Chapter 2 Trigonometry
Section 2.1 Angles in Standard Position
Section 2.1
Page 83
Question 1
a) No; angle θ is not in standard position because its vertex is not at the origin.
b) Yes; angle θ is in standard position because its initial arm is on the positive x-axis and
the vertex is at the origin.
c) No; angle θ is not in standard position because its initial arm is not on the positive
x-axis.
d) Yes; angle θ is in standard position because its initial arm is on the positive x-axis and
the vertex is at the origin.
Section 2.1
Page 83
Question 2
a) Diagram F shows 150°.
b) Diagram C shows 180°.
c) Diagram A shows 45°.
d) Diagram D shows 320°.
e) Diagram B shows 215°.
f) Diagram E shows 270°.
Section 2.1
Page 83
Question 3
a) 48° is is quadrant I.
b) 300° is in quadrant IV.
c) 185° is in quadrant III.
d) 75° is in quadrant I.
e) 220° is in quadrant III.
f) 160° is in quadrant II.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 1 of 96
Section 2.1
Page 83
Question 4
a)
b)
c)
d)
Section 2.1
Page 83
Question 5
a) 180° − 170° = 10°. The reference angle for 170° is 10°.
b) 360° − 345° = 15°. The reference angle for 345° is 15°.
The Essay on Production Costs Emc Standards Electromagnetic
Purpose The purpose of this article is to discuss the EMC problem and to briefly illustrate its relevance in the development, manufacturing and sale of commercial escalator components in Australia. The intent audiences of this report are engineers and managers from WM, an escalator development company planning on entering the Australian market. The Problem Electromagnetic interference or EMI ...
c) The reference angle for 72° is 72°.
d) 215° – 180° = 35°. The reference angle for 215° is 35°.
Section 2.1
Page 83
Question 6
a) 180° − 45° = 135°, 180° + 45° = 225°, 360° − 45° = 315°
The three other angles in standard position, 0°
45° are 135°, 225°, and 315°.
b) 180° − 60° = 120°, 180° + 60° = 240°, 360° − 60° = 300°
The three other angles in standard position, 0°
60° are 120°, 240°, and 300°.
c) 180° − 30° = 150°, 180° + 30° = 210°, 360° − 30° = 330°
The three other angles in standard position, 0°
30° are 150°, 210°, and 330°.
d) 180° − 75° = 105°, 180° + 75° = 255°, 360° − 75° = 285°
The three other angles in standard position, 0°
75° are 105°, 255°, and 285°.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 2 of 96
Section 2.1
Page 83
Question 7
Reference
Angle
Quadrant
72°
IV
56°
II
18°
III
35°
IV
a)
b)
c)
d)
Section 2.1
Page 83
Angle in
Standard Position
360° − 72° = 288°
180° − 56° = 124°
180° + 18° = 198°
360° − 35° = 325°
Question 8
To complete the table, refer to the special triangles shown.
θ
sin θ
1
2
cos θ
tan θ
3
2
1
3
or
3
3
45°
1
2
or
2
2
60°
3
2
1
2
or
2
2
1
2
30°
Section 2.1
Page 84
1
3
Question 9
180° − 20.4° = 159.6°
The angle measured in standard position is 159.6°.
Section 2.1
Page 84
Question 10
a) The coordinates of the other three trees are found using symmetries of the diagram:
flowering dogwood (−3.5, 2), river birch (−3.5, −2), white pine (3.5, −2).
b) For the red maple,
2
tan θ =
3.5
⎛2⎞
θ = tan −1 ⎜
⎟
The Term Paper on Ref Angle 960 Degrees 1256
Table of Contents 1. Angles pg. 2 and 32. DMS and DD pg. 43. Reference Angles pg. 54. Sine, Cosine, Tangent pg. 6 and 75. Reciprocal Functions pg. 86. The Unit Circle and Quadrant Angles pg. 97. Unit Circle pg. 108. Using TI-82 and 83's to find Trig. Function Values pg. 119. Solving Right Triangle's pg. 1210. Law of Sines and Law of Cosines and The Ambiguous Case pg. 1311. Trigonometric Ratios pg. ...
⎝ 3.5 ⎠
θ = 29.744…
The angle in standard position for the red maple is 30°, to the nearest degree.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 3 of 96
Then, the angle in standard position for the flowering dogwood is 180° − 30° or 150°, to
the nearest degree. The angle in standard position for the river birch is 180° + 30° or
210°, to the nearest degree. The angle in standard position for the white pine is 360° −
30° or 330°, to the nearest degree.
c) On the grid, there are 4 vertical units of distance between the red maple and the white
pine. Since each grid mark represents 10 m, the distance between these two trees is 40 m.
Section 2.1
Page 84
Question 11
x
50
3x
=
2 50
x = 25 3
cos 30° =
x
150°
50 cm
30°
By symmetry, the horizontal distance that the tip of the wiper travels in one swipe will be
2x, or 50 3 cm.
Section 2.1
Page 84
Question 12
a) Using the symmetries of the diagram, the coordinates are A′(x, −y), A″(−x, y) and
A′″(−x, −y).
b) A′ is in quadrant IV, so ∠A′OC = 360° − θ.
A″ is in quadrant II, so ∠A″OC = 180° − θ.
A′″ is in quadrant III, so ∠A′″OC = 180° + θ.
Section 2.1
Page 84
Question 13
v2
10
1 v2
=
2 10
v2 = 5
v1
10
3 v1
=
2 10
v1 = 5 3
sin 30° =
sin 60° =
Then, v1 − v2 = 5 3 − 5 .
(
)
The exact vertical displacement of the boom is 5 3 − 5 m.
Section 2.1
Page 85
Question 14
The 72° angle is in quadrant III, so in standard position the angle is 180° + 72 ° or 252°.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 4 of 96
Section 2.1
Page 85
Question 15
An angle of 110° will be in quadrant II. Using a protractor to determine where this angle
falls on the diagram, it is found that the terminal arm passes through Cu, Ag, and Au.
These elements are copper, silver, and gold, respectively.
Section 2.1
Page 85
The Term Paper on English Sample Question Paper
Question Paper Design SA 2 English Communicative Classes IX & X Code No. 101 The design of the question papers in English Communicative for classes IX & X has undergone a few changes. They are as under: Section A –Reading: 20 marks (Question 1-4) In the existing scheme of the question paper Students answer questions based on four unseen passages carrying five marks each –all the ...
Question 16
a) The terminal arm of the blue angle falls at the end of the 12th day of 20 on the second
⎛ 12 ⎞
ring. Since there are 360° in the circle, the blue angle measures ⎜ ⎟ 360° or 216°.
⎝ 20 ⎠
b) If the angle was in quadrant II, it would be the 8th day of 20. The angle would
⎛8⎞
measure ⎜ ⎟ 360° or 144°.
⎝ 20 ⎠
You can check this using reference angles. For 216°, the reference angle is 216° − 180°
or 36°. So, the same reference angle in quadrant II is 180° – 36° or 144°.
c) In quadrant IV, this reference angle would give and angle of 360° − 36° or 324°. This
represents 2 short of 20 days, or 18 days.
Section 2.1
Page 85
Question 17
a) N20°E is equivalent to 70° in
standard position.
N
N20°E
20°
70°
W
E
S
b) S50°W is equivalent to
180° + 40° or 220° in standard
position.
N
W
S50°W
220°
E
50°
S
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 5 of 96
c) N80°W is equivalent to 90° + 80°
or 170° in standard position.
N
170°
N80°W
80°
W
E
S
d) S15°E is equivalent to 270° + 15°
or 285° in standard position.
N
W
E
285°
S15°E
S
Section 2.1
Page 86
15°
Question 18
harm
, where harm is the height of the fingertips above the centre of rotation.
45
harm = 45sin θ
Then, the height of the arm above the table is given by h = 12 + 45 sin θ.
a) sin θ =
θ
h = 12 + 45 sin θ
(in centimetres)
0°
15°
30°
45°
60°
75°
90°
12.0
23.6
34.5
43.8
51.0
55.5
57.0
b) From the table, it is clear that the increase in h is not constant. For example, from 0°
to 15° the height increases 11.6 cm, while from 30° to 45° the height increases 9.3 cm.
the rate of increase lessens as the measure of θ approaches 90°.
c) If θ were extended 90°, my conjecture is that the height would decrease, reaching 12
The Essay on Flight And Spaceflight Questions From Section
16351 FLIGHT AND SPACEFLIGHT 2 January 22 nd 2001 10 - 12 am Attempt 10 questions from section A and 2 questions from section B Section A (4 marks per question) (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) Sketch the layouts of the following types of wind tunnel indicating the working section, settling chamber and fan. Closed return Open return List two advantages and two disadvantages for ...
cm when θ = 180°, and that the height would decrease slowly for angles just past 90° but
the rate of decrease would be greater towards the horizontal position.
