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26
Chapter
Sine and cosine rules
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Obtuse angles
Area of a triangle using sine
The sine rule
The cosine rule
Problem solving with the
sine and cosine rules
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B
C
D
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Contents:
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IB_PD_SSL(MYP5)
2
SINE AND COSINE RULES
(Chapter 26)
OPENING PROBLEM
A triangular property is bounded by
two roads and a long, straight drain.
Can you find:
a
b
c
120°
d
the area of the property in m2 and in
hectares
the length of the drain boundary
the angle that the Johns Road boundary
makes with the drain boundary?
s
n
oh
J
a
Ro
For
Sale
277 m
Ev
an
sR
oa
324 m d
drain
In Chapter 16 on right angled triangle trigonometry we were concerned only with acute
angles, or angles between 0o and 90o . In this chapter we extend our use of trigonometry to
triangles which are not right angled. These include acute angled triangles in which every
angle is acute, and obtuse angled triangles which contain one obtuse angle.
acute angled triangle
obtuse angled triangle
A
OBTUSE ANGLES
An obtuse angle is one which measures between 90o and 180o .
In the following investigation we examine the sines and cosines of obtuse angles.
INVESTIGATION
TRIGONOMETRIC RATIOS OF OBTUSE ANGLES
By finding the trigonometric ratios of various angles, we are to observe
relationships between the sines and cosines of obtuse and acute angles.
What to do:
1 Use your calculator to find the following correct to 4 decimal places:
sin 150o
sin 169o
sin 138o
a
e
i
b
f
j
cos 150o
cos 148o
cos 105o
c
g
k
sin 120o
cos 116o
cos 91o
d
h
l
cos 120o
sin 98o
sin 178o
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2 Copy and complete:
² the sine of an obtuse angle is always ......
² the cosine of an obtuse angle is always ......
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IB_PD_SSL(MYP5)
SINE AND COSINE RULES
(Chapter 26)
3
a Copy and complete this table, giving values correct to 4 decimal places:
3
µ
sin µ
0o
10o
18o
30o
45o
50o
60o
73o
84o
µ
sin µ
180o
170o
162o
150o
135o
130o
120o
107o
96o
b What do you notice from the above table?
c Copy and complete:
sin(180o ¡ µ) = ::::::
a Copy and complete the following table, giving values correct to 4 decimal places:
4
µ
cos µ
5o
19o
38o
53o
67o
76o
81o
88o
µ
cos µ
175o
161o
142o
127o
113o
104o
99o
92o
b What do you notice from the above table?
c Copy and complete:
cos(180o ¡ µ) = ::::::
From the Investigation above you should have observed that:
sin(180o ¡ µ) = sin µ
and
cos(180o ¡ µ) = ¡ cos µ
You should now be able to write, for example:
sin 140o = sin(180o ¡ 140o )
= sin 40o
and cos 140o = ¡ cos(180o ¡ 140o )
= ¡ cos 40o
EXERCISE 26A
1 Write the following in the form sin µ where µ is acute:
a sin 100o
b sin 170o
d sin 95o
e sin 122o
o
g sin 154
h sin 99o
c
f
i
sin 144o
sin 171o
sin 167:2o
2 Write the following in the form ¡ cos µ where µ is acute:
a cos 108o
b cos 166o
c
o
o
d cos 92
e cos 149
f
o
o
g cos 100:7
h cos 168:5
i
cos 128o
cos 114o
cos 133:5o
3 Find another angle which has the same sine as:
a 40o
b 113o
c 83o
f 57o
g 42:6o
h 131:9o
d
i
0o
162:8o
e
j
142o
15:92o
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4 If sin µ = a where a is unknown, find in terms of a:
a sin(180o ¡ µ)
b cos(180o ¡ µ)
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IB_PD_SSL(MYP5)
4
SINE AND COSINE RULES
B
(Chapter 26)
AREA OF A TRIANGLE USING SINE
We have seen previously that the area of a triangle with
base b and altitude h is given by
Area =
h
1
bh.
2
b
Now consider a triangle where we are given two sides
and the angle between them, called the included angle.
To find the area we draw an altitude from a vertex to
one of the known sides.
