Download X - Gore High School

SLO
The basics of Trigonometry
http://www.youtube.com/watch?v=lJqhTr0SI4k
(The basics and how to label the sides of a triangle)
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Where did it come from?
•Trigonometry was probably invented for
use in astronomy.
•The origins of trigonometry can be traced
to the civilizations of ancient Egypt,
Mesopotamia and the Indus valley, more
than 4000 years ago.
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Words to Learn
Angle of Elevation: angle from horizontal upwards
Angle of Depression: Angle from horizontal downwards
dp: decimal place
Trigonometry: The word comes from Greek: trigōnon "triangle" +
metron "measure“ (sine, cosine and tangent)
Vertical: up/down
Horizontal: left/right
Plane: flat surface
Ascend: To go up
Descend: To go down
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Naming parts of a triangle
The side
opposite the
marked angle
is called the
opposite side.
Take care, the
opposite and
adjacent can
‘swap’ places
The longest side is
called the hypotenuse.
O
P
P
O
S
I
T
E
H
Y
P
O
T
E
N
U
S
E
θ
This letter is
called Theta
(a Greek letter)
ADJACENT
The side next to the marked
angle is called the adjacent side.
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Label the sides
5 of 43
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Using a calculator
Finding the value of sin 65°?
First make sure that your calculator is set to degrees.
Key in:
sin
6
5
=
This is
important
Your calculator should display 0.906307787
Only round answers at the final step
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The inverse of sin
Finding the value of sin–1 0.5
To work this out use the sin–1 key on the calculator.
shift
sin
0
.
5
=
The answer to the above is 30.
sin–1 is the inverse or opposite of sin. It is sometimes called arcsin.
sin
30°
0.5
sin–1
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Your Turn:
Find the following.
1) sin 79° = 0.982
2) cos-1 0.4 =
3) tan 65° =
4) cos 11° = 0.982
2.14
66.42
5) sin-10.6 = 36.87
6) tan 84° =
7) tan 49° =
1.15
8) sin 62° = 0.883
9) tan-11.3=
52.43
10) cos 56° = 0.559
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9.51
SLO
Introduction to SOHCAHTOA
http://www.youtube.com/watch?v=54hXSHe1SQE
(Remembering sohcahtoa)
9 of 43
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SOHCAHTOA
You MUST
Some
Old
Hippy
Caught
Another
Hippy
Tripping
On
Acid
remember this
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What do the letters of SOHCAHTOA mean?
Guess what the letters stand for
Sin
SOHCAHTOA
Adjacent
Opposite
Opposite
Hypotenuse
Tangent
Cos
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Adjacent
Hypotenuse
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From SOHCAHTOA to triangles
(click to show how to form triangles)
SOH
θ
12 of 43
CA H
θ
TOA
θ
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How to use the SOHCAHTOA triangles
(same method as Distance/time triangles in science)
If one letter is
above another
letter, divide
O
SxH
θ
If one letter is
next to another
letter, multiply
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How to use a SOHCAHTOA triangle
Basic Rule: Put finger over what you want to find
E.g. 1) What does the opposite equal?
O
SxH
θ
Opposite = Sin (angle) x Height
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How to use the SOHCAHTOA triangles
Basic Rule: Put finger over what you want to find
E.g. 2) What does the Hypotenuse equal to?
O
Sθ H
Hypotenuse = Opposite ‚ Sin(angle)
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Your turn:
Use the triangle given below to find how you would calculate the given lengths
1) Adjacent
A
Cx H
θ
Cos(Angle) x Hypotenuse
2) Hypotenuse
Adjacent ‚ Cos(Angle)
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Your turn:
Use the triangle given below to find how you would calculate the given lengths
1) Adjacent
O
T xA
θ
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Opposite ‚ Tan(Angle)
2) Opposite
Tan(Angle) x Adjacent
SLO
How to decide which part of SOHCAHTOA to use
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SOH or CAH or TOA
O
A
O
S H
C H
T A
When we use Trigonometry we need to use the correct part
of SOHCAHTOA
Step 1: Identify what sides you have and what sides you have
been asked to find.