Section 2.1
Page 86
Question 19
Let x and y represent the two angles. Supplementary angles have a sum of 180°, so
x + y = 180°.
If the terminal arms of the two angles are perpendicular, then one angle is 90° more than
the other, or y = x + 90°.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 6 of 96
Substituting, x + x + 90° = 180°
Then, x = 45°.
The two angles must be 45° and 135°.
Section 2.1
Page 86
Question 20
a) Let ys represent the height of the seat above the centre of rotation.
y
sin 72° = s
9
ys = 9sin 72°
ys = 8.559…
So, the height of Carl’s seat above the ground is 11 + 8.559… or approximately 19.56 m.
1
⎛4⎞
b) i) If the speed is 4 rev/min, then in 5 s Carl has moved ⎜ ⎟ 5 or of a revolution.
3
⎝ 60 ⎠
So , the second stop is 120° past the 72° position. The angle of the seat that Carl is on, in
standard position, is 72° + 120° or 192°.
ii) The second stop is 12° below the horizontal. So, in this position
y
sin12° = s
9
ys = 9sin12°
ys = 1.871…
In this case, Carls seat is 11 − 1.871… or approximately 9.13 m above the ground.
Section 2.1
Page 86
Question 21
CD
OC
CD
sin θ =
, since the radius is 1.
1
So, option B is correct.
a) sin θ =
BA
OA
BA
sin θ =
, since the radius is 1.
1
So, option D is correct.
b) tan θ =
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 7 of 96
Section 2.1
Page 86
Question 22
P(x, y)
Using the Pythagorean Theorem,
x2 + y 2 = r 2
r
y
θ
x
Section 2.1
a)
Page 86
θ
sin θ
sin (180° − θ)
The Term Paper on History Free Responce Questions
AP US HISTORY FREE RESPONSE QUESTIONS SINCE 1971 I. Colonial Time 1607 – 1775 1. Puritanism bore within itself the seeds of its own destruction. Apply this generalization. (74) 2. In the seventeenth century, New England Puritans tried to create a model society. What were their aspirations, and to what extent were those aspirations fulfilled during the seventeenth century? (83) 3. Between ...
sin (180° + θ)
sin (360° − θ)
Question 23
20°
0.3420
0.3420
−0.3420
−0.3420
40°
0.6428
0.6428
−0.6428
−0.6428
60°
0.8660
0.8660
−0.8660
−0.8660
80°
0.9848
0.9848
−0.9848
−0.9848
b) My conjecture is that sin θ and sin (180° − θ) have the same value. Also,
sin (180° + θ) and sin (360° − θ) have the opposite value to sin θ (i.e., same numeric
value but negative).
c) Similar results will hold true for values of cosine and tangent, but they will be
negative in different quadrants. Refer to the diagram in the previous question and think of
a point on the terminal arm in each quadrant. Cosine will be negative in quadrants II and
III, because cosine involves the adjacent side which is negative in those quadrants.
Tangent will be negative in quadrants II and IV, because either the adjacent side or the
opposite side is negative in those quadrants.
Section 2.1
Page 86
Question 24
a) Substitute V = 110 and θ = 30° into the formula d =
V 2 cos θ sin θ
.
16
1102 cos 30° sin 30°
d=
16
3025 ⎛ 3 ⎞ ⎛ 1 ⎞
d=
⎜
⎟⎜ ⎟
4 ⎜ 2 ⎟⎝ 2 ⎠
⎝
⎠
d=
3025 3
16
The exact distance that Daria hit the ball with this driver was
MHR • Pre-Calculus 11 Solutions Chapter 2
3025 3
ft.
16
Page 8 of 96
b) To get a longer hit Daria should increase the angle of hit. An angle of 45° gives the
greatest distance in the formula for d.
c) An angle of elevation of 45° will probably produce the hit that travels the greatest
distance. For this angle sine and cosine are both the same, so their effect on the formula
will be balanced.
2.2 Trigonometric Ratios of Any Angle
Section 2.2
Page 96
Question 1
a)
b)
c)
d)
Section 2.2
a) sin 60° =
Page 96
Question 2
3
1
, cos 60° = , tan 60° = 3
2
The Essay on Discussion Questions 4
2. How do data from qualitative research differ from data in quantitative research? A quantitative research question is an interrogative sentence that asks a question about the relation that exists between two or more variables. Its purpose is to identify the variables being investigated and to specify the type of relationship, descriptive, predictive, or causal, being investigated. A qualitative ...
2
b) sin 225° = −
1
2
1
2
or −
, cos 225° = −
or −
, tan 60° = 1
2
2
2
2
1
3
1
3
c) sin150° = , cos150° = −
, tan150° = −
or −
2
2
3
3
d) sin 90° = 1, cos 90° = 0, tan 90° is undefined
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 9 of 96
Section 2.2
Page 96
Question 3
a) Use the Pythagorean Theorem to determine the hypotenuse: r = 5.
y
x
y
sin θ = , cos θ = , tan θ =
r
r
x
4
3
4
sin θ = , cos θ = , tan θ =
5
5
3
b) r 2 = x 2 + y 2
r2 = (−12)2 + (−5)2
r2 = 144 + 25
r2 = 169
r = 13
y
x
y
sin θ = , cos θ = ,
tan θ =
r
r
x
5
−5
−12
−5
sin θ =
, cos θ =
, tan θ =
=
13
13
−12 12
c ) r2 = x2 + y2
r2 = (8)2 + (–15)2
r2 = 64 + 225
r2 = 289
r = 17
y
x
y
sin θ = , cos θ = , tan θ =
r
r
x
8
−15
−15
sin θ =
, cos θ = , tan θ =
17
17
8
d) r2 = x2 + y2
r2 = (1)2 + (−1)2
r2 = 2
r= 2
y
sin θ = ,
r
sin θ =
x
cos θ = ,
r
tan θ =
y
x
−1
2
2
−1
1
or −
, cos θ =
or
, tan θ =
= −1
2
2
1
2
2
Section 2.2
Page 96
Question 4
a) The cosine ratio is negative and the sine ratio is positive in quadrant II.
b) The cosine ratio and the tangent ratio are both positive in quadrant I.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 10 of 96
c) The sine ratio and the cosine ratio are both negative in quadrant III.
d) The tangent ratio is negative and the cosine ratio is positive in quadrant IV.
Section 2.2
Page 96
Question 5
a) First calculate r.
r2 = x2 + y2
r2 = (−5)2 + (12)2
r2 = 25 + 144
r2 = 169
r = 13
y
x
sin θ =
cos θ =
r
r
12
5
−5
sin θ =
cos θ =
or −
13
13
13
b) r2 = x2 + y2
r2 = (5)2 + (−3)2
r 2 = 25 + 9
r 2 = 34
r = 34
y
sin θ =
r
−3
3 34
sin θ =
or −
34
34
c ) r2 = x2 + y2
r2 = (6)2 + (3)2
r 2 = 36 + 9
r 2 = 45
r = 45
y
sin θ =
r
3
1
5
sin θ =
=
or
5
45
5
cos θ =
y
x
12
12
tan θ =
or −
−5
5
tan θ =
x
r
5
5 34
or
34
34
cos θ =
cos θ =
cos θ =
y
x
3
−3
tan θ =
or −
5
5
tan θ =
x
r
y
x
31
tan θ = =
62
tan θ =
6
2
25
=
or
5
45
5
d) r2 = x2 + y2
r2 = (−24)2 + (−10)2
r2 = 576 + 100
r2 = 676
r = 26
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 11 of 96
y
r
−10
5
sin θ =
=−
26
13
sin θ =
Section 2.2
x
r
−24
12
cos θ =
=−
26
13
cos θ =
Page 96
y
x
10 5
tan θ =
=
24 12
tan θ =
Question 6
a) The angle is in quadrant II, so sin 155° is positive.
b) The angle is in quadrant IV, so cos 320° is positive.
c) The angle is in quadrant II, so tan 120° is negative.
d) The angle is in quadrant III, so cos 220° is negative.
Section 2.2
Page 96
a) Given sin θ =
r2 = x2 + y2
132 = x2 + 52
169 = x2 + 25
x2 = 144
Question 7
5
, y = 5 and r = 13. Use the Pythagorean Theorem to determine x.
13
x = ±12
b) Determine the reference angle.
5
sin θ =
13
⎛5⎞
θ = sin −1 ⎜ ⎟
⎝ 13 ⎠
θ = 22.619…
Then, to the nearest degree, in quadrant I, θ = 23° and in quadrant II, θ = 180° − 23° or
157°.
Section 2.2
Page 96
Question 8
x
2
= − , so x = −2 and r = 3 for an angle in quadrant II.
