Year 11
Mathematics
IAS 1.7
Right-Angled Triangles
Robert Lakeland & Carl Nugent
Contents
•
Achievement Standard . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
•
The Theorem of Pythagoras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
•
Calculating Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
•
Calculating Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
•
Three-Dimensional Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
•
Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
•
Limits of Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
•
Measurement Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
•
Practice Internal Assessment 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
•
Practice Internal Assessment 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
•
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
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2
IAS 1.7 – Right-Angled Triangles
NCEA 1 Internal Achievement Standard 1.7 – Trigonometry
This achievement standard involves applying right-angled triangles in solving measurement problems.
•
Achievement
Apply right-angled triangles
in solving measurement
problems.
•
Achievement with Merit
Apply right-angled triangles,
using relational thinking,
in solving measurement
problems.
•
Achievement with Excellence
Apply right-angled
triangles, using extended
abstract thinking, in solving
measurement problems.
◆
This achievement standard is derived from Level 6 of The New Zealand Curriculum, Learning Media. The following achievement objectives taken from the Shape and Measurement threads of the Mathematics and Statistics learning area are related to this achievement standard:
❖
use trigonometric ratios and Pythagoras’ theorem in two and three dimensions
❖
recognise when shapes are similar and use proportional reasoning to find an unknown length
❖
select and use appropriate metric units for length and area
❖
measure at a level of precision appropriate to the task.
◆
Apply right-angled triangles involves:
❖
selecting and using a range of methods in solving measurement problems
❖
demonstrating knowledge of measurement and geometric concepts and terms
❖
communicating solutions which would usually require only one or two steps.
◆
Relational thinking involves one or more of:
❖
selecting and carrying out a logical sequence of steps
❖
connecting different concepts and representations
❖
demonstrating understanding of concepts
❖
forming and using a model;
and also relating findings to a context, or communicating thinking using appropriate mathematical statements.
◆
Extended abstract thinking involves one or more of:
❖
devising a strategy to investigate or solve a problem
❖
identifying relevant concepts in context
❖
developing a chain of logical reasoning, or proof
❖
forming a generalisation;
and also using correct mathematical statements, or communicating mathematical insight.
◆
Problems are situations set in a real-life context which provide opportunities to apply knowledge or understanding of mathematical concepts and methods. For assessment, situations may involve non right-angled triangles which can be divided into right-angled triangles.
◆
The phrase ‘a range of methods’ indicates that evidence of the application of at least three different methods is required.
◆
Students need to be familiar with methods related to:
❖
Pythagoras’ theorem
❖
trigonometric ratios (sine, cosine, tangent)
❖
similar shapes
❖
measuring at a level of precision appropriate to the task.
IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
3
IAS 1.7 – Right-Angled Triangles
The Theorem of Pythagoras
©
Using the Casio 9750GII. From the
MENU select EQUA then the Solver,
deleting any existing equations
Pythagoras’ Theorem
Pythagoras’ Theorem gives a relationship between
the sides of a right-angled triangle. We always
label the sides a, b and h with the longest side (the
hypotenuse) labelled h as in the diagram below.
h
a
The Theorem of Pythagoras is
Solver
DEL
Yes
8
F3
F2
F1
MENU
Enter in Pythagoras’ theorem h2 = a2 + b2
H
=
ALPHA F ) D
x2
SHIFT
.
ALPHA
+
x2
b
EQUA
B
ALPHA
log
x2
A
x, ø, T
EXE
Enter the known values for H,
A or B and place the cursor next
to the unknown (in this case H)
and select F6 to solve.
h2= a2 + b2
We use the Theorem of Pythagoras when we have a right-angled triangle and we know two lengths of
the triangle and need to find the third length.
The hypotenuse is always opposite the
right angle symbol. Always label it h.
On the TI-84 Plus select the MATH menu.
Press 0 or scroll down until the cursor is
on Solver and press ENTER. To delete
any existing equation press the up arrow
and then the CLEAR key.
Enter h2 = a2 + b2 as a2 + b2 – h2 as it needs to = 0.
a
b
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hypotenuse (h)
right angle symbol
–
You can use the equation function
on your graphics calculator to solve
Pythagoras problems (see right).
Once you have the equation entered
you can solve multiple problems.