Step 2: Find the SOHCAHTOA triangle with these 2 sides.
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SOH or CAH or TOA
E.g. for the following triangle do you use sin, cos or tan to
find the length x?
Step 1: Label what is given and what is to find
Find: Hypotenuse
x
390
Step 2:
Which part of
SOH CAH TOA
uses H and A
27cm
A
Given: Adjacent
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C H
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SOH or CAH or TOA
E.g. for the following triangle do you use sin, cos or tan to
find the length x?
Step 1: Label what is given and what is to find
Find: Opposite
Step 2:
Which part of
SOH CAH TOA
uses O and A
x
390
O
27cm
Given: Adjacent
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T A
Your Turn:
For the following triangle do you use sin, cos or tan to find
the length x?
19cm
O
x
250
S H
67cm
A
470
x
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C H
Your Turn:
For the following triangle do you use sin, cos or tan to find
the length x?
x
O
8cm
250
470
S H
O
x
T A
83cm
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Questions to do from the books
SOH or CAH or TOA
Achieve
Gamma
IAS 1.7
24 of 43
P402 Ex28.01 Q1
Merit
Excellence
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SLO
Finding lengths of sides
Note
Trigonometry is for right angled triangles only
Trigonometry must involve angles
http://www.youtube.com/watch?v=D38S1sgErjQ
(Finding sides using SOHCAHTOA)
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E.g. 1: Find the length of the side marked X
[1] Label given side and side
to find.
380
[2] Use information from the
diagram to choose which
part of SOHCAHTOA to
use
48m
[4] check answer reasonable
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O
X
O
S
[3] Use SOHCAHTOA triangle
to find length (Cover up
what you want to find)
H
H
X  Sin38 48
X = 29.6m
Opposite is smaller than
Hypotenuse therefore
good answer
E.g. 2: Find the length of the side marked X in the diagram below.
[1] Label given side and side
to find.
[2] Use information from the
diagram to choose which
part of SOHCAHTOA to
use
48m
A
570
H
X
A
C H
[3] Use SOHCAHTOA triangle
to find length (Cover up
what you want to find
[4] Check answer reasonable
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X = 48 ‚ cos57
X = 88.1 m
Hypotenuse bigger than
adjacent, therefore good answer.
Your Turn: Find the length of the side marked X
[1] Label given side and side
to find.
17m
290
[2] Use information from the
diagram to choose which
part of SOHCAHTOA to
use
[3] Use SOHCAHTOA triangle
to find length (Cover up
what you want to find
[4] Check answer reasonable
Hard to check this one.
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A
O
X
O
T A
X = Tan29 x 17
X = 9.4 m
Your Turn: Find the length of the side marked X
15m
X
[1] Label given side and side
to find.
H
O
340
[2] Use information from the
diagram to choose which
part of SOHCAHTOA to
use
S
[3] Use SOHCAHTOA triangle
to find length (Cover up
what you want to find
X = 15 ‚ sin34
X = 26.8 m
[4] Check answer reasonable
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O
H
Hypotenuse is longer than
opposite therefore good answer.
Your Turn:
Find x to 2 decimal places.
O
12 cm
x
S H
x = 12 × sin 56°
x = 9.95 cm
56°
x
50°
6.5 cm
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O
T A
x = tan50 x 6.5
x = 7.75 cm
Your Turn:
Find x to 2 decimal places.
x
32°
455 m
O
S H
x
O
T A
31 of 43
x = 455 ‚ sin32
x = 858.62 m
x = 7 ‚ tan43
x = 7.51m
Questions to do from the books
Achieve
Gamma
P402 Ex28.01
Q3 – 5
IAS 1.7
P8 Q19 – 40
Merit
Excellence
P12 Q41 – 49
Some harder Achieve questions to follow but first practice
the basics
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Applications of Trigonometry
A 5 m long ladder is resting against a wall. It makes an
angle of 70° with the ground.