3
r
Use the Pythagorean Theorem to determine y.
r2 = x2 + y2
32 = (−2)2 + y2
9 = 4 + y2
y= 5
a) cos θ =
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 12 of 96
Then, sin θ =
sin θ =
y
r
5
3
and
tan θ =
y
x
tan θ =
5
5
=−
−2
2
y3
= , so y = 3 and r = 5 for an angle in quadrant I.
r5
Use the Pythagorean Theorem to determine x.
r2 = x2 + y2
52 = x2 + 32
25 = x2 + 9
x=4
x
y
Then, cos θ =
and
tan θ =
r
x
4
3
cos θ =
tan θ =
5
4
b) sin θ =
y
4
= − , so x = 5 and y = −4 for an angle in quadrant IV.
5
x
Use the Pythagorean Theorem to determine r.
r2 = x2 + y2
r2 = (5)2 + (−4)2
r2 = 25 + 16
r = 41
y
x
Then, sin θ =
and
cos θ =
r
r
−4
4 41
5
5 41
sin θ =
or −
cos θ =
or
41
41
41
41
c) tan θ =
y
1
= − , so y = −1 and r = 3 for an angle in quadrant III.
r
3
Use the Pythagorean Theorem to determine x.
r 2 = x2 + y2
32 = x2 + (−1)2
9 = x2 + 1
x=− 8
x
y
and
Then, cos θ =
tan θ =
r
x
−8
22
−1
1
2
cos θ =
or −
tan θ =
or
or
3
3
4
−8
22
d) sin θ =
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 13 of 96
y −1
=
, so x = −1 and y = −1 for an angle in quadrant III.
x −1
Use the Pythagorean Theorem to determine r.
r 2 = x2 + y2
r2 = (–1)2 + (–1)2
r= 2
y
x
and
cos θ =
Then, sin θ =
r
r
−1
2
−1
2
sin θ =
or −
cos θ =
or −
2
2
2
2
e) tan θ =
Section 2.2
Page 97
Question 9
a) The diagram shows the two possible positions of θ, 0° ≤ θ
1
cos θ = .
2
1
cos θ = is part of the 30°-60°-90° right
2
2
y
triangle with sides 1, 2, and 3 .
θ
The reference angle for θ is 60°.
In quadrant I, θ = 60°.
1
In quadrant IV, θ = 360° − 60° or 300°.
b) The diagram shows the two possible positions of θ, 0° ≤ θ
1
cos θ = −
.
2
1
, refer to the 45°-45°-90°
2
right triangle with sides 1, 1, and 2 .
The reference angle for θ is 45°.
In quadrant II, θ = 180° − 45° or 135°.
In quadrant III, θ = 180° + 45° or 225°.
For cos θ = −
2
y
θ
−1
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 14 of 96
c) The diagram shows the two possible positions of θ, 0° ≤ θ
1
tan θ = −
.
3
1
For tan θ = −
, refer to the 30°-60°-90°
3
1
3
right triangle with sides 1, 2, and 3 .
θ
The reference angle for θ is 30°.
−3
−1
In quadrant II, θ = 180° − 30° or 150°.
In quadrant IV, θ = 360° − 30° or 330°.
d) The diagram shows the two possible positions of θ, 0° ≤ θ
3
sin θ = −
.
2
3
For sin θ = −
, refer to the 30°-60°-90°
2
right triangle with sides 1, 2, and 3 .
The reference angle for θ is 60°.
θ
In quadrant III, θ = 180° + 60° or 240°. In
−3
−3
quadrant IV, θ = 360° − 60° or 300°.
2
e) The diagram shows the two possible positions of θ, 0° ≤ θ
tan θ = 3 .
For tan θ =
3
−1
θ
1
−3
3 , refer to the 30°-60°-90°
right triangle with sides 1, 2, and 3 .
The reference angle for θ is 60°.
In quadrant I, θ = 60°.
In quadrant IV, θ = 180° + 60° or 240°.
f) The diagram shows the two possible positions of θ, 0° ≤ θ
tan θ = −1.
For tan θ = −1, refer to the 45°-45°-90°
right triangle with sides 1, 1, and 2 .
1
The reference angle for θ is 45°.
−1
In quadrant II, θ = 180° − 45° or 135°.
1
In quadrant IV, θ = 360° − 45° or 315°.
−1
2
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 15 of 96
Section 2.2
θ
0°
90°
180°
270°
360°
Section 2.2
Page 97
sin θ
0
1
0
−1
0
Page 97
Question 10
cos θ
1
0
−1
0
1
tan θ
0
undefined
0
undefined
0
Question 11
a) Given P(−8, 6), use x = −8, y = 6, and the Pythagorean Theorem to determine r.
r2 = x2 + y2
r2 = (−8)2 + 62
r2 = 64 + 36
r2 = 100
r = 10
y
x
y
sin θ =
cos θ =
tan θ =
r
r
x
3
4
6
−8
6
3
sin θ =
or
cos θ =
or −
tan θ =
or −
10
5
10
5
−8
4
b) Given P(5, −12), use x = 5, y = −12, and the Pythagorean Theorem to determine r.
r 2 = x2 + y2
r2 = 52 + (−12)2
r2 = 25 + 144
r2 = 169
r = 13
y
x
y
sin θ =
cos θ =
tan θ =
r
r
x
−12
5
−12
sin θ =
cos θ =
tan θ =
13
13
5
Section 2.2
Page 97
Question 12
a)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 16 of 96
b) tan θ R =
4
9
c) θ = 180° − 24°
θ = 156°, to the nearest degree
⎛4⎞
θ R = tan −1 ⎜ ⎟
⎝9⎠
θ R ≈ 24°
Section 2.2
Page 97
Question 13
a)
24
7
⎛ 24 ⎞
θ R = tan −1 ⎜ ⎟
⎝7⎠
θ R ≈ 74°
b) tan θ R =
c) θ = 360° − 74°
θ = 286°, to the nearest degree
Section 2.2
Page 97
Question 14
a) Given P(2, 4), then x = 2 and y = 4. Use the Pythagorean Theorem to determine r.
r2 = x2 + y2
r2 = 22 + 42
r2 = 20
r = 20
y
Then, sin θ =
r
4
4
25
sin θ =
=
or
5
20
25
b) Given Q(4, 8), then x = 4 and y = 8.
Then, r2 = x2 + y2
r2 = 42 + 82
r2 = 80
r = 80
y
sin θ =
r
8
8
25
sin θ =
=
or
5
80
45
c) Given R(8, 16), then x = 8 and y = 16.
Then, r2 = x2 + y2
r2 = 82 + 162
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 17 of 96
r2 = 320
r = 320
y
sin θ =
r
16
16
25
sin θ =
=
or
5
320
85
d) The sine ratio is the same because the points P, Q, and R are all on the terminal arm of
angle θ. The right triangles formed by x, y, and r in each case are all similar triangles so
the ratio of corresponding sides are equal.
Section 2.2
Page 97
Question 15
a) Given P(k, 24) and r = 25,
y
sin θ =
r
24
sin θ =
25
⎛ 24 ⎞
θ = sin −1 ⎜ ⎟
⎝ 25 ⎠
θ = 73.739…
The sine ratio is positive in quadrants I and II, so θ is approximately 74° or
180° − 74° = 106°.
b) Use the Pythagorean Theorem to determine x.
r2 = x2 + y2
252 = x2 + 242
x2 = 625 − 576
x2 = 49
x = ±7
Then, in quadrant I, x = 7, y = 24, and r = 25:
y
x
y
sin θ =
cos θ =
tan θ =
r
r
x
7
24
24
sin θ =
cos θ =
tan θ =
25
25
7
Then, in quadrant II, x = −7, y = 24, and r = 25:
y
x
sin θ =
cos θ =
tan θ =
r
r
7
−7
24
cos θ =
sin θ =
=−
tan θ =
25
25
25
MHR • Pre-Calculus 11 Solutions Chapter 2
y
x
24
24
=−
−7
7
Page 18 of 96
Section 2.2
Page 97
Question 16
x
1
, so given cos θ = , x = 1, and r = 5.
r
5
y
tan θ = , so given tan θ = 2 6 , y = 2 6 .
x
y
Then, sin θ =
r
26
sin θ =
5
cos θ =
Section 2.2
Page 97
Question 17
At the equator, where the angle of dip is 0°, a point on the terminal arm is (1, 0).
So, x = 1, y = 0, and r = 1.
y
x
y
Then, sin θ =
cos θ =
tan θ =
r
r
x
0
1
0
sin 0° = or 0
cos 0° = or 1
tan 0° =
=0
1
1
1
At the North and South Poles, where the angle of dip is 90°, a point on the terminal arm
is (0, 1).
