values for the magnitude of angle XZY in the triangle XYZ, given that
Y = 25◦ , y = 5 cm and z = 6 cm.
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Chapter 13 – Trigonometry
399
Solution
1 Observe that the information given is two
sides and a non-included angle. Therefore
the sine rule is used.
5
y
.
=
The known ratio is
sin(Y )
sin 25◦
2 The angle to find is XZY. The opposite side
is XY has size 6 cm. Remember:
sin (180 − ) = sin . There are two solutions
for the equation.
sin Z = 0.5071 . . .
They are Z = 30.47◦ and Z = 149.53◦ .
E
6
5
=
◦
sin 25
sin Z
sin 25◦
sin Z
∴
=
6
5
6 × sin 25◦
∴ sin Z =
5
= 0.5071 . . .
∴ Z = sin−1 (0.5071)
Z1
5 cm
PL
30.47°
Z2
149.53°
25°
Y
6 cm
5 cm
∴ Z = 30.4736 . . . or 180 − 30.4736 . . .
∴ Z = 30.47◦ or Z = 149.53◦ correct
to two decimal places.
M
X
Exercise 13B
1 Find the value of x in each of the following.
a
b
Y
SA
70°
X
Z
65°
x cm
x cm
40°
10 cm
Z
x cm
5.6 cm
12 cm
d Y
X
38°
x cm
92°
X
Y
Y
6 cm
Z
c
100°
37°
X
28°
Z
2 Find the value of for each of the following triangles.
A
b
a
C
7 cm
A
72°
9.4 cm
θ°
8 cm
B
C
θ°
42°
8.3 cm
B
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400
Essential Further Mathematics – Module 2 Geometry and trigonometry
c
d
C
C
θ°
10 cm
8 cm
108°
θ°
A
8 cm
B
A
38°
B
9 cm
3 Solve the following triangles (i.e. find all sides and angles).
b A = 75.3◦ , b = 5.6, B = 48.25◦
d A = 23◦ , a = 150, B = 40◦
E
a a = 12, B = 59◦ , C = 73◦
c A = 123.2◦ , a = 11.5, C = 37◦
e B = 140◦ , b = 20, A = 10◦
4 Solve the following triangles (i.e. find all sides and angles).
PL
a b = 17.6, C = 48.25◦ , c = 15.3
b B = 129◦ , b = 7.89, c = 4.56
c A = 28.25◦ , a = 8.5, b = 14.8
5 A landmark A is observed from two points B and C, which are 400 m apart. The magnitude
of angle ABC is found to be 68◦ and the magnitude of angle ACB is 72◦ . Find the distance of
A from C.
M
6 AB is a tower 60 m high on top of a hill. The magnitude
of ACO is 49◦ and the magnitude of BCO is 37◦ .
a Find the magnitude of angles ACB, CBO and CBA.
b Find the length of BC.
c Find the height of the hill, i.e. the length of OB.
B
C
O
P
7 P is a point at the top of a lighthouse. Measurements of the
length of AB and angles PBO and PAO are taken and are as
shown in the diagram. Find the height of the lighthouse.
SA
A
A
27.6° 46.2°
34 m B
O
8 A and B are two points on a coastline. They are 1070 m apart; C is a point at sea. The angles
CAB and CBA have magnitudes of 74◦ and 69◦ respectively. Find the distance of C from A.
9 Find:
a AX
b AY
Y
X
88°
A
32°
50 m
89°
20°
B
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Chapter 13 – Trigonometry
401
The cosine rule
The cosine rule is used to find unknown quantities in a triangle when one of the following
situations arises:
two sides and an included angle are given
three sides are given
The cosine rule
The cosine rule states that for a triangle ABC
2
2
B
a
c
E
a = b + c − 2bc cos A or, equivalently,
b2 + c2 − a 2
cos A =
2bc
2
C
A
b
M
PL
The symmetrical results also hold, i.e.:
b2 = a 2 + c2 − 2ac cos B
c2 = a 2 + b2 − 2ab cos C
The result will be proved for an acute-angled triangle. The proof for obtuse-angled triangles is
similar.