What is the distance between the
base of the ladder and the wall?
5m
We are given the hypotenuse and we want
to find the length of the side adjacent to
the angle, so we use:
A
70°
x
C H
x = 5 × cos 70°
= 1.71 m
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Your Turn:
A ladder 4.7 m long is leaning against a wall. The angle between
the wall and ladder is 27°.
Draw a diagram and find the height the ladder extends up the
wall. (Click for diagram)
27°
Ladder (4.7 m)
Wall
(x)
A
A
C H
H
x = cos27 x 4.7
x = 4.19 m
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Your Turn:
A building that is 5.6 m high is to have a rope hooked from it’s
top to make an angle of 70° with the ground. How long will
the rope need to be? (Click for diagram)
O
H
rope
5.6m
building
O
S H
70 o
Rope = 5.6  sin(70)
= 5.96 m
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Your Turn:
To find the height of the school flagpole, the angle from a
person’s eye-level to the top of the flagpole is measured. This
angle is 23°. If the person is standing level with the base of
the flagpole and 3.5 m from it, and their eye is 1.63 m above
the ground, how tall is the flagpole? (Click for diagram)
23°
A = 3.5 m
x=0
O
T A
1.63 m
3.5 m
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flag = 1.63 + 3.5 x tan(23)
= 3.1 m
Your Turn:
When crossing a river which is 4.2 metres wide with parallel
river banks, a boat forms an angle of 55° with the bank
where it starts. How far down the river will the boat reach
the opposite bank? (Click for diagram)
4.2 m
55 o
Starting point
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A
O
O
T A
A = 4.2 ÷ tan(55)
= 2.9 m
Your Turn:
7.
A metal brace is to be used to support a
frame with an angle of 70° on a base of 1.2 m
as shown.
(a) How long should the metal brace be?
(b) How high up the frame will the brace meet
the vertical frame?
70°
1.2 m
(b)
(a)
A
O
C H
T A
d = 1.2 ÷ cos(70)
Metal brace length = 3.5m
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ht = 1.2 x tan(70)
ht = 3.3m
Finding side lengths
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Your Turn:
Find the height of the tree
x
23
1.72m
12m
O
tan  
A
x
tan 23 
12
x  12 tan 23
x  5.09m(3sf )
 height of the tree  1.72  5.09
 6.8m(2sf )
The tree is approximately 7m tall
40
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Your Turn:
When throwing the discus, Anna’s throw is in line with the centre of the
throwing circle and a distance of 25.6 m straight out in the middle of the
allowable sector. Bernice’s throw is on an angle of 12o from the middle
and is a perpendicular distance of 24.8 m.`
Who threw the greatest throw and by how much? (click for clue)
Step 1: label sides:
Step 2:
C H
24.8 m
d
A 12° H
Step 3:
d = 24.8 ‚ cos(12) = 25.35m
Anna wins by (25.6 – 25.35) = 0.25 m
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distance
perpendicular
middle line
A
A
B
Questions to do from the books
Achieve
Gamma
P402 Ex28.01
Q3 – 5
IAS 1.7
P8 Q19 – 34
42 of 43
Merit
P11 Q35 – 49
Excellence
SLO
Finding angles
http://www.youtube.com/watch?v=6wG3_x40YBA
(Finding angles using SOHCAHTOA)
43 of 43
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SOHCAHTOA to find angles
O
A
O
S H
C H
T A
Step 1: Label the 2 sides you are given using the angle you
have been asked to find.
Step 2: Find the SOHCAHTOA triangle with the 2 given sides.
Step 3: Finger over S or C or T (this is the angle).
Step 4: Use inverse button and SOHCAHTOA triangle to find
angle (you will need to use brackets).
44 of 43
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E.g. 1: Find the size of angle x
METHOD
[1] Label 2 given sides.