So, x = 1, y = 0, and r = 1.
y
x
y
Then, sin θ =
cos θ =
tan θ =
r
r
x
1
0
1
sin 90° = or 1
cos 90° = or 0
tan 90° = or undefined
1
1
0
Section 2.2
Page 98
Question 18
a) sin 151° = sin 29° is a true statement, because in quadrant I and II the sine ratio is
positive and 29° is the reference angle for 151°.
b) cos 135° = sin 225° is a true statement, because in quadrant II the cosine ratio is
negative, in quadrant III the sine ratio is negative, and 45° is the reference angle for both
angles.
c) tan 135° = tan 225° is a false statement, because in quadrant II the tangent ratio is
negative and in quadrant III the tangent ratio is positive.
d) sin 60° = cos 330° is a true statement, because using the special 30°-60°-90° reference
3
triangle both have a value of
.
2
e) sin 270° = cos 180° is a true statement, because both have a value of −1.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 19 of 96
Section 2.2
Page 98
θ
0°
sin θ
0
1
2
45°
cos θ
1
tan θ
0
3
2
1
3
or
3
3
1
2
or
2
2
30°
1
2
or
2
2
1
2
0
1
−
2
3
2
1
3
2
60°
90°
120°
1
2
or
2
2
135°
180°
210°
−
−
270°
300°
330°
360°
Section 2.2
1
2
or −
2
2
3
2
−1
3
−
2
240°
315°
−
1
2
0
1
−
2
150°
225°
Question 19
−
1
2
or −
2
2
1
2
0
−
Page 98
3
undefined
−3
1
2
or −
2
2
−
−
−
1
3
2
−1
−1
−
3
2
1
2
or −
2
2
1
−
2
0
1
2
1
3
undefined
−3
1
2
or
2
2
−
3
2
1
1
3
or −
3
3
0
1
3
or
3
3
−1
−
1
3
or −
3
3
0
Question 20
a) The measure for ∠A is 45°, for ∠B is 135°, for ∠C is 225°, and for ∠D is 315°.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 20 of 96
b) Use the special 45°-45°-90° reference triangle with sides
1
1
,
, and 1.
2
2
⎛1 1⎞
,
Then, A is on the terminal arm of 45°: A ⎜
⎟.
⎝ 2 2⎠
⎛ 1 1⎞
B is on the terminal arm of 135°: B ⎜ −
,
⎟.
2 2⎠
⎝
1⎞
⎛1
C is on the terminal arm of 225°: C ⎜ −
,−
⎟.
2
2⎠
⎝
1⎞
⎛1
D is on the terminal arm of 315°: D ⎜
,−
⎟.
2⎠
⎝2
Section 2.2
a)
Angle
0°
15°
30°
45°
60°
75°
90°
105°
120°
135°
150°
165°
180°
Page 98
Sine
0
0.2588
0.5
0.7071
0.8660
0.9659
1
0.9659
0.8660
0.7071
0.5
0.2588
0
Question 21
Cosine
1
0.9659
0.8660
0.7071
0.5
0.2588
0
−0.2588
−0.5
−0.7071
−0.8660
−0.9659
−1
Tangent
0
0.2679
0.5774
1
1.7321
3.7321
undefined
−3.7321
−1.7321
−1
−0.5774
−0.2679
0
b) As θ increases from 0° to 90° the sine ratio increases from 0 to 1, while the cosine
ratio decreases from 1 to 0. As θ increases from 90° to 180° the sine ratio decreases from
1 to 0, while the cosine ratio decreases from 0 to −1. As θ increases from 0° to 90° the
tangent ratio increases from 0 to undefined, then from 90° to 180° the tangent ratio starts
with large negative values but increases to 0.
c) The values of sine and cosine seem to be related. For example, sin 30° = cos 60°.
However in quadrant II, the sign is different. For example, −sin 120° = cos 150°. For
0° ≤ θ ≤ 90°, cos θ = sin (90° − θ).
For 90° ≤ θ ≤ 180°, cos θ = −sin (θ – 90°).
d) In quadrant I, sine, cosine, and tangent values are all positive. In quadrant II, sine
values are positive but cosine and tangent values are negative.
e) In quadrant III, the sine and cosine values will be negative, but the tangent values will
be positive. In quadrant IV, the cosine values will be positive, but the sine and tangent
values will be negative.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 21 of 96
Section 2.2
Page 98
Question 22
a) Choose a point on the line y = 6x. When x = 1, y = 6.
Use the Pythagorean Theorem to determine r.
r2 = x2 + y2
r2 = 12 + 62
r = 37
y
x
Then, sin θ =
cos θ =
r
r
6
6 37
1
37
sin θ =
or
cos θ =
or
37
37
37
37
y
x
6
tan θ = = 6
1
tan θ =
b) 4y + 3x = 0, for x ≥ 0
3
So,
y=− x
4
Since the slope is negative, the angle is in quadrant
IV. The point (4, −3) lies on the line and r = 5.
x
y
cos θ =
tan θ =
r
x
4
3
−3
cos θ =
tan θ =
=−
5
4
4
4
−3
5
y=−
3
4
x
34
Then, tan θ + cos θ = − +
45
−15 + 16
=
20
1
=
20
Section 2.2
Page 98
Question 23
As θ increases from 0° to 90°:
• the value of x will decrease from 12 to 0
• the value of y will increase from 0 to 12
• the value of sin θ will increase from 0 to 1
• the value of cos θ will decrease from 1 to 0
• the value of tan θ will increase from 0 to undefined
Section 2.2
Since cos θ =
Page 99
Question 24
x
and cos θ = a, then x = a, r = 1, and y = 1 − a 2 .
r
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 22 of 96
tan θ =
tan θ =
y
x
1 − a2
a
Section 2.2
Page 99
Question 25
OA = OB = 1 They are both radii.
Therefore, ∆OAB is isosceles with ∠OAB = ∠ABO = 60°. So, in actual fact ∆OAB is
equilateral and AB = 1.
CB = 2
⎛1 3⎞
Since OA is the terminal arm of 60°, the coordinates of A are ⎜ ,
⎜ 2 2 ⎟ . The coordinates
⎟
⎝
⎠
of C are (−1, 0).
Then, use the Pythagorean Theorem in the right triangle with hypotenuse
AC.
2
⎛ 3 ⎞ ⎛ 3 ⎞2
AC = ⎜
⎜ 2 ⎟ +⎜ 2⎟
⎟ ⎝⎠
⎝
⎠
39
AC2 = +
44
2
AC = 3
2
AC = 3
Then, in ∠ABC:
BC2 = 22
AC2 + AB2 = 3 + 1
2
2
BC2 = 4
AC + AB = 4
So the sides of ∆ABC satisfy the Pythagorean Theorem, with ∠CAB = 90° because BC is
the hypotenuse.
Section 2.2
Page 99
Question 26
For any angle that is not in quadrant I, the reference angle is the acute angle made
between the terminal arm and the x-axis. Use the reference angle measure to determine
the numeric value of the trigonometric ratios but then adjust the sign to negative if
necessary for the quadrant that the terminal arm is in.
Section 2.2
Page 99
Question 27
Point P(−5, −9) is in quadrant III. The reference triangle, formed with the x-axis, has
sides −5 and −9, and hypotenuse 106 . Use the reference triangle to determine the
measure of the acute angle made by the x-axis and the terminal arm. Then, add 180° to
determine the actual angle measure.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 23 of 96
Section 2.2
Page 99
Question 28
y
r
The y-coordinate for any point in quadrant I will be the same as the y-coordinate of the
reflection of that point in quadrant II. Similarly, the y-coordinates of points in quadrant
III match one other point in quadrant IV. So for any non-quadrantal angle between 0° and
360° there is always one other matching angle that has the same sine ratio.
sin θ =
Section 2.2
Page 99
Question 29
x
1
So, given cos θ = − , x = −1 and
2
r
r = 2. This means that the angle is either in
quadrant II or III.
y
3
, y = − 3 . So the
But since sin θ = = −
2
r
angle must be in quadrant III.
The reference angle is 60°, so the angle must
be 180° + 60° or 240°.
cos θ =
Section 2.2
Page 99
2
−1
θ
−3
2
Question 30
Answers will vary. Example:
Step 1
A: (4.18, 2.67)
6
4
2
-10
-5
r
A
5
10
-2
-4
-6
-8
Step 2
a) x = 4.18, y = 2.67
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 24 of 96
y
; sin A = 0.534
5
x
y
c) cos A = , cos A = 0.836; tan A = , tan A = 0.6388…
5
x
b) sin A =
Step 3 Animation will vary.
