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Achievement Standard 90291 Mathematics
2.8
Solve trigonometry problems requiring
modelling of practical situations
Internally assessed
2 credits
Cosine and Sine Rules
C
Conventions
Capital letters are usually used to show the angles at
the corners of a triangle. The length of the side opposite
each corner is denoted by the same letter using lower case.
The Cosine Rule
a
b
B
A
This rule enables non right-angled triangles to be solved.
a2 = b2 + c2 – 2bc cos A or cos A =
c
b2 + c 2 – a2
2bc
Example
Q. A fishing boat sets out to sea. It is 6.2 km from point A on the shore and 6.7 km from another point, B.
The angle between A, the boat, and B is 75°.
1. How far apart are the two points A and B?
A. 1. The diagram shows the points A and B. By the cosine rule:
B
A
2. Failing to catch any fish, the captain moves the boat to a point which is now 6.4 km
from point A, and 6.8 km from B. What is the angle between A, the boat, and B now?
AB2= 6.22 + 6.72 – 2 × 6.2 × 6.7 cos 75° = 61.8273
∴ AB= 7.86 km (2 dp)
2. By the cosine rule for unknown angles:
6.42 + 6.82 – 7.862
cos x=
= 0.2921
2 × 6.4 × 6.8
∴ x= 73.0°
7.86
A
[taking square roots]
75°
6.2
6.4
x
6.7
B
6.8
[taking inverse cos]
Note: Always work to more decimal places than the answer will require, and leave rounding to a sensible
accuracy as the last step. (Premature rounding causes errors.)
The
Sine and Area Rules
The sine rule states:
a
b
c
=
=
sin A sin B sin C
The area rule states that:
or
sin A sin B sin C
=
=
a
b
c
1
Area = ab sin C
2
1
1
Alternatively, the area is equal to bc sin A or ac sin B.
2
2
SAM
REV PLE
ISIO
N
NO
TES
A
b
c
B
C
a
Example
Q. Peter and Kathy tie two ropes to the top of a pole. They stake the other end of each rope into the ground so
the ropes are tight and in the same plane with the pole. Peter’s rope is 5.8 m long and makes an angle of 26°
with the ground. Kathy’s rope makes an angle of 32° with the ground.
106
Achievement Standard 90291 (Mathematics 2.8)
1. Find the length of Kathy’s rope.
2. Find the area of the triangle formed by the two ropes and the line along the ground which joins the two
stakes.
A. The diagram shows Kathy’s rope with length L.
L
5.8
1.
=
[using the sine rule]
sin 26° sin 32°
5.8
∴ L=
sin 26°
[multiplying by sin 26°]
sin 32°
∴ L= 4.8 m (1 dp)
1
2. Area A= × 5.8 × 4.8 × sin 122°
2
5.8
L
26°
32°
[simplifying]
1
[using area = ab sin C where 3rd angle = 122°]
2
= 11.8 m2 (1dp)
Harder problems may involve 3-dimensional settings, in which bearings may be used.
Example
let PB= x
x2 + 1.32= 4.52 [Pythagoras]
x= 4.308[solving for x]
In ∆RPB,
∠RPB= 55° [angles on a line]
In ∆RPB,
10
A
P
102 + 8.3152 – 4.3082
cos R=
= 0.9055
2 × 10 × 8.315
∴ R= 25.1°
So the rescue boat must travel 8.3 km on a bearing of 025°.
P
ocean
surface
125°
B
R
4.5
1.3
RB2= 102 + 4.3082 – 2 × 10 × 4.308 cos 55°
= 69.1395
∴ RB= 8.315 km (3 dp)
1.3 km
[AP is vertical so ∠APB = 90°]
km
4.5
A. In ∆APB,
km
N
A
Q. An aircraft, A, flying at an altitude of 1.3 km above P receives a distress signal from a boat, B, 4.5 km away from the plane on
a bearing of 125° from P. A rescue boat, R, is 10 km due south
of P. How far and in what direction must the rescue boat travel
in order to reach boat B?