9m
Using the shaded right angled triangle we see that
h
sin 48o =
9
48°
12 m
) h = 9 £ sin 48o
9m
Since area = 12 bh we see that
area =
1
2
o
£ 12 £ 9 £ sin 48
hm
48°
2
¼ 40:1 m
12 m
Summary:
Given the lengths of two sides of a triangle and the
included angle between them, the area of the triangle is
side
a half of the product of two sides and the sine of
the included angle.
included
angle
side
A
Notation:
A
If we have a triangle ABC, we label the corresponding
angles A, B and C and the opposite side lengths a, b
and c.
B
Using this notation:
Area =
1
ab sin C
2
1
bc sin A
2
or
b
c
B
a
C
C
1
ac sin B.
2
or
Example 1
Self Tutor
Area = 12 ac sin B
Find the area of triangle ABC.
=
A
1
2
£ 15 £ 11 £ sin 28o
¼ 38:7 cm2
11 cm
28°
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IB_PD_SSL(MYP5)
SINE AND COSINE RULES
(Chapter 26)
5
EXERCISE 26B
1 Find the area of:
a
b
c
31 km
46°
7.8 cm
7 cm
70°
d
115°
33 km
10 cm
6.4 cm
e
48 m
f
2.85 m
87°
58°
10.1 cm
122°
8.3 cm
32 m
2.43 m
2 Find the area of a parallelogram with sides 5:4 cm and 8:8 cm and one interior angle
72o . Hint: Divide the parallelogram into two congruent triangles.
3 If triangle ABC has area 84 cm2 , find the value of x:
A
12 cm
B
51°
x cm
C
P
4
q
18 cm
15 cm
This triangle has not been drawn to scale.
Q
If its area is 100 cm2 , find the two possible
values of µ.
R
C
THE SINE RULE
The sine rule is a set of equations which connects the lengths of the sides of any triangle
with the sines of the angles of the triangle. It is used to solve problems involving triangles
where angles and sides opposite those angles are to be related.
The triangle does not have to be right angled for the sine rule to be used.
THE SINE RULE
In any triangle ABC with sides a, b and c units in length,
and opposite angles A, B and C respectively:
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a
b
c
=
=
.
sin A
sin B
sin C
or
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sin B
sin C
sin A
=
=
a
b
c
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A
A
C
C
c
a
B
B
IB_PD_SSL(MYP5)
6
SINE AND COSINE RULES
(Chapter 26)
1
2 bc sin A
Proof: The area of any triangle ABC is given by
= 12 ac sin B = 12 ab sin C.
sin B
sin C
sin A
=
=
.
a
b
c
Dividing each expression by 12 abc gives
GEOMETRY
PACKAGE
FINDING SIDES
Example 2
Self Tutor
Find the length of AC
correct to two decimal
places.
Using the sine rule,
12
b
=
o
sin 58
sin 39o
A
A
b cm
12 cm
12 cm
39°
58°
39°
C
58°
C
B
12 £ sin 58o
sin 39o
)
b=
)
b ¼ 16:17
) AC is 16:17 cm long.
B
As you are finding
side lengths, use the
sine rule with side
lengths on top.
EXERCISE 26C
1 Find the value of x:
a
b
c
7 cm
51°
110°
18 cm
x cm
42°
46°
53°
d
4.2 km
x cm
44°
x km
e
f
213 km
xm
113°
47°
49°
x km
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83 m
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99°
27°
xm
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IB_PD_SSL(MYP5)
SINE AND COSINE RULES
Example 3
(Chapter 26)
Self Tutor
C
Find the length of
side AB in:
7
78°
14 cm
44°
A
B
x cm
The 14 cm side is opposite the third angle of the triangle. We need to find this
angle before we can use the sine rule.
b = 180o ¡ 78o ¡ 44o = 58o
Now CAB
C
So, by the sine rule:
14
x
=
sin 78o
sin 58o
78°
14 cm
14 £ sin 78o
sin 58o
)
x=
)
x ¼ 16:1
58°
A
44°
B
x cm
) AB is 16:1 cm long.