27m O
[2] Choose from
sin/cos/tan.
x
A
42m
O
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[3] Finger over S, C or T
T A
[4] Calculate angle using
inverse button ( -1 ):
remember brackets.
X = tan-1(27 ‚ 42)
X = 32.70
E.g. 2: Find the size of angle x
METHOD
[1] Label 2 given sides.
x
H 61m
[2] Choose from
sin/cos/tan.
O
52m
O
[3] Finger over S, C or T
[4] Calculate angle using
inverse button ( -1 ):
remember brackets.
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S H
X = sin-1(52 ‚ 61)
X = 58.50
E.g. 3: Find the size of angle x
METHOD
[1] Label 2 given sides.
34m
H
[2] Choose from
sin/cos/tan.
x
A
17m
A
[3] Finger over S, C or T
[4] Calculate angle using
inverse button ( -1 ):
remember brackets.
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C H
X = cos-1(17 ‚ 34)
X = 600
Your Turn:
8 cm
Find θ to 2 decimal places.
5 cm
θ
We are given the lengths of the sides opposite and the adjacent
to the angle, so we use:
O
T A
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θ = tan–1 (8 ‚ 5)
= 57.99° (to 2 d.p.)
Your Turn
1) Find angle A
16.1mm
O
2) Find angle B
23.4 mm
H
A
O
S H
A = sin-1(16.1 ‚ 23.4)
A = 43.5° (1 d.p.)
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B
2.15m
A
4.07 m
H
A
C H
B = cos-1(2.15 ‚ 4.07)
B = 58.1° (1 d.p.)
Your Turn:
A vertical mast is held by a 48 m long wire. The wire is attached to
a point 32 m up the mast.
Draw a diagram and find the angle the wire makes with the mast.
(click for diagram)
A
C H
H
A
32 m
A
48 m
A = cos-1(32 ‚ 48)
A = 48.2° (1 d.p.)
50 of 43
Your Turn:
A swimming pool is to be dug out of the ground. The side view
cross-section is a trapezium shape, with a length of 25 m. The
pool is to be 1.2 m deep at one end and 2.8 m deep at the other
end. If digging starts at the shallow (1.2 m) end, what angle
should be used to ensure that the pool will be 2.8 m deep when
the 25 m length is dug?
Step 1: Label sides
Step 2:
25 m
1.2 m
O
T A
Step 3:
x = tan–1(0.064)
= 3.7° (to 2 s.f.)
51 of 43
xo
A
2.8 m
O
Finding angles
52 of 43
Angles of elevation
53 of 43
Angles of depression
54 of 43
Questions to do from the books
Angles
Achieve
Gamma
P404 Ex 28.02 Q3
IAS 1.7
P15 Q50 – 59
55 of 43
Merit
P16 Q60 – 67
Excellence
3D Shapes
Most 3D questions involve Pythagoras,
Trigonometry and 2 steps
56 of 43
Pythagoras Revision
Find the length of the side marked P
1230 m
P² = 1230² + 960² = 2 434 500
p
P = 1560 (to 3 s.f.)
57 of 43
960 m
Pythagoras Revision
Find the length of the side marked y
13.2 m
y² = 32.7² – 13.2² = 895.05
y = 895.05
= 29.9 (to 3 s.f.)
58 of 43
32.7 m
y
Questions to do from the books
(Pythagoras)
Achieve
Gamma
P386 Ex27.01
Q4–15
Merit
Excellence
P387 Ex27.01
Q16 – 17
P394 Ex27.03
P389 Ex27.02
IAS 1.7
59 of 43
P4 Q1 - 6
P4 7 – 18
P412 Ex 29.01
Q8 – 11
The following questions are in 3D
and involve a combination of
Pythagoras and Trigonometry
60 of 43
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E.g. Calculate the size of:
a) length x,
G
O
7m
H
b) length w, c) angle CHE and d) angle GCH
length x
2 = 52 + 62
x
A
B
x = √(52 + 62)
x = √61
F
x = 7.8 m (1 d.p.)
w
Make sure
length w
2 = 72 + 7.82
you use
w
D
C
whole
w = √(72 + 7.82)
Ax
5m
answer for
w
=
√110
O
x in
w
=
10.5
m
(1
d.p.)