In triangle ACD:
C
b2 = x 2 + h 2 (Pythagoras’ theorem)
x
a
b
h
cos A = and therefore x = b cos A
b
x
In triangle BCD, a 2 = (c − x)2 + h 2 (Pythagoras’ theorem) A
B
D
Expanding gives:
c
a 2 = c2 − 2cx + x 2 + h 2
= c2 − 2cx + b2
∴ a 2 = b2 + c2 − 2bc cos A
(as x 2 + h 2 = b2 )
(as x = b cos A)
Rearranging gives:
b2 + c2 − a 2
2bc
SA
13.3
cos A =
Example 6
Applying the cosine rule—two sides and included angle given
B
For triangle ABC, find the length of AB in centimetres
correct to two decimal places.
5 cm
c
67°
A
10 cm
Solution
Strategy: Apply the cosine rule—two sides and an included angle are given.
c 2 = 52 + 102 − 2 × 5 × 10 cos 67◦
= 85.9268 . . .
∴ c = 9.269 . . .
The length of AB is 9.27 cm correct to two decimal places.
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402
Essential Further Mathematics – Module 2 Geometry and trigonometry
Example 7
Applying the cosine rule—three sides given
B
Find the magnitude of angle ABC for triangle ABC.
Give your answer correct to two decimal points.
6 cm
12 cm
A
Solution
Strategy: Apply the cosine rule—three sides are given.
E
a 2 + c2 − b 2
2a c
122 + 62 − 152
=
2 × 12 × 6
= −0.3125
∴ B = 108.2099 . . .
C
15 cm
PL
cos B =
The magnitude of angle ABC is 108.21◦ correct
to two decimal places.
Exercise 13C
2 Find the magnitudes of angles ABC and ACB.
1 Find the length of BC.
B
M
B
10 cm
A
15°
8 cm
5 cm
C
15 cm
A
C
10 cm
3 For triangle ABC with:
A = 60◦ b = 16 c = 30,
a = 14 B = 53◦ c = 12,
a = 27
b = 35 c = 46,
a = 17 B = 120◦ c = 63,
a = 31
b = 42 C = 140◦ ,
a = 10
b = 12 c = 9,
a = 11
b=9
C = 43.2◦ ,
a=8
b = 10 c = 15,
SA
a
b
c
d
e
f
g
h
find a.
find b.
find the magnitude of angle ABC.
find b.
find c.
find the magnitude of angle BCA.
find c.
find the magnitude of angle CBA.
4 A section of an orienteering course is as shown. Find the
length of leg AB.
B
4 km
A
20°
6 km
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Chapter 13 – Trigonometry
403
5 Two ships sail from point O. At a particular time their
positions A and B are as shown. Find the distance between
the ships at this time.
N
A
B
6 km
30° 4 km
E
O
6 ABCD is a parallelogram. Find the length of the diagonals:
a AC
b BD
5 cm
B
48°
PL
A
7 A weight is hung from two hooks in a ceiling by strings
of length 54 cm and 42 cm, which are inclined at 70◦ to
each other. Find the distance between the hooks.
C
4 cm
D
54 cm
42 cm
70°
M
8 a Find the length of diagonal BD.
b Use the sine rule to find the length of CD.
SA
9 Two circles of radius 7.5 cm and 6 cm have a
common chord of length 8 cm.
a Find the magnitude of angle AO B.
b Find the magnitude of angle AOB.
B
4 cm
A
5 cm
92°
C
88°
6 cm
D
A
cm
6 cm
7.5
O 8 cm
O'
B
A
10 Two straight roads intersect at an angle of 65◦ .
m
A point A on one road is 90 m from the
90
C
65°
O
intersection and a point B on the other road is
70 m
70 m from the intersection, as shown on the diagram.
B
a Find the distance of A from B.
b C is the midpoint of AB. Find the distance of C from the intersection.
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404
13.4
Essential Further Mathematics – Module 2 Geometry and trigonometry
Area of a triangle
It is known that the area of a triangle is given by the
formula
B
c
h
A
A = 12 bh
C
b
E
Area = 12 × base length ×height
By observing that h = c sin A the following formula can be found.
Area of a triangle
Area = 12 bc sin A
Example 8
PL
The area is given by half the product of the length of two sides and the sine of the angle
included between them.
Determining the area of a triangle, A = 1 ac sin B
2
Find the area of triangle ABC. Give your answer
correct to two decimal places.
7.2 cm
B
140°
6.5 cm
A
Solution
M
Strategy: Apply the formula for the area of a triangle, area = 12 ac sin B.
Area = 2 × 7.2 × 6.5 × sin 140◦
1
= 15.04 cm2
SA
The area of triangle ABC is 15.04 cm2 ,
correct to two decimal places.