Step 4
a) The sine and cosine increase and decrease between 1 and −1. The tangent can take any
value.
b) The sine and cosine ratios are both 0.707 at 45° and at 225°.
c) The signs of the ratios change. In quadrant II, the sine ratios are positive, but the
cosine and tangent are negative. In quadrant III, the tangent ratios are positive but the
sine and cosine are negative. In quadrant IV, the cosine ratios are positive but the sine
and tangent are negative.
d) The sine ratio divided by the cosine ratio is equal to the tangent ratio. Yes, this is true
for all angles.
2.3 The Sine Law
Section 2.3
Page 108
Question 1
10
a
=
sin 35° sin 40°
10sin 35°
a=
sin 40°
a = 8.923…
The unknown side a = 8.9, to the nearest tenth.
a)
65
b
=
sin 48° sin 75°
65sin 48°
b=
sin 75°
b = 50.008…
The unknown side b = 50.0, to the nearest tenth.
b)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 25 of 96
sin θ sin 50°
=
12
65
12sin 50°
sin θ =
65
⎛ 12sin 50° ⎞
θ = sin −1 ⎜
⎟
65
⎝
⎠
θ = 8.130…
The measure of angle θ is 8°, to the nearest degree.
c)
sin A sin 62°
=
25
32
25sin 62°
sin A =
32
⎛ 25sin 62° ⎞
∠A = sin −1 ⎜
⎟
32
⎝
⎠
∠A = 43.614…
The measure of ∠A is 44°, to the nearest degree.
d)
Section 2.3
Page 108
Question 2
a) ∠C = 180° − (88° + 35°)
∠C = 57°
c
b
=
sin C sin B
44
c
=
sin 57° sin 88°
44sin 57°
c=
sin 88°
c = 36.923…
The length of AB is 36.9 mm, to the nearest tenth of a millimetre.
c
a
=
sin C sin A
45
c
=
sin 118° sin 52°
45sin118°
c=
sin 52°
c = 50.421…
The length of AB is 50.4 m, to the nearest tenth of a metre.
b)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 26 of 96
Section 2.3
Page 108
Question 3
sin C sin B
=
c
b
sin C sin 62°
=
28
31
28sin 62°
sin C =
31
⎛ 28sin 62° ⎞
∠C = sin −1 ⎜
⎟
31
⎝
⎠
∠C = 52.892…
The measure of ∠C is 53°, to the nearest degree.
a)
sin A sin B
=
a
b
sin A sin 98°
=
15
17.5
15sin 98°
sin A =
17.5
⎛ 15sin 98° ⎞
∠A = sin −1 ⎜
⎟
⎝ 17.5 ⎠
∠A = 58.081…
The measure of ∠A is 58°, to the nearest degree.
b)
Section 2.3
Page 108
Question 4
a) This is an ambiguous case.
First find the acute measure of ∠C:
sin C sin B
=
c
b
sin C sin 67°
=
13
12
13sin 67°
sin C =
12
⎛ 13sin 67° ⎞
∠C = sin −1 ⎜
⎟
⎝ 12
⎠
∠C = 85.721…
The acute measure of ∠C is 86°, to the nearest degree.
Then, the obtuse measure of ∠C is 180° − 85.721° = 94.279…°.
The obtuse measure of ∠C is 94°, to the nearest degree.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 27 of 96
Then, find the measure of ∠A.
∠A = 180° − (85.721…° + 67°) or ∠A = 180° − (94.279…° + 67°)
∠A = 27.279…°
∠A = 18.721…°
The measure of ∠A is 27° or 19°, to the nearest degree.
Now find the measure of side a.
a
b
a
b
=
or
=
sin A sin B
sin A sin B
12
12
a
a
=
=
sin 27.279…° sin 67°
sin 18.721…° sin 67°
12sin 27.279…°
12sin18.721…°
a=
a=
sin 67°
sin 67°
a = 5.974…
a = 4.184…
Summary:
Acute case: ∠C = 86°, ∠A = 27°, side a is 6.0 m, to the nearest tenth of a metre.
Obtuse case: ∠C = 94°, ∠A = 19°, side a is 4.2 m, to the nearest tenth of a metre.
b) ∠C = 180° − (42° + 84°)
∠C = 54°
Find the length of a, side BC:
a
b
=
sin A sin B
50
a
=
sin 42° sin 84°
50sin 42°
a=
sin 84°
a = 33.640…
The length of side a is 33.6 m, to the nearest tenth of a metre.
Find the length of c, side AB:
c
b
=
sin C sin B
50
c
=
sin 54° sin 84°
50sin 54°
c=
sin 84°
c = 40.673…
The length of side c is 40.7 m, to the nearest tenth of a metre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 28 of 96
c) ∠B = 180° − (22° + 39°)
∠B = 119°
Find the length of a, side BC:
a
b
=
sin A sin B
29
a
=
sin 22° sin 119°
29sin 22°
a=
sin119°
a = 12.420…
The length of side a is 12.4 mm, to the nearest tenth of a millimetre.
Find the length of c, side AB:
c
b
=
sin C sin B
29
c
=
sin 39° sin 119°
29sin 39°
c=
sin119°
c = 20.866…
The length of side c is 20.9 mm, to the nearest millimetre.
d) ∠B = 180° − (48° + 61°)
∠B = 71°
Find the length of a, side BC:
a
b
=
sin A sin B
21
a
=
sin 48° sin 71°
21sin 48°
a=
sin 71°
a = 16.505…
The length of side a is 16.5 cm, to the nearest tenth of a centimetre.
Find the length of c, side AB:
c
b
=
sin C sin B
21
c
=
sin 61° sin 71°
21sin 61°
c=
sin 71°
c = 19.425…
The length of side c is 19.4 cm, to the nearest centimetre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 29 of 96
Section 2.3
Page 108
Question 5
a)
∠C = 180° − (57° + 73°)
∠C = 50°
24
x
=
sin 73° sin 50°
24sin 73°
x=
sin 50°
x = 29.960…
The length of AC is 30.0 cm, to the
nearest tenth of a centimetre.
b)
∠A = 180° − (38° + 56°)
∠A = 86°
63
x
=
sin 56° sin 86°
63sin 56°
x=
sin 86°
x = 52.356…
The length of AB is 52.4 cm, to the
nearest tenth of a centimetre.
c)
∠C = 180° − (50° + 50°)
∠C = 80°
x
27
=
sin 80° sin 50°
27 sin 80°
x=
sin 50°
x = 34.710…
The length of AB is 34.7 m, to the
nearest tenth of a metre.
d)
15
x
=
sin 23° sin 78°
15sin 23°
x=
sin 78°
x = 5.991…
The length of BC is 6.0 cm, to the
nearest tenth of a centimetre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 30 of 96
Section 2.3
Page 108
Question 6
a) Given ∠A = 39°, a = 10 cm, and b = 14 cm:
b sin A = 14 sin 39°
b sin A = 8.810…
Then, b sin A
b) Given ∠A = 123°, a = 23 cm, and b = 12 cm:
∠A is obtuse and a > b so there is one solution.
c) Given ∠A = 145°, a = 18 cm, and b = 10 cm:
∠A is obtuse and a > b so there is one solution.
d) Given ∠A = 124°, a = 1 cm, and b = 2 cm:
∠A is obtuse and a
Section 2.3
Page 108
Question 7
h
b
h = b sin A
Then, from the diagram, a > b sin A, or
a > h and b > h.
a) sin A =
h
b
h = b sin A
Then, from the diagram, b sin A
b) sin A =
h
b
h = b sin A
Also, from the diagram, a = b sin A.
c) sin A =
h
b
h = b sin A
From the diagram, a ≥ b > b sin A.
d) sin A =
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 31 of 96
Section 2.3
Page 109
Question 8
a) The diagram shows the given information.
This is an ambiguous case, since h = 5.6 sin 31° = 2.884…
Because h is less than 5.6 and 3.9, two solutions are possible.
B
3.9 cm
5.6 cm
h
31°
A
C
A
∠B = 180° − (31° + 48°)
∠B = 101°
or
sin A sin 31°
=
5.6
3.9
5.6sin 31°
sin A =
3.9
⎛ 5.6sin 31° ⎞
∠A = sin −1 ⎜
⎟
3.9
⎝
⎠
∠A = 47.692…
Or, ∠A = 180° − 47.692…°
∠A = 132.308…°
∠B = 180° − (31° + 132°)
∠B = 17°
Determine b for each measure of ∠B:
3.9
3.9
b
b
=
=
sin101° sin 31°
sin17° sin 31°
3.9sin101°
3.9sin17°
b=
b=
sin 31°
sin 31°
b = 7.433…
b = 2.213…
In ∆ABC, b = 7.4 cm, to the nearest tenth of a centimetre, and ∠A = 48° and ∠B = 101°,
both to the nearest degree, or
b = 2.2 cm, to the nearest tenth of a centimetre, and ∠A = 132° and ∠B = 17°, both to the
nearest degree.
b)
R
15 cm
h = 20 sin 43° = 13.639…
Since this is less than 15 and less than 20,
two solutions are possible, as shown in the
diagram.