E
6 mA
calculation
angle CHE
O
T A
CHE = tan-1(5 ‚ 6)
CHE = 39.8° (1 d.p.)
61 of 43
angle GCH
O
T A
GCH = tan-1(7 ‚ 7.8)
GCH = 41.9° (1 d.p.)
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Two step problems
E.g. Find angle α
Clue: Draw separate triangles
(Click for first half of answer)
G
10m
A
8m
D
F
6m
a 2  b2  h2
(8) 2  (10) 2  h 2
164  h 2
h  12.8m(1dp )
AG = 12.8m (1dp)
62 of 43
A
12.8m
G
O
tan  
A
6
tan  
12.8
 6 
  tan 1 

12.8


  25.1(1dp)
Your Turn:
B
Find the angle θ
15cm
A
G
F
10cm
15cm
H
E
10cm
H
O
A
C
12
tan  
18.03
12cm
1  12 


tan


G
 18.03 
  33.6(1dp)
tan  
a 2  b2  h2
(10) 2  (15) 2  EG 2
100  225  EG 2
E
 18.03cm(2dp)
63 of 43
G

To find EG
EG  325
12cm
D
(Click for half the answer)
E
C

18.0cm
Your Turn:
A ramp is to be used to drive a wheel of a vehicle up a height of 0.7 m. The angle the
ramp forms with the ground is 15° and the ramp is to be 0.5 m wide. A diagonal
brace is to be used as shown
How long will this brace be? (click for first half of answer)
Side face can be used to find
slant length, d
d
0.7 m
H
Step 1: Label sides
O
Step 2:
15 o
0.5 m
O
h, b
t
g
n
e le
S H
Step 3: d = 0.7 ÷ sin(15)
Slant length = 2.7 m (to 2 s.f.)
Now find brace length
c
bra
slant length
d
Use the right-angle triangle on slant
b² = 2.7² + 0.5² = 7.5648
b = 8.73 = 2.75
Brace length = 2.8 m (to 2 s.f.)
64 of 43
Your Turn:
A chair has a square seat with 4 legs 65 cm long and 80 cm apart. A bar
joining two of the adjacent legs is bolted 2 cm from the floor. If the bar
is to form an angle of 36o with a line parallel to the floor, find:
(a) how far up the leg the bolt hole should be drilled
(b) the length of the bar between bolt holes.
bar
65 cm
b
H
O d
36
36o
o
80 cm
A
80 cm
(a) Step 1: Label sides
Step 2: TOA
Step 3: d = 80 x tan (36) = 58.1 (to 3 s.f.)
Distance up leg = 58.1 + 2 = 60.1 cm
65 of 43
36 o
A
80 cm
(b) Step 1: Label sides
Step 2: CAH
Step 3: b = 80 ÷ cos (36)
Length of bar = 98.9 cm (to 3 s.f.)
Your Turn:
A chair has a square seat with 4 legs 65 cm long and 80 cm apart. A bar parallel to the
floor joining two diagonally opposite legs is bolted 2 cm from the floor. Another bar is
slanted to join diagonally opposite legs 2 cm from the floor and 2 cm from the seat on
the other leg.
What angle is there between the horizontal bar and the slanted bar?
(Click for halfway workings)
First step: we need Floor Parallel Bar length
ar
db
nte
sla
65 cm
r
floor parallel ba
Floor bar’s length = d.
d² = 80² + 80²
= 12 800
d = 12 800
= 113 cm (3 s.f.)
d cm
80 cm
80 cm
Floor bar length =113 cm (3 s.f.)