P
B
x
55° 4.308
10
[cosine rule for sides]
R
B
y
[taking square roots]
[cosine rule for angles]
SAM
REV PLE
ISIO
N
NO
TES
[taking inverse cosine]
Problems may also involve relative velocity, which is the velocity of one object in relation to another.
Example
Q. A plane travels at an air velocity of 160 km h–1 on a bearing of 140°. A wind of velocity 65 km h–1 is blowing
from the east. Find the velocity of the plane relative to the ground (air velocity is the velocity of the plane
N
relative to the air).
A.
x2= 1602 + 652 – 2 × 160 × 65 × cos 50° [cosine rule]
= 16 455.018
x= 128.3 km h–1 (1 dp)
1602 + 128.32 – 652
cos q=
2 × 160 × 128.3
= 0.9216
q= 22.8°
Thus the plane makes an angle of
140° + 22.8° = 162.8° with North.
140o
[cosine rule]
qo
[taking inverse cosine]
x
velocity of plane
relative to ground
E
50o
160
velocity of plane
relative to air
50o
Velocity of plane relative to ground is 128 km h–1 on a bearing of 163°.
65
velocity of air
relative to ground
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Solve trigonometry problems requiring modelling of practical situations 107
Questions
Cosine and Sine Rules
1. Three boys Alan, Adam and Aidan are on a beach. Each has a stride of about 1 metre.
Alan stays in one spot while Adam and Aidan walk in straight lines which make an angle of 70° at Alan (see diagram). Adam walks
50 paces and Aidan walks 60 paces.
a. How far apart are Adam and Aidan at this time?
Adam
50
Aidan
70°
60
Alan
b.
2. Julia, a Year 12 student at a high school, is told to find the
perimeter of her school field ABC. The diagram shows the
measurements she took. What is the perimeter of the field?
50 m
Adam
Adam and Aidan hold a string of length 50 m and once again walk away from Alan. When Adam has gone 55 paces and
Aidan 61 paces, the string is tight. What is the angle between
the lines at Alan now?
55
Aidan
61
Alan
A
150 m
100 m
120°
C
160°
B
180 m
A
3. A is at the top of a high hill. The hill meets a plane, DC, at B and
makes an angle of 75° with it.
At C, one hundred metres away from B, is a surveyor who finds
that he reads an angle of 35° with A as shown in the diagram.
a. Find the length AB.
b. Find the height AD of the hill.
 ESA Publications (NZ) Ltd, Freephone 0800-372 266
75°
SAM
P
NCE LE
A
QU
EST
ION
S
D
B
35°
100 m
C
108
Achievement Standard 90291 (Mathematics 2.8)
Achievement Standard 90291 (Mathematics 2.8)
Solve trigonometry problems requiring modelling of
practical situations
2.8 Cosine and Sine Rules (page 107)
502 + 602 – 2 × 50 × 60 × cos 70° = 63.6 m (A)
1. a.
 552 + 612 – 502 
 = 50.7° (A)
b. cos–1 
 2 × 55 × 61 
2. Perimeter is AB + BC + AC=
using the cosine rule for sides
using the cosine rule for angles
1502 + 1002 – 2 × 150 × 100 × cos 120° +
+ 1502 + 1802 – 2 × 150 × 180 × cos 160° + 1002 + 1802 – 2 × 100 × 180 × cos 80°
using cosine rule for sides (the other angle is 80° because the
sum of adjacent angles at a point is 360°)
= 733 m (3 sf) (A)
3. a. ∠BAC = 40°
exterior angle ∠ABD (75°) is equal to the sum of the two opposite interior
angles ∠BDC and ∠ACB).
Therefore, by the sine rule
100
AB
100
AB
sin 35° × 100
=
∴
=
∴ AB =
= 89.2 m (A)
sin BAC
sin ACB
sin 40°
sin 35°
sin 40°
AD
b. AD = AB sin ABD = 89.2 × sin 75° = 86.2 m (A)
sin 75° =
in triangle ABD
AB
SAM
P
NCE LE
A
AN
SW
ERS
 ESA Publications (NZ) Ltd, Freephone 0800-372 266