2 Find x correct to 3 significant figures:
a
b
c
48°
xm
x cm
24.2 m
12.6 cm
47°
36°
73°
69°
15.7 m
xm
d
e
98°
f
81°
xm
24.8 km
x km
62°
32°
3.71 km
39°
124.6 m
33°
41°
x km
3 In triangle ABC find:
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a a if A = 72o , B = 49o and b = 16 cm
b b if A = 37o , C = 69o and c = 21 cm
c c if B = 18o , C = 54o and a = 8:2 cm.
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IB_PD_SSL(MYP5)
8
SINE AND COSINE RULES
(Chapter 26)
FINDING ANGLES GIVEN AN ACCURATE DIAGRAM
If an accurate diagram of a triangle with sufficient information is given, then we can use the
sine rule to determine the size of an unknown angle.
Example 4
Self Tutor
Find µ in these figures which are accurately drawn to scale:
a
b
q
q
12 cm
8 cm
48°
48°
16 cm
10 cm
a
b
By the sine rule:
By the sine rule:
o
sin 48o
sin µ
=
16
12
sin 48
sin µ
=
10
8
10 £ sin 48o
8
¶
µ
10 £ sin 48o
¡1
) µ = sin
8
16 £ sin 48o
12
¶
µ
16 £ sin 48o
¡1
) µ = sin
12
) sin µ =
) sin µ =
) µ ¼ 68:3 o or 111:7 o
) µ ¼ 68:3 o
fas µ is clearly acuteg
) µ ¼ 82:2o or 97:8o
) µ ¼ 97:8o
fas µ is clearly obtuseg
4 The following figures are drawn accurately to scale.
Find µ correct to 1 decimal place.
a
b
q°
12 cm
19 cm
c
51°
q°
49 m
9m
28°
11 m
108°
95 m
q°
d
e
f
72.6 m
q°
31°
113 m
89°
q°
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27°
130.7 m
8.9 cm
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87 m
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14.8 cm
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IB_PD_SSL(MYP5)
SINE AND COSINE RULES
(Chapter 26)
9
FINDING ANGLES WITHOUT AN ACCURATE DIAGRAM
Sometimes information is given in sentence form without a diagram. In these situations we
need to draw a diagram to help us solve the problem. However, we must be careful because
the diagram we draw may be misleading. The following example illustrates such a case.
Example 5
Self Tutor
Find the measure of angle C in triangle ABC if AC is 7 cm, AB is 11 cm and
angle B measures 25o .
This is called the
“ambiguous case”!!
By the sine rule,
sin 25o
sin C
=
11
7
A
11 £ sin 25o
7
µ
¶
11 £ sin 25o
¡1
) C = sin
7
7 cm
) sin C =
11 cm
C
25°
B
) C ¼ 41:6o or 180o ¡ 41:6o
) C ¼ 41:6o or 138:4o
) C measures 41:6o if angle C is acute
or C measures 138:4o if angle C is obtuse.
In this case there is insufficient information to determine the actual shape of the
triangle. We must not be misled by the diagram we have drawn into believing the
angle must be acute.
The reason for the two answers in the above example can be shown by a simple construction.
Step 2: From A draw the arc of
a circle of radius 7 cm.
Step 1: Draw AB of length
11 cm and construct an
angle of 25o at B.
Cx
41.6°
7 cm
Cz
7 cm
25°
A
B
138.4°
25°
A
11 cm
B
11 cm
5 Find the two possible measures of angle C in triangle ABC given that AC is 8 cm,
AB is 12 cm, and angle B measures 27o .
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6 Triangle PQR has PR = 7 m, PQ = 8 m, and angle PQR is 45o . Find the two possible
measures of angle PRQ.
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IB_PD_SSL(MYP5)
10
SINE AND COSINE RULES
(Chapter 26)
Sometimes there is information given in the question which enables us to reject one of the
answers.
Example 6
Self Tutor
Find the measure of angle L in triangle KLM given that angle LKM
measures 56o , LM = 16:8 m and KM = 13:5 m.
sin 56o
sin L
=
13:5
16:8
By the sine rule,
L
13:5 £ sin 56o
16:8
µ
¶
13:5 £ sin 56o
¡1
) L = sin
16:8
) sin L =
16.8 m
56°
K
) L ¼ 41:8o
) L ¼ 41:8o
M
13.5 m
We reject L = 138:2o as 138:2o + 56o > 180o
) ]L ¼ 41:8o .
or 180o ¡ 41:8o
or 138:2o
which is impossible.