ALPHA
x2
h
+
ALPHA
x2
ENTER
x
7.65 cm
(a)
(h)
We begin by labelling the sides of the
triangle a, b and h.
h2 = a2 + b2 Substitute
x2 = 7.652 + 11.32 x2 = 186.2125
Square root
x =
x = 13.645 97...
x = 13.6 cm (1 dp)
186.2125
x2
Enter the values for the known variables and enter
an initial value for the unknown and select
SOLVE
ALPHA
ENTER
to solve.
Find the length y
in the right-angled
triangle.
(a)
9.4 cm
y
(h)
8.6 cm
(b)
11.3 cm
(b)
Pythagoras
APPS
Example
Example
Find the length x
in the right-angled
triangle.
ALPHA MATH
We begin by labelling the sides of the
triangle a, b and h.
Pythagoras
h2 = a2 + b2 Substitute
9.42 = y2 + 8.62 88.36 = y2 + 73.96
y2 + 73.96 = 88.36
y2 = 14.4 Square root y =
y = 3.8 cm (1 dp)
14.4
IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
In
8
IAS 1.7 – Right-Angled Triangles
Example
©
Example
Find the distance the ladder reaches up the wall,
labelled w in the diagram.
w
43˚
6.5 m
A 9.8 m ladder is leaning against a
wall. The ladder makes an angle
of 74˚ with the ground. How far is the base from the wall?
We begin by drawing a diagram,
adding the information we have and
labelling the sides of the triangle opp,
hyp and adj.
We begin by labelling the sides of the
triangle opp, hyp and adj.
9.8 m
hyp
43˚
6.5 m
adj
opp
w
opp
hyp
74˚
adj
g
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Because we are required to find w (opposite)
and have information on the adjacent we use the
trigonometric ratio that involves these two, i.e.
tangent.
opp
tan A =
adj
w
Substitute
tan 43˚ =
6.5
Multiplying by 6.5
w = 6.5 x tan 43˚
Because we are required to find g (adjacent) and
have information on the hypotenuse we use the
trigonometric ratio that involves these two, i.e. cosine.
adj
cos A =
hyp
Substitute
g
9.8
g = 9.8 x cos 74˚
cos 74˚ =
Multiplying by 9.8
= 2.7 m (2 sf)
= 6.1 m (2 sf)
Achievement – Find the lengths of the unknown sides in the right-angled triangles below.
19.
23.4 mm
y
20.
43˚
z
36.4 m
36˚
22.
21.
w
29˚
13.4 km
v
31˚
12.8 m
IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
In
13
IAS 1.7 – Right-Angled Triangles
46. A hill has an angle of climb of
8.5˚ to the horizontal. How far
would it be necessary to walk
to increase one’s altitude by
200 m?
47. The angle of pitch of an A
frame house roof is 63.0˚ to
the horizontal. The vertical
rise of the roof is 7.40 m.
Find the slanted length of
the roof, and the width of
the house at ground level.
©
48. Two support cables each 262 m long run from the top of a radio tower
to the ground. Each cable makes an angle of 37˚ with the ground.
a) Find, to the nearest metre, the height of the
tower.
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Another cable is run
from the top of the
tower to the ground
and makes an angle of
54˚ with the ground.
c) Calculate the distance between the two cables on
the ground to the nearest metre.
b) Find the length of this
cable to the nearest
metre.
49. An escalator has an angle of elevation of 36˚ and
climbs a vertical height of 9.6 m. What is the
horizontal movement of the escalator?
IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
In
17
IAS 1.7 – Right-Angled Triangles
Three-Dimensional Figures
©
Find Lengths and Angles in Three-Dimensional Figures
To solve length and angle problems in threedimensions we identify the two-dimensional
right-angled triangle that contains our unknown
and has two side measurements or a side and an
angle measurement. We then use our knowledge of
Pythagoras and/or trigonometric ratios to find the
required angle or length.
B
C
D
2.85 m
A
w
F
x
6.4 m
E
H
7.2 m
G
To find the length x or the angle GFH we use the
right-angled triangle FGH.
F
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x
G
Angle GFH
7.2 m
H
To find the length w or the angle FHC we use the
right-angled triangle FCH.
C
w
Angle FHC
F
H
x
Once you have identified the two-dimensional
triangle from the 3D figure, draw it separately,
carefully marking in all the appropriate lengths and
angles.
IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
In
21
IAS 1.7 – Right-Angled Triangles
©
Excellence – Devise a strategy to solve each problem. Explain your approach with the aid of a diagram
72. A nail 9 cm long is lying inside an empty baked
beans can which has radius 3.75 cm and height
10.5 cm. See the diagram below.
