Level 1 Engineering Mathematics, Tutorial sheet 3
You should consider or attempt these problems before your tutorial in
Week 4 (14th - 18th October)
1. Rearrange the following expressions to obtain x:
√
√
3
(a) 2x2 + 8x + 6 = x + 3 (b) x−1
=x+1
(c) x − 1 − 3 = 0
1
1
2
3
(d) x1 − 12 = x
(e) x−1
− x+1
= 1 (f) x+1
= 2x−1
2. Given the following length of a line and the angle of the line to the
horizontal (measured clockwise from the right hand side of the horizontal)
find the height reached by the line, and the distance travelled to the right.
(a) 10, 45◦
(c) 15, 90◦
(b) 6, 30◦
(d) 20, 60◦
3. For each of the following triangles, labelled by convention, find all the
angles and sides.
(a) a = 50, c = 40, B = 66◦
(c) a = 14.2, b = 10.1, B = 63◦
(b) a = 10, b = 9, c = 5
(d) a = 7.9, b = 11.2, B = 27◦
4. Solve the following trigonmetric equations in the range 0◦ ≤ θ < 360◦ .
(a) 8 sin2 θ − 2 sin θ − 1 = 0 (b) 2 tan θ sin θ − 3 sin θ − 2 cos θ = 0
5. Solve the following trigonmetric equations in the range −180◦ ≤ θ < 180◦ .
(a) 8 cos2 θ + 2 sin θ = 7
(b) 6 cos θ − sec θ = 1
6. A function of the form A sin(ωt) represents a wave with amplitude A
and angular frequency ω. The frequency f in Hz of the wave obeys the
relationship 2πf = ω and the period T is given by T = f −1 . For each of the
following waves, find A, ω, T and f .
(a) √
5 sin(t)
(c) 2 sin(6πt)
(b) sin(2πt)
(d) 3 sin(3t)
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Level 1 Engineering Mathematics
Extra Questions
This question is a little extra to be attempted for some more exercise.
X 1.
For any regular n-sided polygon it is possible to draw a circle around it
which touches each point of the polygon (this is called circumscribing).
1. Find the area of the regular 3-sided regular polygon (equilateral triangle) that can be drawn within a circle of radius 10;
2. Find the perimeter of the regular 6-sided regular polygon (hexagon)
that can be drawn within a circle of radius 10;
3. Find a formula for the area of an n-sided regular polygon that can be
drawn inside a circle of radius r;
4. Find a formula for the perimeter of an n-sided regular polygon that
can be drawn inside a circle of radius r;
5. Show that the formulae produced above can be used to find an approximate value for π.
Figure 1: A regular hexagon inscribed in a circle.
2
Sketch Solutions
Approach your tutor or lecturer if you do not understand any part of the
solutions for further explanation.
1. In each case, we use the principle of performing the same operation on
both sides to manipulate the expression to the form we want. You may wish
to include more steps then we have here.
(a)
⇒ 2x2 + 8x + 6 = (x + 3)2 ⇒ 2x2 + 8x + 6 = x2 + 6x + 9
⇒ x2 + 2x − 3 = 0 ⇒ (x − 1)(x + 3) = 0 ⇒ x = 1, x = −3
(b)
⇒ 3 = (x + 1)(x − 1) ⇒ 3 = x2 − 1 ⇒ x2 = 4 ⇒ x = ±2
(c)
⇒
√
x−1=3⇒x−1=3⇒x=4
(d) either take a common denominator
⇒
2−x
= x ⇒ 2 − x = 2x2
2x
or multiply by 2x both sides (or by 2, then x)
⇒
2
− 1 = 2x ⇒ 2 − x = 2x2
x
(e) Similarly, we take a common denominator for illustration
⇒
√
(x + 1) − (x − 1)
2
=1⇒ 2
= 1 ⇒ x2 − 1 = 2 ⇒ x2 = 3 ⇒ x = ± 3
(x − 1)(x + 1)
x −1
(f) This time we’ll multiply by (x + 1) and then (x − 1), you could do it
together of course.
⇒2=
3(x + 1)
⇒ 2(2x − 1) = 3(x + 1) ⇒ 4x − 4 = 3x + 3 ⇒ x = 7
2x − 1
2. Some diagrams will be helpful here. In each case, you can form a right
angled triangle by dropping a vertical down from the end of the line onto a
horizontal line from the start of the line (except case (c)). You will notice
3
that in each case the vertical line will be the opposite to the angle given, and
the horizontal line will be the adjacent. The size given pertains to the last
side, which is always opposite the right angle and so this is the hypotenuse.
This means that in each case, where the size given is H and the angle is
θ then the horizontal side x will satisfy
cos θ =
x
⇒ x = H cos(θ).