7 Find the value of µ given that these triangles are not drawn accurately:
a
b
c
q°
55°
17.6 m
14.9 m
31 cm
q°
39°
6.6 km
2.5 km
q°
36 cm
16°
There may be two
possible solutions.
8 In triangle ABC, find the measure of:
a angle A if a = 14:7 cm, b = 17:5 cm and ]ABC = 66o
b angle B if b = 44:9 cm, c = 30:6 cm and ]ACB = 42o
c angle C if a = 6:7 km, c = 4:9 km and ]BAC = 68o .
D
THE COSINE RULE
THE COSINE RULE
In any triangle ABC with sides a, b and c units, and opposite angles A, B and C respectively:
C
b C
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a2 = b2 + c2 ¡ 2bc cos A
b2 = a2 + c2 ¡ 2ac cos B
c2 = a2 + b2 ¡ 2ab cos C.
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IB_PD_SSL(MYP5)
SINE AND COSINE RULES
(Chapter 26)
11
The cosine rule can be used to solve problems involving triangles given:
² two sides and the included angle between them
² three sides.
Example 7
Find, correct to 2 decimal places,
the length of BC.
11 cm
42°
A
a cm
11 cm
42°
C
13 cm
13 cm
C
By the cosine rule:
a2 = 112 + 132 ¡ 2 £ 11 £ 13 £ cos 42o
) a2 ¼ 77:4605
) a ¼ 8:801
B
A
Self Tutor
B
) BC is 8:80 cm in length.
If we are given the lengths of the three sides of a triangle, the angles of the triangle
can be found using:
cos A =
(b2 + c2 ¡ a2 )
(2bc)
cos B =
(a2 + c 2 ¡ b2 )
(2ac)
cos C =
Example 8
(a2 + b2 ¡ c2 )
(2ab)
Self Tutor
A triangle has sides of length 5 cm, 7 cm and 8 cm. Find the measure of its
smallest angle.
The smallest angle is opposite the shortest side.
A
magenta
) cos B =
(82 + 72 ¡ 52 )
(2 £ 8 £ 7)
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) cos B ¼ 0:785 714
) B ¼ 38:2o
So, the smallest angle is about 38:2o .
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C
a = 8 cm
B
5
(a2 + c2 ¡ b2 )
(2ac)
b = 5 cm
c = 7 cm
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There is only
one possible
answer here.
IB_PD_SSL(MYP5)
12
SINE AND COSINE RULES
(Chapter 26)
EXERCISE 26D
1 Find the length of the remaining side in the given triangle:
a
b
c
A
A
6m
B
32°
7 cm
Q
P
B
98°
60°
10 cm
10 cm
7m
C
C
d
R
e
Q
8m
C
4.5 km
63°
R
48°
f
K
14.3 m
12.7 m
5.9 km
M
L
P
2 Find the size of the angle µ in these triangles:
a
b
E
c
7m
10 cm
q°
9 cm
6 cm
D
132°
q°
11 cm
8m
10 m
18 cm
q°
8 cm
3 Find the size of the shaded angle:
a
b
A
c
W
11 cm
P
126 m
R
X
10 cm
145 m
C
17 m
232 m
19 m
13 cm
B
Q
Y
4 Find the size of all angles:
P
8.7 m
6.2 m
R
13.4 m
Q
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5 Find:
a the smallest angle of a triangle with sides 11 cm, 13 cm and 17 cm
b the largest angle of a triangle with sides 4 cm, 7 cm and 9 cm.
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IB_PD_SSL(MYP5)
SINE AND COSINE RULES
E
13
(Chapter 26)
PROBLEM SOLVING
WITH THE SINE AND COSINE RULES
Note: Whenever there is a choice between using the sine rule or the cosine rule, always use
the cosine rule to avoid the ambiguous case.
Example 9
Self Tutor
An aircraft flies 74 km on a bearing 038o and then 63 km on a bearing 160o .
Find the distance of the aircraft from its starting point.