9.0 cm
3.75 cm
10.5 cm
a) What angle does the nail make with the bottom of the can?
73.
A teepee is cone shaped. It has diameter 2.4 m
and slant height 2.2 m. See the diagram below.
A
2.2 m
b) How far up from the base of the can does the nail reach?
2.4 m
a) Could a person 1.85 m tall stand in the middle of the teepee with their head not touching the top? Justify your answer.
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b) Calculate the apex angle of the teepee labelled A in the diagram.
IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
In
46
Page 20
70. r = 1.30 m
71.
IAS 1.7 – Right-Angled Triangles
(3 sf)
steel= 11.5 m (3 sf)
A = 15.6˚
SQ= 5.94 m WQ= 7.87 m
VQ = 6.87 m
SQP = 49.6˚ WQT = 35.0˚
VQU = 41.2˚ Page 26
Page 24 cont...
75. a) Calculate BX as half of BG. 78. a)
Then work with triangle
ABX.
(1 dp)
(3 sf)
(3 sf)
(3 sf)
(1 dp)
(1 dp)
(1 dp)
Page 21
A
H
a) Ht. = 1.84 m
(3 sf)
b) A = 66.1˚
(1 dp)
Would touch the top.
Page 24
74. a) First calculate BD using
Pythagoras. Then work
with triangle FBD.
23.7 mm
B
29.1 mm
D
FDB = 39.2˚
b)First calculate ED. Then
work with triangle FDE.
(1 dp)
F
23.4 mm
E
D
29.3 mm
FDE = 38.6˚
DAH = 61.5˚ (1 dp)
8.163 m
H
C
3.8 m
B
HBC = 65.0˚ (1 dp)
79. a) A
6.5 m
B
5.4 m
A
(1 dp)
7.45 m
D
C
AC = 8.5 m
b)
(2 sf)
E
3.2 m
F
B
4.25 m
Ropes = 5.3 m
c) Same triangle.
(2 sf)
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F
Midpoint of BC
4.67 m
(1 dp)
73. Approach is to
form a triangle
with the centre
A/2
of the teepee. 2.2 m
h
Half the angle
at the apex and
the height can be 1.2 m
calculated.
3.875 m
3.8 m
A
b)
76. a) The projection of plane
ABGH on plane ABCD has
H in front of D. So we work
with triangle ADH.
a) Angle X = 33.6˚(1 dp)
b) h = 5.0 cm
©
A
b) Let K be the midpoint of BC. Then we calculate AK and
KX.
K
XAK = 34.5˚
Page 25
B
7.50 cm diameter
(1 dp)
5.64 m
h
X
AXB = 45.1˚
D
B
4.85 m
X
72. Approach. The triangle is
formed by the nail, diameter
and height it reaches up the
side. Angle X is the angle the
nail makes with the bottom
N
of the can.
9.0 cm
X
4.825 m
7.0 m
H
(1 dp)
HAD = 57.9˚
(1 dp)
A
Angle = 37.1˚
b)The projection of plane
Page 28
ABGH on plane ABFE has 80. x = 12.0 cm
H in front of E. So we work 81. x = 6.0 cm
with triangle AEH.
82. x = 3.5 cm
H
y = 3.6 cm
4.67 m
83. x = 8.0 cm
y = 17 cm
E
A
7.45 m
84. x = 12.0 cm
HAE = 32.1˚
(1 dp)
y = 10.9 cm
85.
x = 10.2 cm
77. a) The rays at right angles to
y = 28.8 cm
the line of intersection pass
through Z and X from the
Page 29
midpoint of AD labelled E.
86. x = 7.5 m
X
y = 13.0 m
6.86 m
87. x = 9.0 m
Midpoint y = 15.0 m
Z
of AD
E
5.68 m
88. x = 28.0 cm
XEZ = 50.4˚
(1 dp)
y = 24.2 cm
z = 6.9 cm
b)The rays at right angles to
the line of intersection pass 89. x = 2.8 m
through Z and X from the
y = 17.2 m
midpoint of AB labelled F. 90. x = 26.8 mm
X
y = 10.7 mm
6.86 m
z = 14.4 mm
Midpoint
91. x = 3.6 m
of AB
Z
2.89 m
F
y = 8.0 m
XFZ = 67.2˚
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(2 sf)
(2 sf)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
(1 dp)
IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
In