20 cm
h
43°
P
P′
Q
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 32 of 96
sin P sin 43°
=
20
15
20sin 43°
sin P =
15
⎛ 20sin 43° ⎞
∠P = sin −1 ⎜
⎟
15
⎝
⎠
∠P = 65.413…
So ∠P = 65°, or ∠P = 180° − 65° = 115°.
Then, ∠R = 180° − (65° + 43°) = 72°, or ∠R = 180° − (115° + 43°) = 22°.
Now determine r for each value of ∠R:
15
15
r
r
or
=
=
sin 72° sin 43°
sin 22° sin 43°
15sin 72°
15sin 22°
r=
r=
sin 43°
sin 43°
r = 20.917…
r = 8.239…
Therefore, in ∆PQR, ∠P = 65°, ∠R = 72°, and PQ = 20.9 cm, or
∠P = 115°, ∠R = 22°, and PQ = 8.2 cm.
c)
Z
8.5 cm
53°
X
12.3 cm
Y
h = 12.3 sin 53° = 9.823…
Since 8.5
Section 2.3
Page 109
Question 9
C
120 cm
a
26°
h
A
B
h = 120 sin 26°
h = 52.604…
a) There is one oblique triangle if a ≥ 120 cm.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 33 of 96
b) There is one right triangle if a = 52.6 cm.
c) The are two oblique triangles if 52.6 cm
d) There is no such triangle if a
Section 2.3
Page 109
Question 10
a)
b) Determine the measure of the third angle, ∠B.
∠B = 180° − (49° + 64°)
∠B = 67°
BM
500
Then,
=
sin 49° sin 67°
500sin 49°
BM =
sin 67°
BM = 409.943…
The hot air balloon is 409.9 m from Maria, to the nearest tenth of a metre.
Section 2.3
Page 109
Question 11
C
H
Let C be the position of the coast guard ship and H the foot of the perpendicular from C
to the shoreline.
sin D sin 20°
=
500
250
500sin 20°
sin D =
250
⎛ 500sin 20° ⎞
∠D = sin −1 ⎜
⎟
250
⎝
⎠
∠D = 43.160…
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 34 of 96
Then, in ∆CDH:
DH
250
DH = 250 cos 43.160…°
cos 43.160…° =
DH = 182.361…
Since CA = CD, ∆ACH ≈ ∆DCH and AH + HD.
So AD = 2(182.361…) or 364.722…
The length of shoreline that is illuminated by the spotlight is 364.7 m, to the nearest tenth
of a metre.
Section 2.3
Page 109
Question 12
Let A and B be the points where the chains
attach to the beam and C the chandelier.
sin A sin 62
=
3.6
4.8
3.6sin 62
sin A =
4.8
⎛ 3.6sin 62 ⎞
∠A = sin −1 ⎜
⎟
⎝ 4.8 ⎠
∠A = 41.468…
The second chain makes an angle of 41° with the beam, to the nearest degree.
Section 2.3
Page 109
Question 13
In ∆ABD, ∠ABD = 180° − 40° = 140°.
Then, ∠BDA = 180° − (140° + 26°) = 14°.
BD
3.9
=
sin 26° sin14°
3.9sin 26°
BD =
sin14°
BD = 7.066…
Then, in ∆BCD
CD
sin 40° =
7.066…
CD = (7.066…) sin 40°
CD = 4.542…
The height of the monument is 4.5 m, to the nearest tenth of a metre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 35 of 96
Section 2.3
Page 110
Question 14
a)
M
21°
3°
T
X
h
N
66 m
S
b) Let N be at the base of the hotel, S be the base of the statue, and T the top of the
statue.
In ∆MNS, ∠MSN = 21° + 3°, because of equal alternate angles.
MN
tan 24° =
66
MN = 66 tan 24°
MN = 29.385…
Then, in ∆MTX, ∠MTX = 21°, because of equal alternate angles.
MX
tan 21° =
66
MX = 66 tan 21°
MX = 25.335…
The height of the statue is MN − MX = 29.385 − 25.335 = 4.05, or 4.1 m to the nearest
tenth.
c) In ∆MNS,
66
cos 24° =
MS
66
MS =
cos 24°
MS = 72.245…
The line of sight distance from Max to the foot of the statue is 72.2 m, to the nearest
metre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 36 of 96
Section 2.3
Page 110
Question 15
a) Let d represent the distance between the two hydrogen atoms.
d
0.958
=
sin104.5° sin 37.75°
0.958sin104.5°
d=
sin 37.75°
d = 1.514…
The distance between the two hydrogen atoms is 1.51 Å, to the nearest hundredth.
b) Given 1 Å = 0.01 mm, the distance between two hydrogen atoms is 1.51(0.01)
or 0.0151 mm.
Section 2.3
Page 110
Question 16
T
5.1 m
5.1 m
V
W
For the greatest wingspan, ∠T = 132°.
Since the triangle is isosceles, ∠V = ∠W = 24°.
VW
5.1
=
sin132° sin 24°
5.1sin132°
VW =
sin 24°
VW = 9.318…
For the least wingspan, ∠T = 127°.
Since the triangle is isosceles, ∠V = ∠W = 26.5°.
VW
5.1
=
sin127° sin 26.5°
5.1sin127°
VW =
sin 26.5°
VW = 9.128…
Then, to the nearest tenth of a metre, the greatest wingspan is 9.3 m and the least is 9.1 m.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 37 of 96
Section 2.3
Page 110
Question 17
a)
B′
360 m
cairn
C
B
500 m
35°
A
500 sin 35° = 286.788…
Since this is less than 360 m, there are two
possible triangles.
Determine possible measures of ∠B:
sin B sin 35°
=
500
360
⎛ 500sin 35° ⎞
∠B = sin −1 ⎜
⎟
360
⎝
⎠
∠B = 52.809…
So, ∠B ≈ 52.8° or
∠B ≈ 180° − 52.8° = 127.2°
Then, ∠C ≈ 92.2° or ∠C ≈ 17.8°.
Note: for accuracy in the next step, angles
are recorded to the nearest tenth.
b)
x
360
x
360
or
=
=
sin 92.2° sin 35°
sin17.8° sin 35°
360sin 92.2°
360sin17.8°
x=
x=
sin 35°
sin 35°
x = 627.178…
x = 191.866…
The possible distances between Armand’s first and second stops are 191.9 m or 627.2 m,
to the nearest tenth of a metre.
c)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 38 of 96
Section 2.3
Page 111
Question 18
First, use the oblique triangle to determine the distance, d, from the cable operator to the
top of the tower.
d
30
=
sin 34.5° sin 0.9°
30sin 34.5°
d=
sin 0.9°
d = 1081.800…
Then, use the Pythagorean Theorem in
the smaller right triangle to determine
the height, h, of the tower.
d2 = h2 + 6272
2
⎛ 30sin 34.5° ⎞
2
h2 = ⎜
⎟ − 627
⎝ sin 0.9° ⎠
2
⎛ 30sin 34.5° ⎞
2
h= ⎜
⎟ − 627
⎝ sin 0.9° ⎠
h = 881.568….
The height of the structure is 881.6 + 30 or 911.6 m, to the nearest tenth of a metre.
Section 2.3
Page 111
Question 19
Statements
sin C =
h
b
h = b sin C
sin B =
Reasons
h
c
h = c sin B
b sin C = c sin B
sin C sin B
=
c
b
sin C ratio in ACD
sin B ratio in ABD
Solve each ratio for h.
Equivalence property or
substitution
Divide both sides by bc.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 39 of 96
Section 2.3
Page 111
Question 20
Given ∠A = ∠B, prove a = b.
Using the sine law:
a
b
=
sin A sin B
If ∠A = ∠B, then sin A = sin B.
a
b
So,
=
sin A sin A
b sin A
a=
sin A
a=b
Section 2.3
Page 111
Question 21
First, determine the height, h, of ∆ABC.
sin B sin102°
=
4.6
8.5
⎛ 4.6sin102° ⎞
∠B = sin −1 ⎜
⎟
8.5
⎝
⎠
∠B = 31.961…
Then, ∠C ≈ 180° − (102° + 32°) = 46°.
h
sin 46° =
4.6
h = 4.6sin 46°
h = 3.308…
A
102°
4.6 km
h
B
8.5 km
C
Now calculate the area.
1
A = bh
2
1
A = (8.5)(3.308…)
2
A = 14.063…
The area of the oil spill was 14.1 km2, to the nearest tenth.