N
N
By the cosine rule:
b2 = a2 + c2 ¡ 2ac cos B
) b2 = 632 + 742 ¡ 2 £ 63 £ 74 £ cos 58o
) b2 ¼ 4504:032 78
) b ¼ 67:1
142° B 160°
74 km
58°
63 km
38°
A
b km
) the aircraft is about 67:1 km from its starting point.
C
EXERCISE 26E
In this exercise give all answers correct to 2 decimal places.
1 Two observation points, P and Q, are connected by a straight road 8 km long. From
each observation point it is possible to see a lighthouse L. If the angle PQL is 63o and
the angle QPL is 54o , find the distance of the lighthouse from both P and Q.
2 Find the total area of the two adjacent paddocks
PQR and QRS.
Q
675 m
44°
R
360 m
P
88°
750 m
S
3 Emma paddles her canoe 350 metres in a straight line from the shore to a buoy. She
then turns through an angle of 65o and paddles another 240 m in a straight line to her
campsite. How far is she from her starting point?
4 Pike Road and Fisher St intersect at an
angle of 116o . To go from home H to
school S, Jonte travels 400 m along Pike
Road, then 615 m along Fisher St. How
much shorter would his journey to school
be if he could go directly from H to S?
H
Pike Rd
S
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Fisher St
5 Two fishing boats A and B leave a port P.
o
A sails 12 km in the direction 064 , and B sails 15 km in the direction 059o , before they
anchor. Calculate the distance and bearing of A from B when the boats are anchored.
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Y:\HAESE\IB_PDSL_MYP5\IB_PDSSL_M5_26\POPUP\013IB_PDM5_26.CDR Tuesday, 23 October 2007 4:16:14 PM PETERDELL
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14
SINE AND COSINE RULES
(Chapter 26)
6 The diagram given is of a side view of a shed.
The shed has triangular roof trusses as shown.
10 m
Find the angle of the roofline at the front of the
shed.
5m
12 m
front
back
7 Mount X is 9 km from Mount Y on a bearing of 146o . Mount Z is 14 km away from
Mount X and on a bearing of 072o from Mount Y. Find the bearing of X from Z.
10 cm
8
Find the measure of angle PQR in
the rectangular box shown.
Q
5 cm
P
8 cm
R
9 Solve the Opening Problem on page 2.
REVIEW SET 26A
1 Write the following in the form sin µ where µ is acute:
a sin 137o
b sin 93o
2 Find the area of:
3 Find the length of PQ, correct
to 2 decimal places:
R
2.8 km
2m
28°
37°
4.6 km
60°
Q
P
4 Find the measure of angle C in triangle ABC, given that angle BAC measures 63o ,
BC = 9:2 cm and AC = 7:8 cm.
5 Find the length of the remaining
side in the given triangle:
x cm
16 cm
48°
18 cm
6
A
a
b
c
9 cm
q°
7 cm
B
Use the cosine rule to find cos µ.
Find µ.
Hence, find the area of triangle ABC.
5 cm
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Y:\HAESE\IB_PDSL_MYP5\IB_PDSSL_M5_26\POPUP\014IB_PDM5_26.CDR Thursday, 15 November 2007 12:12:26 PM PETERDELL
IB_PD_SSL(MYP5)
SINE AND COSINE RULES
(Chapter 26)
15
7 An aircraft flies 200 km on a bearing of 079o and then 150 km on a bearing of 192o .
Find the distance of the aircraft from its starting point.
REVIEW SET 26B
1 Find another angle which has the same sine as:
a 86:7o
b 172o
R
2
S
5m
a
b
4m
142°
Find the area of ¢RST.
Find the length of RS, correct to
3 significant figures.
T
A
3
4m
D
a
Find the length of side AC in radical form,
given that cos 60o = 12 .
b
Find the length of BC, correct to 2
decimal places.
40°
60°
70° B
5m
C
4 a
Redraw the figure without the mountain.
Fully label it.
b
From a boat at A, the angle of elevation
to D is 42o . From B it is 47o . Find the
size of angle ADB.
AB is 500 m. Use triangle ABD to find
the length of BD correct to one decimal
place.
c
D
hm
A
42° 47°
B
sea
C
500 m
Use ¢BCD to find h and hence find the height of the mountain.
d
5 Find the smallest angle of a triangle with sides 5:2 cm, 7:6 cm and 8:4 cm.
11
6
Is the given triangle acute angled or obtuse
angled?