H
Similarly, the vertical size y will be given by
sin(θ) =
y
⇒ y = H sin(θ).
H
Therefore, in each case :
(a)
x = 10 cos 45 = 7.071, y = 10 sin 45 = 7.071
(b)
√
√
3
= 3 3 ≈ 5.196
2
1
y = 6 sin 30 = 6 × = 3
2
(c) You might think that the formulae won’t work here, and certainly they
are a little pointless. We can see that x = 0 and y = 15 from a quick sketch,
but you can check the formulae will agree with this.
(d)
1
x = 20 cos 60 = 20 × = 10
2
√
√
3
y = 20 sin 60 = 20 ×
= 10 3 = 17.321
2
x = 6 cos 30 = 6 ×
3. Drawing diagrams for all questions is highly recommended, sketch diagrams only.
(a) Using the cosine rule we obtain
b2 = a2 + c2 − 2ac cos B ⇒ b2 = 2473.053 ⇒ b = 49.730
Now using the sine rule
sin A
sin B
sin A
sin 66
50 sin 66
=
⇒
=
⇒ sin A =
= 0.9185
a
b
50
49.730
49.730
4
which yields a solution of A = 66.708◦ which is not ambiguous - why? Similarly
sin C
sin B
sin C
sin66
40 sin 66
=
⇒
=
⇒ sin C =
= 0.7348
c
b
40
49.730
49.730
so that C = 47.291◦ .
(b) Here we have only sides, no angles, only the cosine rule can help
cos A =
b 2 + c 2 − a2
92 + 52 − 102
=
= 0.0667
2bc
2(9)(5)
so A = 86.177◦ . Similarly
b2 = a2 + c2 − 2ac cos B ⇒ cos B =
and so
cos B =
a2 + c 2 − b 2
2ac
102 + 52 − 92
⇒ B = 63.896◦
2(10)(5)
and so C = 29.927◦ (recall that A + B + C = 180◦ .
(c) We attempt the sine rule
sin A
sin B
sin A
sin 63
14.2 sin 63
=
⇒
=
⇒ sin A =
= 1.253
a
b
14.2
10.1
10.1
It isn’t possible to solve this, and we conclude that the figures given are
impossible. There cannot be any such triangle.
(d) We start with the sine rule
sin A
sin B
sin A
sin 27
7.9 sin 27
=
⇒
=
⇒ sin A =
= 0.320
a
b
7.9
11.2
11.2
so that A = 18.676◦ which is unambigous, why? We can now show that
C = 180 − A − B = 134.324◦ . We only have to find c now, either by the sine
rule or the cosine rule. We shall use the latter
c2 = a2 + b2 − 2ab cos C = 7.92 + 11.22 − 2(7.9)(11.2) cos 134.324 = 311.495
so that c = 17.649.
4. (a) Let s = sin θ ⇒ s2 = sin2 θ:
1
1
8s2 − 2s − 1 = 0 ⇒ (4s + 1)(2s − 1) = 0 ⇒ s = , s = −
2
4
5
Solve each of these in turn
sin θ =
1
⇒ θ = 30◦
2
using our CAST diagram finds another solution at θ = 150◦ . Also
1
sin θ = − ⇒ θ = −14.477◦
4
which is not in our desired range, the same angle in our range is
θ = −14.477 + 360 = 345.523◦
and our CAST diagram yields another solution at θ = 194.477◦ . Thus
θ = 30◦ , 150◦ , 194.477◦ , 345.523◦
(b) Divide by cos θ throughout
⇒ 2 tan θ tan θ − 3 tan θ − 2 = 0
Let t = tan θ ⇒ t2 = tan2 θ
1
2t2 − 3t − 2 = 0 ⇒ (2t + 1)(t − 2) = 0 ⇒, t = − , t = 2
2
Solve each of these in turn
1
tan θ = − ⇒ θ = −26.565◦
2
which is 334.435◦ is our desired range, and our CAST diagram finds another
solution θ = 153.435◦ . Now
tan θ = 2 ⇒ θ = 63.435◦
and our CAST diagram finds another solution at θ = −116.565◦ , which is
243.435◦ in our range. Thus
θ = 63.435◦ , 153.435◦ , 243.435◦ , 334.435◦
5. Very similar to 4, but with a different range.
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(a) Recall that sin2 θ + cos2 θ = 1
8(1 − sin2 θ) + 2 sin θ = 7 ⇒ 8 sin2 θ − 2 sin θ − 1 = 0
which is the same as 5(a), but in a different range to solve. The same
techniques, or transforming the old solutions show that
θ = −165.523◦ , −14.477◦ , 30◦ , 150◦
(b) Our equation may be written
6 cos θ −
1
=1
cos θ
and multiplying by cos θ on both sides gives
6 cos2 θ − 1 = cos θ ⇒ 6 cos2 θ − cos θ − 1 = 0
Applying our trick of c = cos θ ⇒ c2 = cos2 θ we obtain
1
1
6c2 − c − 1 = 0 ⇒ (3c + 1)(2c − 1) = 0 ⇒ c = − , c =
3
2
And we solve each in turn
cos θ =
1
⇒ θ = 60◦
2
and our CAST diagram also gives us θ = −60◦ . Also
1
cos θ = − ⇒ θ = 109.471◦
3
and our CAST diagram shows that θ = −109.471◦ . Thus
θ = −109.471◦ , −60◦ , 60◦ , 109.471◦
6. Clearly we can read A and ω directly, and then we may rearrange our
equation for f to obtain
ω
f=
2π
and thus obtain T which is given by
T =
1
f
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(a)
A = 5, ω = 1, f =
1
= 0.159Hz, T = 6.283s
2π
(b)
A = 1, ω = 2π = 6.283, f = 1Hz, T = 1s
(c)
A=
√
2, ω = 6π = 18.850, f = 3, T =
1
= 0.333s
3
(d)
3
2π
= 0.477Hz, T =
= 2.094s
2π
3
All answers are given accurate to three decimal places.
X.1. This question is just for some fun, but the solutions are included anyway.
A = 3, ω = 3, f =
1. Consider figure 2 that shows an
Figure 2: An equilateral triangle inscribed in a circle.
equilateral triangle inscribed in a circle. It is possible to split the
triangle into three triangles, look at 4AOB, 4BOC and 4COA. Each
of this triangles has the same dimensions, and further, we can split them
down the middle to produce two right angled triangles. For example,
we split 4COA into 4AOM and 4M OC.
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Consider 4M OC. Now we already know some of the angles in this
triangle. For example, clearly 6 OM C = 90◦ , and as there are 360◦
in the entire circle, there are 360
= 120◦ in the angle 6 AOC and thus
3
6 M OC = 60◦ .
Note also that the side OC is a radius and thus is of length 10. So
using right angled trigonometry we can obtain
MC
MC
=
OC
10
⇒ M C = 10 sin 60.
sin 60 =
Similarly
MO
MO
=
OC
10
⇒ M O = 10 cos 60.
cos 60 =
Finally note that the area of triangle 4OCM is given by
1
M O × M C = 50 sin 60 cos 60
2
and that the total triangle is made up of 6 such triangles, so the area
is
300 sin 60 cos 60 = 129.904
2. Consider figure 3
that shows a regular hexagon inscribed in a circle. Once more we can
split the polygon into triangles surrounding the centre of the circle O
but this time there will 6 such triangles. Triangle 4AOB is one such
triangle and once again we note that the angle 6 AOB will be found by
dividing 360 by the number of sides, giving 60◦ in this case. Therefore,
when we split this triangle into two we obtain two right-angled triangles.
Consider 4AOM , now 6 AOM = 30◦ and 6 OM A = 90◦ and also AO
is a radius and hence of length 10. We note that AM is a portion of
the required perimeter and that
AM
AM
=
OA
10
⇒ AM = 10 sin 30 = 5
sin 30 =
but this is one twelth of the total perimeter and so the desired result
is 60.
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Figure 3: A regular hexagon inscribed in a circle.
3. Consider the same triangle in the previous example, but suppose that
it was in an n-sided polygon in a circle of radius r. Then 4OAM is
still right-angled, but now 6 AOM = 360
÷ 2 = 360
.
n
2n
Now
sin
360
AM
360
=
⇒ AM = r sin
2n
r
2n
and
360
OM
360
=
⇒ OM = r cos
2n
r
2n
so the area of triangle 4OAM is given by
cos
1
360
360
× r sin
× r cos
2
2n
2n
1
360
360
= r2 sin
cos
2
2n
2n
and we could use the double angle identity for sin 2θ to make this even
tidier in fact (if you saw that!).
1
360
= r2 sin
.
4
n
Now there are 2n of these triangles in the polygon so
1
360
A = nr2 sin
2
n
10
4. Looking at the reasoning above we see that the perimeter will be 2n
multiplied by the length AM and so
P = 2n × r sin
360
2n
⇒ P = 2nr sin
360
2n
5. Consider that as we add a lot of sides to the polygon, it begins to
closely resemble the circle it is drawn in. Therefore the perimeter will
be very close to the circumference of the circle.
For large n
360
2n
360
⇒ r ≈ n sin
.
2n
2πr ≈ 2nr sin
Some values of n and the approximation to π that they yield are show
in table 1.
n
π (approx)
3
2.598076211
10
3.090169944
100
3.141075908
1000 3.141587486
Table 1: Approximations to π
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