Download Answers to activity questions

Answers to activity questions
Check your skills
1.
(a)
Use Pythagoras’ Theorem to calculate
x.
h2 = a 2 + b2
x 2 = 302 + 102
30 cm
x 2 = 900 + 100
x 2 = 1000
x
x = 1000
x = 31.6cm
10cm
(b)
A rectangle has a length of 6.72m and
width of 4.83m. Find the length of the
diagonal (to 2 d.p.).
Let the length of the diagonal be d.
h2 = a 2 + b2
d 2 = 6.722 + 4.832
l
na
go
D ia
d 2 = 45.1584 + 23.3289
4.83m
d 2 = 68.4873
d = 68.4873
d = 8.28cm
6.72m
2.
(a)
Calculate the value of x
in this triangle.
x
opp
adj
x
Tan 40° =
12.5
12.5 × Tan 40° = x
Tanθ ° =
40°
x = 10.49 (to 2 d.p.)
12.5m
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(b)
Calculate the value of x
in this triangle.
7.3km
49°
x
(c)
Calculate the value of θ
in this triangle.
2m
1.
θ
θ = Sin −1 0.7083
θ = 45.10° or 45°5'48"
0.85m
3.
opp
hyp
7.3
Sin49° =
x
x × Sin49° = 7.3
7.3
x=
Sin49°
x = 9.67km
opp
Sinθ ° =
hyp
0.85
Sinθ ° =
1.2
Sinθ ° = 0.7083
Sinθ ° =
A kite on the end of a 30m string is flying at an angle of elevation of 72°. What is the height of
the kite directly above the ground?
opp
hyp
h
Sin72° =
30
30 × Sin72° = h
x = 28.53m
30
m
Sinθ ° =
h
72°
Pythagoras’ Theorem
1.
Use Pythagoras’ Theorem to find the missing length.
(a)
h2 = a 2 + b2
x 2 = 262 + 202
x 2 = 676 + 400
x
26
x 2 = 1076
x = 1076
x = 32.8
20
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(b)
h2 = a2 + b2
x
182 = x 2 + 12.52
324 = x 2 + 156.25
324 − 156.25 = x 2
12.5 cm
18 cm
x = 167.75
x = 12.95cm
(c)
Find the hypotenuse when a=6.1
and b=3.4
h2 = a2 + b2
x 2 = 6.12 + 3.42
x 2 = 37.21 + 11.56
x 2 = 48.77
(d)
x = 48.77
x = 6.98
h2 = a2 + b2
Find the missing side given
a=23.5 and h=40
402 = 23.52 + x 2
1600 = 552.25 + x 2
1600 − 552.25 = x 2
x = 1047.75
x = 32.37cm
(e)
Let the length of the linking side be l.
6m
x
6m
8m
(f)
h2 = a 2 + b2
h2 = a2 + b2
l 2 = 62 + 82
x 2 = l 2 + 62
l 2 = 36 + 64
x 2 = 102 + 62
l 2 = 100
x 2 = 136
l = 100
l = 10m
x = 136
x = 11.66m
This is an equilateral triangle. Half the triangle
to obtain a right angled triangle.
h2 = a2 + b2
h
102 = h 2 + 52
10 mm
Find the height of the triangle.
Also calculate the area.
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100 = h 2 + 25
100 − 25 = x 2
x = 75
x = 8.66mm
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(g)
A orienteering participant runs
650m north an then turns and
runs 1.4 km east. How far from
the starting point is the runner?
h 2 = a 2 + b2
650m = 0.65km
x 2 = 2.3825
x 2 = 0.652 + 1.42
x 2 = 0.42252 + 1.962
x = 2.3825
x = 1.54km
**A Real Challenge**
A square has a diagonal of 20 cm,
what is the side length?
20
cm
(h)
202 = x 2 + x 2
400 = 2 x 2
400
x2 =
2
x = 200
x = 14.14cm
x
x
2.
h2 = a 2 + b2
Do the following triangles contain a right angle?
(a)
10, 24, 26
Remember the
hypotenuse is the
longest side.
(b) n 7, 8, 10
2, 4.8, 5.2
a 2 + b 2 = 7 2 + 82
a 2 + b 2 = 22 + 4.82
a 2 + b 2 = 49 + 64
a 2 + b 2 = 4 + 23.04
a 2 + b 2 = 113
a 2 + b 2 = 27.04
a 2 + b 2 = 676
h 2 = 102
h 2 = 5.22
h 2 = 262
h 2 = 100
h 2 = 27.04
As a 2 + b 2 ≠ h 2 the
triangle does not
contain a right
angle.