Heron’s formula
Heron’s formula gives a way of determining the area of a triangle given the lengths of three
sides.
A
c
B
b
a
C
Heron’s formula
The area of a triangle with side lengths a, b and c is given by
a+b+c
A = s(s − a)(s − b)(s − c) where s is the semi-perimeter and s =
2
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Chapter 13 – Trigonometry
405
Example 9
Determining the area of a triangle using Heron’s formula
Find the area of the triangle with sides 6 cm, 4 cm and 4 cm. Give your answer correct to two
decimal places.
Solution
s(s − a )(s − b )(s − c)
6+4+4
s=
=7
2
A =
3 Substitute in the formula.
A =
4 Write down your answer correct to
two decimal places.
The area of the triangle is 7.94 cm correct to
to two decimal places.
E
1 Three sides given, so use Heron’s formula.
a+b+c
.
2 Find the value of s =
2
In this case a = 6, b = 4 and c = 4
7×1×3×3=
63 = 7.937 . . .
PL
Exercise 13D
1 Find the area of each of the following triangles.
a
b
C
4 cm
m
c
70°
6
5.1 cm
B
M
A
c
M
8.2 cm
d
3.5 cm
N
130°
X
6.2 cm
72.8°
Z
Y
A
25°
5 cm
5 cm
B
C
L
SA
2 Find the area of an equilateral triangle with side length:
a 6.2 cm
b 3.7 cm
3 In triangle XYZ, XY = 9 cm and YZ = 13 cm. The area of the triangle is 31 cm2 . Find two
possible values for the magnitude of ∠ XYZ .
4 In triangle ABC, A = 108.6◦ , b = 9 cm and c = 8 cm. Find the area of the triangle.
5 Find the areas of each of the following triangles in cm2 giving your answers correct to two
decimal places.
a
5 cm
10 cm
8 cm
b
2 cm
11 cm
c
8 cm
8 cm
10 cm
10 cm
6 The area of a triangle ABC is 6 cm2 ; AB = 3 cm and AC = 5 cm.
a Find two possible values for the magnitude of angle BAC.
b Find two possible lengths for BC.
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Essential Further Mathematics – Module 2 Geometry and trigonometry
Key ideas and chapter summary
y
Sine and cosine may be defined for any
angle through the unit circle. For an
P(cos θ°, sin θ°)
angle of ◦ , a point P on the unit circle
θ°
x
is defined as illustrated. The angle is
(0, 0)
measured in an anticlockwise direction
from the positive direction of the x axis.
cos ( ◦ ) is defined as the x coordinate
of the point P and sin ( ◦ ) is defined as the y coordinate of P. Your
calculator gives approximate values for these coordinates.
Trigonometric ratios
For a right-angled triangle the naming of sides with respect to an angle
◦ is as shown.
B
opposite
opp
◦
sin ( ) =
hyp hypotenuse adjacent
adj
hypotenuse
opposite
cos ( ◦ ) =
hyp hypotenuse
sin ( ◦ )
opp
O θ°
C
tan ( ◦ ) =
=
adjacent
cos ( ◦ )
adj
From the unit circle note that
e.g. sin 45◦ = sin 135◦
sin ◦ = sin(180 − )◦
and cos( ◦ ) = − cos(180 − ◦ ) e.g. cos(45◦ ) = − cos(135◦ )
PL
M
Obtuse angles with
sine and cosine
E
Sine and cosine
(cos(180 – θ)°, sin(180 – θ)°)
SA
Review
406
Labelling convention
Sine rule, when
to use
y
(180 – θ)°
(cos(θ°), sin(θ°))
θ°
x
O
Interior angles are denoted by upper-case letters and
the length of the side opposite an angle is denoted by
the corresponding lower-case letter.
c
The magnitude of angle BAC is denoted by A.
The length of side BC is denoted by a.
A
B
a
b
The sine rule is used to find unknown quantities in a triangle when
one of the following situations arises:
r one side and two angles are given
r two sides and a non-included angle are given
In the first of the two cases a unique triangle is defined but for the
second it is possible for two triangles to exist.