Step 1: Label sides
Step 2:
Step 3: x = tan–¹(61 ÷ 113) = 28.4°
66 of 43
O
T A
b cm
61 cm
O
113 cmA
80 cm
Angles in a cuboid
67 of 43
Lengths in a square-based pyramid
68 of 43
Questions to do from the books
3D shapes
Achieve
Merit
Gamma
P414 Ex29.02 Q6 – 11
P420 Ex29.05
IAS 1.7
P21 Q72 – 73
P26 Q78 – 79
69 of 43
Excellence
P19 Q68 – 70
P24 Q74 – 77
Revision
http://www.youtube.com/watch?v=E1MAUC5TLOI
(Revision of Trigonometry + Pythagoras)
http://www.youtube.com/watch?v=hG-1RO0TghA
(Good revision of Trigonometry(sides and angles).
Achieve: watch first 17 minutes,
Ignore: 17.10 to 21.05
Merit: watch 21.05 onwards
70 of 43
Achieve Trigonometry Revision
(Angles and sides)
71 of 43
Find the length of the side marked x:
Step 1: Label sides
Step 2:
16 cm
A
C H
H
39 o
xcm
x = 16 x cos (39)
= 12.4 (to 3 s.f.)
72 of 43
A
Find the size of the angle, Theta:
o
Step 1: Label sides
Step 2:
O
S H
Step 3:
Θ
= sin–¹(32.8 ÷ 45.8)
= 45.7 (to 3 s.f.)
73 of 43
H 45.8 m
O
32.8 m
Find the angle marked k:
k
Step 1: Label sides
Step 2:
O
S H
Step 3:
k = sin–¹(0.4)
= 23.6 (to 1 d.p.)
74 of 43
o
H
O
2.8
7
Find the length of the side marked c:
Step 1: Label sides
Step 2:
O
S H
Step 3:
c = sin(23) x 8
= 3.1 mm (to 2 s.f.)
75 of 43
23
O
c
H
o
8 mm
Find the length of the side marked d:
Step 1: Label sides
Step 2:
7
O
A
T A
Step 3:
d = 7 x tan(43)
= 6.5 (to 2 s.f.)
76 of 43
dO
43 o
Find the angle marked y:
Step 1: Label sides
Step 2:
A
C H
75 m
H
y
Step 3:
y = cos-1 (50 ÷ 75)
= 48.2 (to 1 d.p.)
77 of 43
o
A
50 m
Find the length of the side marked y:
Step 1: Label sides
Step 2:
O
S H
50.0 m
O
H
Step 3:
y = sin(75) x 50.0
= 48.3 m (to 3 s.f.)
78 of 43
y
75
o
Find the angle marked p:
9 cm
o
p
Step 1: Label sides
Step 2:
O
T A
Step 3:
p = tan–¹(9 ÷ 16)
= 29.4 (to 1 d.p.)
79 of 43
A
16 cm
O
Find the angle marked t:
Step 1: Label sides
Step 2:
O
S H
Step 3:
t = sin–¹(8 ÷ 18)
= 26.4 (to 1 d.p.)
80 of 43
8 mm
t
O
H
o
18 mm
Find the angle marked k:
k
Step 1: Label sides
Step 2:
O
S H
Step 3:
k = sin–¹(0.4)
= 23.6 (to 1 d.p.)
81 of 43
o
H
O
2.8
7
108 m
Find the size of x:
xm
Step 1: Label sides
Step 2:
O
T A
Step 3:
x = 108 x tan(18)
= 35.1 (to 3 s.f.)
82 of 43
O
Ao
18
Find the size of Y:
y
Step 1: Label sides
Step 2:
A
C H
Step 3:
y = cos(34) x 75
= 62.2 (to 3 s.f.)
83 of 43
34 o
A
75
H
Find the size of h:
Step 1: Label sides
Step 2:
ho
75 m
A
O
T A
Step 3:
h = tan–¹(59 ÷ 75)
= 38.2 (to 3 s.f.)
84 of 43
O
59 m