Centre for Education
in Mathematics and Computing
Euclid eWorkshop # 4
Solutions
c
2014
UNIVERSITY OF WATERLOO
Euclid eWorkshop #4
S OLUTIONS
S OLUTIONS
1.
1
(a) Here sin 2θ = − . Thus 2θ = 210◦ and θ = 105◦ .
2
(b)
cos2 (θ) = 1 − sin2 (θ)
2(2 sin2 (θ) − 1) = 8 sin θ − 5
4 sin2 (θ) − 8 sin θ + 3 = 0
(2 sin θ − 1)(2 sin θ − 3) = 0
1 3
sin θ = , but| sin θ| ≤ 1
2 2
π 5π
So θ = ,
6 6
2. Let θ = ∠AM C. Using the cosine law in 4ABM gives
49 = 9 + 25 − 30 cos(180◦ − θ)
15 = −30 cos(180◦ − θ)
1
cos(180◦ − θ) = −
2
cos(θ) = − cos(180◦ − θ)
1
=
2
Using the cosine law in 4AM C gives
AC 2 = 9 + 36 − 36 cos(θ)
= 27
√
AC = 3 3.
3. Use the sine law:
BN
100
MN
sin 108◦
=
. But
= tan 32◦ . So M N = 100 ·
· tan 32◦ ≈ 141 m.
◦
◦
sin(108 )
sin(25 )
BN
sin 25◦
5π
4. Since the area of the rectangle is
, its height is 5. Since the cosine graph is symmetrical about the y-axis,
√
3√
π
π
3
10 3
P O = OQ = . But cos
=
. So k =
.
6
6
2
3
5. Since the minimum point has a y coordinate of -2, the amplitude is a = 2. Also since the minimum occurs at
3π
3π
1
π
x =
(rather than
where it is found for sin(x)), k = 2. Therefore sin(2x) = and x =
. Thus
4
2
2
12
π
D = ( , 1).
12
a
6. We let the side of the triangles opposite θ be a and side leg adjacent to θ be b. Then tan θ = , a − b = 3 and
b
1
40
40
2
4 ab = 89 − 9 = 80 and b = a . Thus a −
= 3 or a − 3a − 40 = 0 which gives a = 8 or −5. Now
2
a
C ENTRE FOR E DUCATIONS IN M ATHEMATICS C OMPUTING
2
Euclid eWorkshop #4
S OLUTIONS
since a is positive, a = 8 and b = 5 and tan θ =
8
.
5
7. Using Pythagorean theorem, we find that F A = 2, AC =
√
2 and F C = 2. The cosine law in 4F AC gives
F C 2 = F A2 + AC 2 − 2 · F A · AC · cos(∠F AC)
√
4 = 4 + 2 − 2 · 2 · 2 cos(∠F AC)
1
cos(∠F AC) = √ .
2 2
√
8. Using item 4 from the toolkit, the height of each small equilateral triangle is 23 . Let O be the vertex immediately
to the right of √
W , and consider right-angled triangle W OT . Now OT is the height of√four of the small
√ triangles,
thus OT = 2 3. Also W O = 1. By the Pythagorean theorem,√we have W T = 1 + 4 · 3 = 13. Again
13 3
using item 4 from the toolkit, we have the area of 4W AT =
.
4
W O
A
T
9. The cosine law states
a2 = 64 + b2 − 16b(cos 60◦ )
= b2 − 8b + 64
= (b − 4)2 + 48
a2 − (b − 4)2 = 48
(a + b − 4)(a − b + 4) = 48
But 48 = 24 · 2 = 12 · 4 = 8 · 6, where we have only considered the even-even factorings of 48 since the 2
brackets on the left side must have the same parity (evenness or oddness). Thus (a,b) = (13,15), (8,8), (7,5) or
(7,3).
C ENTRE FOR E DUCATIONS IN M ATHEMATICS C OMPUTING
3