As a 2 + b 2 = h 2 the
triangle must contain a
right angle.
a 2 + b 2 = 102 + 242
a 2 + b 2 = 100 + 576
h 2 = 676
As a 2 + b 2 = h 2 the
triangle must contain a
right angle.
(d)
(c)
1.4, 4.8, 5
(e)
6, 6, 8
(f)
5, 6, 7
a 2 + b 2 = 1.42 + 4.82
a 2 + b2 = 62 + 62
a 2 + b 2 = 52 + 6 2
a 2 + b 2 = 1.96 + 23.04
a 2 + b 2 = 36 + 36
a 2 + b 2 = 25 + 36
a 2 + b 2 = 25
a 2 + b 2 = 72
a 2 + b 2 = 61
h 2 = 52
h 2 = 82
h2 = 72
h 2 = 25
h 2 = 64
h 2 = 49
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As a 2 + b 2 = h 2 the
triangle does contain a
right angle.
As a 2 + b 2 ≠ h 2 the
triangle does not
contain a right
angle.
As a 2 + b 2 ≠ h 2 the
triangle does not
contain a right angle.
The Trigonometric Ratios – Finding Sides
1.
For the following triangles, identify the adjacent, opposite and hypotenuse for each triangle.
(a)
(b)
(c)
5
θ°
α°
a
12
z
y
c
13
b
60°
a – adjacent
b – opposite
c – hypotenuse
x
5 – adjacent
12 – opposite
13 – hypotenuse
x – adjacent
y - opposite
z - hypotenuse
(e)
(d)
(f)
245
β°
4mm
45.3
γ°
25
h
450
34mm
k
30mm
δ°
g
30mm – adjacent
34mm – opposite
45.34mm – hypotenuse
g – adjacent
h – opposite
25 – hypotenuse
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245 – adjacent
450 – opposite
k – hypotenuse
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2.
For the following triangles, find the value of the missing side(s).
(a)
(b)
(c)
245
63°
22.5°
16m
f
25
h
b
58°
opp
hyp
h
Sin58° =
25
25 × Sin58° = h
x = 21.2
Sinθ ° =
f = 480.8
(e)
(d)
14.7m
(f)
b
e
24°45'
m
33°
adj
hyp
14.7
Cos33° =
m
m × Cos33° = 14.7
14.7
m=
Cos33°
m = 17.5m
Cosθ ° =
opp
adj
b
Tan 22.5° =
16
16 × Tan 22.5° = b
b = 6.63m
Tanθ ° =
opp
Tanθ ° =
adj
f
Tan63° =
245
245 × Tan63° = f
m
5c
72
adj
hyp
b
Cos 24°45 ' =
725
725 × Cos 24°45 ' = b
b = 658.4cm
Cosθ ° =
68.52°
500m
d
opp
hyp
500
Sin68.52° =
e
e × Sin68.52° = 500
500
e=
Sin68.52°
e = 537.3m
Sinθ ° =
Using Pythagoras
e2 = 5002 + d 2
537.32 = 5002 + d 2
288691.3 − 250000 = d 2
d 2 = 38691.3
d = 38691.3
d = 196.7
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(g)
(h)
(i)
9°
36°36'
34
5m
m
h
x
h
57°
57°
62.5 km
24 cm
Half the shape is a right angled
triangle.
adj
Cosθ ° =
hyp
345
Cos9° =
x
x × Cos9° = 345
345
x=
Cos9°
x = 349.3mm
opp
hyp
62.5
Sin36°36 ' =
h
h × Sin36°36 ' = 62.5
62.5
h=
Sin36°36 '
h = 104.8km
opp
adj
h
Tan57° =
12
12 × Tan57° = h
h = 18.48cm
Sinθ ° =
Tanθ ° =
Area:
1
A = bh
2
A = 0.5 × 24 ×18.48
A = 221.76cm 2
3.
4.
Express the following directions as a true bearing.
(a)
S15°E = 165°T
(b)
NE = 45°T
(c)
N45°W = 315°T
(d)
10° W of N = 350°T
(e)
SSW = 202.5°
(f)
215° = 215°T
A person standing on top of a 25 m cliff sees a rower out to sea. The angle of depression of
the boat is 15.3°. How far is the boat from the base of the cliff?