Section 2.3
Page 111
Question 22
a)
h = 50 sin 40°
h = 32.139…
There are two possible solutions for ∆ABC if
32.1 cm
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 40 of 96
b)
h = 125.7 sin 56°
h = 104.210…
There are no possible solutions for ∆ABC if
a
c)
h = 73.7 sin 57°
h = 61.810…
There is only one possible solution for ∆ABC
when a = 61.8 cm.
Section 2.3
Page 111
Question 23
First, determine whether the light from
D will reach the pathway. Let h
represent the shortest distance from D to
the pathway.
h = 150 tan 21°
h = 57.579…
Since this distance is less than 60 m, the
light from D will reach the pathway.
X
60 m
21°
A
50 m
B
50 m
C
50 m
D
Check whether the light from the next streetlight will reach the pathway.
h = 200 tan 21°
h = 76.772…
Since this distance is more than 60 m, the light from the streetlight at 200 m will not
reach the pathway.
Focus on ∆ADX, where X is the point on the pathway that is 60 m from D.
sin X sin 21°
=
150
60
150sin 21°
sin X =
60
⎛ 150sin 21° ⎞
∠X = sin −1 ⎜
⎟
60
⎝
⎠
∠X = 63.626…
Then, ∠D = 180° − (21° + 63.6°) = 95.4°
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 41 of 96
AX
60
=
sin 95.4° sin 21°
60sin 95.4°
AX =
sin 21°
AX = 166.682…
The farthest point on the pathway that is lit is 166.7 m from A, to the nearest tenth of a
metre.
Section 2.3
Page 112
Question 24
a)
No side length is given, so in the sine law there will
always be two unknowns.
b)
No angle measure is given, so in the sine law there
will always be two unknowns.
c)
The side opposite the given angle cannot be
matched with another side and angle.
d)
No angle measure is given and only one side is
given, so in the sine law there will be three
unknowns.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 42 of 96
Section 2.3
Page 112
Question 25
Use the sine ratio twice:
a
b
sin A =
and
sin B =
c
c
a
b
c=
c=
sin A
sin B
So, since both equal c:
a
b
=
sin A sin B
Section 2.3
Page 112
Question 26
AB
8
=
sin 72° sin 36°
8sin 72°
AB =
sin 36°
AB = 12.944…
Using the sine law, the length of each equal side is 12.9 cm, to the nearest tenth of a
centimetre.
5 +1
AB
b)
=2
8
1
⎛ 5 +1⎞
AB = 8 ⎜
⎜2⎟
⎟
⎝
⎠
a)
AB = 4( 5 + 1)
Using the golden ratio, the exact length of each equal side is 4( 5 + 1) cm.
CD
8
=
sin 36° sin 72°
8sin 36°
CD =
sin 72°
CD = 4.944……
CD = 4.9 cm, to the nearest tenth of a centimetre.
c)
DE
4.944…
=
sin 36° sin 72°
(4.944…) sin 36°
DE =
sin 72°
DE = 3.055…
DE = 3.1 cm, to the nearest tenth of a centimetre.
d)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 43 of 96
e) The spiral is created by drawing arcs
of a circle on one of the equal sides with
centre at one of the equal angles in the
next step. For example, the first arc is
AB with centre at D. Then, the second
arc is BC with centre at E.
D
E
C
Section 2.3
Page 112
B
Question 27
Answers may vary. Example:
The sine law can be used to solve an oblique triangle in two cases:
You are given two angles (which means
you easily know three angles) and one side.
↓↓
For example, you can solve the triangle
shown. Use angle sum of a triangle to find
the measure of ∠B and then use the sine
law twice to determine lengths a and b.
C
You are given two sides and one angle that
is opposite one of the given sides.
↓↓
For example, you can solve the triangle
shown. Use the sine law to find the
measure of ∠A. Then, use angle sum of a
triangle to determine ∠B. Use the sine law
again to calculate b.
C
54°
b
80°
a
b
72°
A
3 cm
Section 2.3
B
Page 112
A
8 km
7 km
B
Question 28
Step 1
Step 2 a) BD is too short to reach AC, so a triangle cannot be formed.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 44 of 96
b) No triangle is formed when BD is less than the perpendicular distance from B to AC.
Step 3
a) Yes, a triangle is formed when the
circle just touches the line.
b) Exactly one triangle is formed when
BD is the perpendicular distance from B
to AC.
Step 4
a) Yes, when the circle cuts the line in
two places, two triangles can be formed.
b) Two triangles can be drawn when BD
is greater than the perpendicular distance
from B to AC but BD must be shorter
than BA.
Step 5
a) One triangle can be formed when the string is longer than AB but less than AC.
b) When the radius BD is greater than AB but less than AC, one triangle is formed. If the
radius is also greater than AC then no triangle is formed.
Step 6 The results will apply so long as ∠A is acute.
Section 2.4 The Cosine Law
Section 2.4
a)
Page 119
Question 1
c2 = 142 + 92 − 2(14)(9) cos 17°
c2 = 36.011…
c = 36.011…
c = 6.000…
The length of side AB is 6.0 cm to the nearest tenth of a
centimetre.
l2 = 132 + 292 − 2(13)(29) cos 41°
l2 = 440.094…
l = 440.948…
l = 20.998…
The length of MN is 21.0 mm to the nearest tenth of a
millimetre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 45 of 96
f2 = 302 + 212 − 2(30)(21) cos 123°
f2 = 36.011…
c = 2027.245…
c)
c = 45.024…
The length of DE is 45.0 m to the nearest tenth of a metre.
Section 2.4
Page 119
Question 2
102 = 172 + 112 − 2(17)(11) cos J
100 = 289 + 121 − 374 cos J
374 cos J = 289 + 121 − 100
310
cos J =
374
⎛ 310 ⎞
∠ J = cos −1 ⎜
⎟
⎝ 374 ⎠
∠ J = 34.016…
∠J = 34°, to the nearest degree.
a)
182 = 10.42 + 21.92 − 2(10.4)(21.9) cos L
324 = 108.16 + 479.61 − 455.52 cos L
455.52 cos L = 587.77 − 324
263.77
cos L =
455.52
⎛ 263.77 ⎞
∠L = cos −1 ⎜
⎟
⎝ 455.52 ⎠
∠L = 54.616…
∠L = 55°, to the nearest degree.
b)
142 = 92 + 62 − 2(9)(6) cos P
196 = 81 + 36 − 108 cos P
108 cos P = 117 − 196
79
cos P = −
108
⎛ 79 ⎞
∠P = cos −1 ⎜ −
⎟
⎝ 108 ⎠
∠P = 137.010…
∠P = 137°, to the nearest degree.
c)
312 = 202 + 132 − 2(20)(13) cos C
961 = 400 + 169 − 520 cos C
520 cos C = 569 − 961
d)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 46 of 96
392
520
⎛ 392 ⎞
∠C = cos −1 ⎜ −
⎟
⎝ 520 ⎠
∠C = 138.924…
∠C = 139°, to the nearest degree.
cos C = −
Section 2.4
Page 120
Question 3
a) Use the cosine law to solve for p.
p2 = 292 + 282 − 2(29)(28) cos 52°
p2 = 625.165…
p = 625.165…
p = 25.003…
p = 25.0 km, to the nearest tenth of a
kilometre.
Use the sine law to solve for ∠Q.
sin Q
sin 52°
=
28
25.003…
28sin 52°
sin Q =
25.003…
⎛ 28sin 52° ⎞
∠Q = sin −1 ⎜
⎟
⎝ 25.003… ⎠
∠Q = 61.939…
∠Q = 62°, to the nearest degree.
Use the angle sum of a triangle to determine ∠R.
∠R = 180° − (52° + 62°)
∠R = 66°
b) Use the cosine law to solve for ∠S.
9.12 = 6.82 + 52 − 2(6.8)(5) cos S
82.81 = 46.24 + 25 − 68 cos S
68 cos S = 71.24 − 82.81
11.57
cosS = −
68
⎛ 11.57 ⎞
∠S = cos −1 ⎜ −
⎟
⎝ 68 ⎠
∠S = 99.796…
∠S = 100°, to the nearest degree.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 47 of 96
Use the sine law to solve for ∠T.
sin T sin 99.796…°
=
6.8
9.1
6.8sin 99.796…°
sin T =
9.1
⎛ 6.8sin 99.796…° ⎞
∠T = sin −1 ⎜
⎟
9.1
⎝
⎠
∠T = 47.421…
∠T = 47°, to the nearest degree.
Use the angle sum of a triangle to determine ∠R.
∠R = 180° − (100° + 47°)
∠R = 33°
Section 2.4
Page 120
Question 4
a)
b)
BC2 = 242 + 342 − 2(24)(34) cos 67°
BC2 = 1094.326…
BC2 = 1094.326…
BC = 33.080…
BC = 33.1 cm, to the nearest tenth of a centimetre.
