Exercise 3G
Let A be the origin and let AC be the positive x-axis.
By Pythagoras’ theorem;
BC2 = 102 – 62
= 64
BC = 8 cm
x 3 3 6
24 = 10 + 6 + 8
y 4 0 4
x 1 96
=
y 24 72
4
=
3
2
tan θ =
5
θ = 21.801…
= 21.8° (3 s.f.)
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2
Let A be the origin and let AF be the positive x-axis.
x 0 5 10 7.5 5 2.5
50 = 15 + 10 + 5 + 5 + 10 + 5
y 7.5 15 12.5 10 5 0
0 50 50 37.5 50 12.5
= + + + + +
112.5 150 62.5 50 50 0
x 1 200
=
y 50 425
4
=
8.5
6
tan θ =
1.5
θ = 75.963…
= 76.0° (3 s.f.)
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3 a
Since the Let D be the origin and let DC be the positive x-axis.
y 3 x 8
tan θ= = so =
x 8 y 3
Therefore x = 8
x 0 −8
b ( M + kM ) =M + kM
y 3 6
x 1 −8k
=
y 1 + k 3 + 6k
−8k
1+ k
=
3 + 6k
1+ k
8
tan θ =
15
8k
1+ k = 8
3 + 6k 15
6−
1+ k
8k
1+ k 8
=
6 + 6k − 3 − 6k 15
1+ k
8k 8
=
3 15
1
k=
5
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4 From question 1:
x 4
=
y 3
Where A is the origin and AC is the positive x-axis.
x 4 0
1.75M= M + 0.75M
y 3 0
x 1 4
=
y 1.75 3
16
7
=
12
7
26
tan θ = 7
44
7
θ = 30.579...
= 30.6° (3 s.f.)
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5 From question 2:
x 4
=
y 8.5
Let A be the origin and let AF be the positive x-axis.
x 4 5
1.15=
M M + 0.15M
y 8.5 0
x 1 4.75
=
y 1.15 8.5
95
23
=
170
23
170
tan θ = 23
95
23
θ = 60.802...
= 60.1° (3 s.f.)
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6
Let E be the origin and let EC lie along the positive x-axis.
By symmetry the centre of mass of the lamina is at the point (2, 2).
x 2 4
1.1M= M + 0.1M
y 2 0
x 1 2.4
=
y 1.1 2
24
= 11
20
11
24
tan θ = 11
68
11
θ = 19.440...
= 19.4° (3 s.f.)
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7 a
Let F be the origin and let FE lie on the positive x-axis.
x 12 0 3 1.5 8 3 5 5.5 4 8 8 4
W = W + W + W + W + W + W
y 40 6 40 12 40 8 40 4 40 2 40 0
x 1 120
=
y 40 200
3
=
5
Res(↑) T1 + T2 = W (1)
Taking moments about the centre of mass gives:
3T1 = 5T2
5
T1 = T2 (2)
3
Substituting (2) into (1) gives:
5
T2 + T2 = W
3
8
T2 = W
3
3
T2 = W
8
Therefore:
5
T1 = W
8
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7 b
5
tan θ =
1
θ = 78.690...
= 78.7° (3 s.f.)
8 a
Let A be the origin and let AB lie on the positive x-axis.
x 6 3 3 7.5 10 9 3 7.5 10 6 6 3 4 0
W = W + W + W + W + W + W + W
y 42 0 42 2 42 −3 42 −8 42 −3 42 −4 42 −2
231 11
x 42 2
=
=
y − 110 − 55
42 21
Res(↑) T1 + T2 sin 30 =
W
1
T1 + T2 = W
2
Taking moments about D gives:
11
9 × T1 = 9 − × W
2
7
9T1 = W
2
7
T1 = W
18
Therefore:
11
T2 = W
9
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8 b
55
tan θ = 21
11
2
θ = 25.463...
= 25.5° (3 s.f.)
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