15.3°
opp
adj
25
Tan15.3° =
d
d × Tan15.3° = 25
Angle of Depression
Tanθ ° =
25m
by alternate angles =
15.3°
d
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25
Tan15.3°
d = 91.4m
d=
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5.
A student wishes to find the height of a building. From a distance of 50m on perfectly level
ground, the angle of elevation to the top is 24.6°. Find the height of the building?
opp
adj
h
Tan 24.6° =
50
50 × Tan 24.6° = h
h = 22.9m
Tanθ ° =
h
24.6°
50m
6.
A plane is flying at altitude of 5000m. The pilot observes a boat at an angle of depression of
12°, calculate the horizontal distance which places the plane directly above the boat.
opp
adj
5000
Tan12° =
d
d × Tan12° = 5000
Tanθ ° =
d
5000m
12°
5000
Tan12°
d = 23523m or 23.5km
d=
7.
A walker decides to take a direct route to a landmark. They walk 1.7 km at a bearing of 78°T.
How far did they walk in a northerly and easterly direction?
N
e
n
78°
adj
hyp
n
Cos 78° =
1.7
1.7 × Cos 78° = n
n = 0.353km or 353m
opp
Sinθ ° =
hyp
e
Sin78° =
1.7
1.7 × Sin78° = e
e = 1.663km
Cosθ ° =
km
7
.
1
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8.
Find the perimeter of this trapezium.
opp
adj
h
Tan51° =
16
16 × Tan51° = h
h = 19.8cm
Tanθ ° =
28 cm
s
h
51°
adj
hyp
16
Cos51° =
s
s × Cos51° = 16
16
s=
Cos51°
s = 25.4cm
Cosθ ° =
44-28 = 16cm
44 cm
Perimeter = 44 + h + 28 + s
= 44 + 19.8 + 28 + 25.4
= 117.2cm or 1.172m
9.
A kite is attached to a 45m line. On a windy day, the kite flies at an angle of elevation of 28°.
Calculate the height of the kite above the ground.
opp
hyp
h
Sin 28° =
45
45 × Sin 28° = h
h = 21.13m
Sinθ ° =
45m
h
28°
10.
A plane flying at an altitude of 10 000m is flying away from a person. The angle of elevation
of the plane is 76° when initially observed. After 1 minute 15 seconds, the plane is at an
angle of elevation of 29°. Ignoring the height of the person, what is the speed of the plane in
km/hr?
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10000m
76°
dc
di
29°
df
Initially, the plane is di away from the observer.
After 1minute 15 seconds, the plane is d f away from the observer.
The distance covered in 1 minute 15 seconds is d f − d i .
opp
adj
10000
Tan76° =
di
opp
adj
10000
Tan 29° =
df
di × Tan76° = 10000
d f × Tan 29° = 10000
Tanθ ° =
Tanθ ° =
10000
Tan76°
di = 2493m
d f − di
= 18040 − 2493
= 15547 m or 15.547km
10000
Tan29°
d f = 18040m
di =
df =
1 minute 15 seconds is 1.25 minutes. 1.25 minutes ÷ 60 =0.0208333 hours.
distance covered (km)
time taken(hr)
15.547 km
=
0.0208333hr
= 746km / hr (rounded)
speed (km / hr ) =
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The Trigonometric Ratios – Finding Angles
1.
Find the size of the angle in the questions below to 2 d.p.
(a)
Sin θ = 0.2345
θ = 13.56°
(b)
Cos θ = 0.5736
θ = 55°
(c)
Tan θ = 2.4604
θ = 67.88°
(d)
Cos β = 0.63
β = 50.95°
(e)
Tan µ = 0.4998
µ = 26.56°
(f)
Sin σ = 0.9455
σ = 71°
2.
Find the size of the angle in the questions below in degrees, minutes and seconds.
(a)
Sin θ = 0.2345
θ = 13°33’44”
(b)
Cos θ = 0.5736
θ = 54°59’54”
(c)
Tan θ = 2.4604
θ = 67°52’53”
(d)
Cos β = 0.63
β = 50°57’
(e)
Tan μ = 0.4998
μ = 26°33’21”
(f)
Sin σ = 0.9455
σ = 70°59’48”
3.
For the following triangles, find the size of the missing angle.