The figure has not been drawn accurately.
q
13
18
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7 A cyclist rides for 42:3 km on a bearing of 168o and then for 53:9 km on a bearing
of 300o . Find the distance of the cyclist from her starting point.
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Y:\HAESE\IB_PDSL_MYP5\IB_PDSSL_M5_26\POPUP\015IB_PDM5_26.CDR Thursday, 15 November 2007 12:13:12 PM PETERDELL
IB_PD_SSL(MYP5)
ANSWERS
16
EXERCISE 26A
1 a
f
2 a
e
i
3 a
g
4 a
sin 80o b sin 10o c sin 36o d sin 85o e sin 58o
sin 9o g sin 26o h sin 81o i sin 12:8o
¡ cos 72o b ¡ cos 14o c ¡ cos 52o d ¡ cos 88o
¡ cos 31o f ¡ cos 66o g ¡ cos 79:3o h ¡ cos 11:5o
¡ cos 46:5o
140o b 67o c 97o d 180o e 38o f 123o
137:4o h 48:1o i 17:2o j 164:08o
a b ¡a
EXERCISE 26B
1 a 32:9 cm2 b 368 km2 c 22:6 cm2 d 651 m2
e 35:5 cm2 f 3:46 m2
2 45:2 cm2 3 x ¼ 18:0 4 µ ¼ 47:8o or 132:2o
EXERCISE 26C
1 a
e
2 a
e
3 a
4 a
e
5 C
7 a
8 a
x ¼ 16:7 b
x ¼ 161 f
x ¼ 11:6 b
x ¼ 109 f
a ¼ 20:2 cm
µ ¼ 36:9 b
µ ¼ 121:1 f
x ¼ 8:24 c x ¼ 4:70 d x ¼ 112
x ¼ 56:3
x ¼ 13:2 c x ¼ 17:9 d x ¼ 19:2
x ¼ 5:44
b b ¼ 21:6 cm c x ¼ 6:98 cm
µ ¼ 71:8 c µ ¼ 114:5 d µ ¼ 50:3
µ ¼ 125:2
¼ 42:9o or 137:1o 6 Pb
RQ ¼ 53:9o or 126:1o
µ ¼ 32:2 b µ ¼ 72:0 or 108:0 c µ ¼ 46:7 or 133:3
A ¼ 50:1o b B ¼ 79:1o or 100:9o c C ¼ 42:7o
EXERCISE 26D
1 a
f
2 a
3 a
4 P
3:71 m b 13:0 cm c 10 cm d 4:42 km e 14:2 m
14:6 m
µ ¼ 40:8 b µ ¼ 83:3 c µ ¼ 118
C ¼ 48:4o b Y ¼ 56:0o c P ¼ 118o
¼ 127:3o , Q ¼ 31:1o , R ¼ 21:6o 5 a 40:3o b 107o
EXERCISE 26E
1
2
5
9
LP = 8:00 km, LQ = 7:26 km
PR ¼ 820:5 m, area ¼ 32:73 ha 3 501:10 m 4 146:72 m
3:22 km, 220:05o 6 24:15o 7 213:83o 8 58:01o
a 38 862:02 m2 , 3:89 ha b 521:01 m c 32:59o
REVIEW SET 26A
1 a sin 43o b sin 87o 2 3:88 km2 3 4:26 m
4 C ¼ 67:9o 5 13:9 cm
6 a cos µ = 57
b µ ¼ 50:7o c 17:4 cm2 7 197:6 km
90
REVIEW SET 26B
1 a 93:3o pb 8o 2 a 6:16 m2 b 8:52 m
3 a AC = 21 m b BC = 3:13 m
4 a
D
h
42° B
A
47°
C
500 m
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b 5o c 3838:7 m d 2807 m
5 37:5o 6 µ ¼ 96:8 ) obtuse angled 7 40:5 km
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Y:\HAESE\IB_PDSL_MYP5\IB_PDSSL_M5_26\POPUP\016IB_PDM5_26.CDR Thursday, 17 January 2008 3:04:27 PM DAVID3
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