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Chapter 13 – Trigonometry
407
The sine rule states that for triangle ABC:
b
c
a
=
=
sin A
sin B
sin C
B
c
A
a
C
b
The cosine rule is used to find unknown quantities in a triangle when
one of the following situations arises:
r two sides and an included angle are given
r three sides are given
Cosine rule
The cosine rule states that for triangle ABC:
E
Cosine rule, when
to use
Area of a triangle
a
c
C
A
b
PL
a 2 = b2 + c2 − 2bc cos A or equivalently
b2 + c2 − a 2
cos A =
2bc
The symmetrical results also hold, i.e.
r b2 = a 2 + c2 − 2ac cos B
r c2 = a 2 + b2 − 2ab cos C
B
The area of a triangle is given by the
formula A = 12 bh.
B
c
a
h
A
b
By observing that h = c sin A the following formula can be found:
M
A = 12 bc sin A
i.e. The area is given by half the product of the length of two sides and
the sine of the angle included between them.
Heron’s formula states that the area of a triangle with side lengths a, b
and c is given by A = s(s − a)(s − b)(s − c) where s is the
a+b+c
semi-perimeter and s =
2
SA
Heron’s formula
Skills check
Having completed this chapter you should be able to:
apply the trigonometric ratios sine, cosine and tangent to right-angled triangles
apply the sine rule to triangles to determine unknown lengths and angles
apply the cosine rule to triangles to determine unknown lengths and angles
determine the area of any triangle given
r the base and height (A = 1 × base × height)
2
r two sides and an included angle (A = 1 ab sin )
2
r the lengths of three sides (Heron’s formula)
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Review
Sine rule
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Essential Further Mathematics – Module 2 Geometry and trigonometry
Multiple-choice questions
E
1 In the right angled triangle XYZ the right angle is at Y and ∠YZX = 35◦ . The length
of X Y = 10 cm.
The length of XZ is given by:
X
10
A 10 cos 35◦ B 10 sin 35◦ C
sin 35◦
10
10 cm
D
E 10 tan 35◦
◦
cos 35
35°
Y
2 The length of AC correct to one decimal place is:
B 22.9 cm
E 40.6 cm
C 23.4 cm
Z
B
102°
11 cm
28°
PL
A 13.2 cm
D 40.4 cm
A
C
M
3 In a triangle ABC, a = 5.2 cm, b = 6.8 cm and c = 7.3 cm. ∠ ACB correct to the
nearest degree is:
C 74◦
D 82◦
E 98◦
A 43
B 63◦
51
. The value of c to the nearest
4 In a triangle ABC, a = 30, b = 21 and cos C =
53
whole number is:
A 9
B 10
C 11
D 81
E 129
5 The length of AC to the nearest metre is:
A 16
D 36
B 18
E 326
C 22
10 m
B
110°
12 m
A
C
6 The length of the radius of the circle shown,
correct to two decimal places, is:
SA
Review
408
A 5.52 cm
D 12.18 cm
B 8.36 cm
E 18.13 cm
C 9.01 cm
130°
10 cm
7 In a triangle XYZ, x = 21 cm, y = 18 cm and ∠YXZ = 62◦ . The size of ∠ XYZ
correct to one decimal place is
B 0.8◦
C 1.0◦
D 49.2◦
E 53.1◦
A 0.4◦
8 In a parallelogram ABCD, AB = C D = 8cm and BC = AD = 12cm. If
∠ BC D = 52◦ , the length of the diagonal AC to the nearest centimetre is:
A 12 cm
B 18 cm
C 16 cm
D 149 cm
E 267 cm
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Chapter 13 – Trigonometry
409
10 The length of AC correct to one decimal place is:
A 6.2 cm
B 16.3 cm
C 19.6 cm
D 40.4 cm
E 20.3 cm
B
11 cm
109°
32°
A
C
B
E
11 The square of the length of side BC is:
A 36
B 85
C 49
D 42
E 43
7
A
60°
C
6
12 For the triangle shown, the value of the cosine of angle ABC is:
8
5
A
B 74
C
25
6
−5
D
E 73
6
PL
A
25 cm
C
C
a
b
A
x
C
b
a
B
SA
M
13 In the triangle ABC, cos x =
b
a
B √
A √
a 2 + b2
a 2 + b2
√
b
a 2 + b2
D
E
a
a
B
8 cm
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Review
9 The area of the triangle ABC, where b = 5 cm, c = 3 cm, ∠ A = 30◦ and
∠ B = 70◦ is:
B 3.75 cm2
C 6.50 cm2
D 7.50 cm2
E 8 cm2
A 2.75 cm2