(a)
(b)
(c)
δ°
3.7m
25 cm
22 cm
4.45m
2800mm
θ°
α°
opp
Sinα ° =
hyp
22
Sinα ° =
25
Sinα ° = 0.88
α ° = 61.64° or 61°38'33"
(d)
α°
adj
Cosθ ° =
hyp
3.7
Cosθ ° =
4.45
Cosθ ° = 0.8315
θ = 33.75° or 33°45'3"
(e)
1200mm
opp
adj
2800
Tanθ ° =
1200
Tanθ ° = 2.333
θ ° = 66.8° or 66°48'5"
Tanθ ° =
(f)
14.2 cm
2.4km
1800m
β°
9.2 cm
3000mm
θ°
2.4km = 2400 m
θ°
θ°
2900mm
This is an isosceles triangle.
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4.
Half the base is 1450mm.
adj
hyp
9.2
Cos β ° =
14.2
Cos β ° = 0.6479
β = 49.62° or 49°37'3"
opp
adj
1800
Tanθ ° =
2400
Tanθ ° = 0.75
θ ° = 36.9° or 36°52'12"
Cos β ° =
Tanθ ° =
opp
adj
3000
Tanθ ° =
1450
Tanθ ° = 2.0690
θ ° = 64.2° or 64°12'14"
Tanθ ° =
Complete the triangle below:
Calculate x using Pythagoras.
2
α°
β°
cm
cm
3
5
2
h = a +b
x
2
x 2 = 52 + 32
x 2 = 25 + 9
x 2 = 34
x = 34
x = 5.83
Calculating α
opp
adj
3
Tanα ° =
5
Tanα ° = 0.6
α ° = 30.96° or 30°57'50"
Tanα ° =
Calculating β
β = 90 − α
β = 90 − 30.96°
β = 59.04°
or
β = 90 − 30°57 '50"
β = 59°2 '10"
5.
A ship sails 400 km north and then 800 km west. How far is it away from the port it started
from and at what true bearing?
Find d using Pythagoras.
N
h 2 = a 2 + b2
400 km
800 km
d
θ°
d 2 = 4002 + 8002
d 2 = 160000 + 640000
d 2 = 800000
d = 800000
d = 894.4km
Finding the true bearing
opp
adj
800
Tanθ ° =
400
Tanθ ° = 2
θ ° = 63.4° or 63°26 '6"
Tanθ ° =
The ship is located 894km away
at a true bearing of 63.4°.
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6.
It is recommended that ramps for wheelchairs have a rise to run ratio of 1:15. What angle
would the ramp surface make to the horizontal if a ramp follows this specification?
Units are not required to answer this question.
1
θ°
15
The surface of the ramp should make an angle of
about 4° to the horizontal to be wheelchair
friendly.
7.
opp
adj
1
Tanθ ° =
15
Tanθ ° = 0.0666
θ ° = 3.8° or 3°48'51"
Tanθ ° =
Two flag poles are in a park on a level surface 30m apart. One pole is 10m tall and the other
is 19m tall. What is the angle of elevation of the top of the tall pole from the top of the smaller
pole?
opp
adj
9
Tanθ ° =
30
Tanθ ° = 0.3
θ ° = 16.7° or 16°41'57"
Tanθ ° =
19-10=9m
θ°
30m
19m
10m
8.
A 10m antenna is supported by guide wires on four sides. Each wire is attached to the
ground 6 metres from the base of the antenna.
Find the length of each
wire
h2 = a 2 + b2
2
2
d = 6 + 10
l
2
d 2 = 36 + 100
10m
d 2 = 136
θ°
d = 136
d = 11.66m
What angle does the wire
make with the ground?
opp
adj
10
Tanθ ° =
6
Tanθ ° = 1.666
θ ° = 59.04° or 59°2 '10"
Tanθ ° =
6m
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9.
Calculate the pitch of this roof. (Assuming symmetry)
This is an isosceles triangle. Half the large
triangle is a right angled triangle.
opp
adj
2.5
Tanθ ° =
4.1
Tanθ ° = 0.6098
θ ° = 31.4° or 31°22 ' 23"
2.5m
Tanθ ° =
θ°
8.2m
10.
In ∆ABC : ∠A = 90°, a = 16.9, b = 6.5, calculate∠B .
The first step is to draw the diagram to match the
information given.
B
θ°
16.9
A
6.5
opp
hyp
6.5
Sinθ ° =
16.9
Sinθ ° = 0.3846
∠B = θ ° = 22.62° or 22°37'12"
Sinθ ° =
C
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