AC2 = 152 + 82 − 2(15)(8) cos 24°
AC2 = 69.749…
AC2 = 69.749…
AC = 8.351…
AC = 8.4 cm, to the nearest tenth of a centimetre.
c)
AB2 = 102 + 92 − 2(10)(9) cos 48°
AB2 = 60.556…
AB2 = 60.556…
AB = 7.781…
AB = 7.8 cm, to the nearest tenth of a centimetre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 48 of 96
d)
Use the sine ratio.
12
sin B =
15
⎛ 12 ⎞
∠B = sin −1 ⎜ ⎟
⎝ 15 ⎠
∠B = 53.130…
∠B = 53°, to the nearest degree.
∆ABC is a right triangle,
because 152 = 92 + 122.
e)
9.62 = 10.82 + 18.42 − 2(10.8)(18.4) cos A
92.19 = 116.64 + 338.56 − 397.44 cos A
397.44 cos A = 455.2 − 92.19
363.01
cos A =
397.44
⎛ 363.01 ⎞
∠A = cos −1 ⎜
⎟
⎝ 397.44 ⎠
∠A = 24.024…
∠A = 24°, to the nearest degree.
f)
Section 2.4
4.62 = 3.22 + 2.52 − 2(3.2)(2.5) cos C
21.16 = 10.24 + 6.25 − 16 cos C
16 cos C = 16.49 − 21.16
4.67
cos C = −
16
⎛ 4.67 ⎞
∠C = cos −1 ⎜ −
⎟
⎝ 16 ⎠
∠C = 106.970…
∠C = 107°, to the nearest degree.
Page 120
Question 5
a)
Use the cosine law to determine ∠F because three sides
are given, but no angle.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 49 of 96
b)
Since two angles and one side are given, you can use
the sine law to find p. First use the angle sum of a
triangle to determine ∠P.
c)
This case will need both, since two sides and the angle
contained between them is given. First use the cosine
law to determine AB. Then, use the sine law to
determine ∠B.
Section 2.4
Page 120
Question 6
a)
c2 = 262 + 412 −2(26)(41) cos 30°
3
c2 = 676 + 1681 − 2(26)(41)
2
2
c = 2357 − 1066 3
c = 2357 − 1066 3 cm
c2 = 62 + 102 − 2(6)(10) cos 45°
⎛ 2⎞
c 2 = 36 + 100 − 120 ⎜
⎜2⎟
⎟
⎝
⎠
c 2 = 136 − 60 2
c 2 = 4(34 − 15 2)
c = 4(34 − 15 2)
c = 2 34 − 15 2 m
Section 2.4
Page 120
Question 7
x2 = 402 + 222 −2(40)(22)cos 116°
x2 = 2855.553…
x = 2855.553…
x = 53.437…
x
116°
22 cm
40 cm
The longest diagonal is 53.4 cm long, to the nearest tenth of a centimetre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 50 of 96
Section 2.4
Page 120
Question 8
Let a represent the length of the tunnel.
a2 = 20002 + 30002 − 2(2000)(3000) cos 67.7°
a2 = 8 446 526.086…
a = 8 446 526.086…
a = 2906.290…
3000 m
A
67.7°
a
2000 m
The length of the tunnel would be 2906 m, to the nearest metre.
Section 2.4
Page 120
Question 9
C
8.56 km
5.93 km
A
10.24 km
B
8.562 = 5.932 + 10.242 − 2(5.93)(10.24) cos A
121.4464 cos A = 66.7489
66.7489
cos A =
121.4464
⎛ 66.7489 ⎞
∠A = cos −1 ⎜
⎟
⎝ 121.4464 ⎠
∠A = 56.659…
∠A = 57°, to the nearest degree.
Next, use the sine law to determine ∠B.
sin B sin 56.659…°
=
5.93
8.56
⎛ 5.93sin 56.659…° ⎞
∠B = sin −1 ⎜
⎟
8.56
⎝
⎠
∠B = 35.362…
∠B = 35°, to the nearest degree.
Use the angle sum of a triangle to determine ∠C.
∠C = 180° − (57° + 35°)
∠C = 88°
Section 2.4
Page 121
Question 10
Let ∠P represent the required angle.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 51 of 96
1.832 = 20.32 + 21.32 − 2(20.3)(21.3) cos P
864.78 cos P = 862.4311
862.4311
cos P =
864.78
⎛ 862.4311 ⎞
∠P = cos −1 ⎜
⎟
⎝ 864.78 ⎠
∠P = 4.223…
She must shoot within an angle of 4.2° to hit the net.
Section 2.4
Page 121
Question 11
Let w represent the width at the base of the bay.
w2 = 652 + 852 − 2(65)(85) cos 7.8°
w2 = 502.236…
7.8°
65 km
w = 502.236…
w = 22.410…
The width of the bay at its base is 22.4 km, to the
nearest tenth of a kilometre.
85 k m
w
Section 2.4
Page 121
Question 12
B
d
A
72 k m
50 km
49°
H
MHR • Pre-Calculus 11 Solutions Chapter 2
Let d represent the distance between the two
aircraft.
d2 = 502 + 722 − 2(50)(72) cos 49°
d2 = 2960.374…
d = 2960.374…
d = 54.409…
The distance between the two aircraft is 54.4 km,
to the nearest tenth of a kilometre.
Page 52 of 96
Section 2.4
Page 121
Question 13
The maximum width is BH,
which is BC + CG + GH.
In ∆ABC:
BC
161.3
=
sin 33.5° sin 73.25°
161.3sin 33.5°
BC =
sin 73.25°
BC = 92.972…
Since ∆GHI ≅ ∆ABC,
GH = 92.972… also.
E
D
B
F
C
A
H
G
I
∆BCD is equilateral, so CD = BC = 92.972… also. Likewise, FG = 92.972…
In ∆CDE:
CE
92.972
=
sin 73° sin 29°
92.972sin 73°
CE =
sin 29°
CE = 183.390…
Since ∆EFG ≅ ∆CDE, EG = 183.390… also.
∠DCE = 180° − (29° + 73°)
∠DCE = 78°
Then, in ∆CEG:
∠ECG = 180° − (60° + 78°)
∠ECG = 42°
Since the triangle is isosceles, ∠EGC = 42° also.
Then, ∠CEG = 180° − (42° + 42°)
∠CEG = 96°
CG
EC
=
sin ∠CEG sin ∠EGC
183.390
CG
=
sin 96° sin 42°
183.390sin 96°
CG =
sin 42°
CG = 272.570…
Finally, BC + CG + GH = 92.97 + 272.57 + 92.97 = 458.51.
So, the maximum width of Moondog is 458.5 cm.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 53 of 96
Section 2.4
Page 121
Question 14
a)
C
A
B
b) ∠C = 42° + 43° (alternate angle + supplement of the exterior angle)
∠C = 85°
2
AB = 82 + 52 −2(8)(5) cos 85°
AB2 = 82.027…
AB = 82.027…
AB = 9.056…
Julia and Isaac are 9.1 km from the base camp, to the nearest tenth of a kilometre.
sin A sin 85°
=
8
9.056…
8sin 85°
sin A =
9.056…
⎛ 8sin 85° ⎞
∠A = sin −1 ⎜
⎟
⎝ 9.056… ⎠
∠A = 61.635…
Then the heading that they must travel, from South, is 137° − 62°, or S75°W. This is a
bearing of 180° + 75° or 255°.
c)
Section 2.4
Page 122
Question 15
x2 = 82 + 72 − 2(8)(7) cos 80°
x2 = 93.551…
x = 93.551…
x = 9.672…
The distance from the spotlight to the point on
the wall where the light is reflected is 9.7 m, to
the nearest tenth of a metre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 54 of 96
Section 2.4
Page 122
Question 16
Answers may vary. Example:
Use the cosine law in each of the smaller
triangles to determine the measure of
each of the three angles that meet in the
middle of the design. The sum of those
three angles should be 360°.
Section 2.4
Page 122
Question 17
The diagram shows the triangular part of the
frame. ∠H is the head tube angle, ∠S is the seat
angle, and ∠P is the angle at the peddle.
Use the cosine law to determine ∠S.
542 = 502 + 402 −2(50)(40) cos S
2916 = 4100 − 4000 cos S
4000 cos S = 4100 − 2916
1184
cosS =
4000
⎛ 1184 ⎞
∠S = cos −1 ⎜
⎟
⎝ 4000 ⎠
∠S = 72.782…
H
50 cm
S
40 cm
5 4 cm
P
Use the sine law to determine ∠H.
sin H sin 72.782…
=
40
54
40sin 72.782…
sin H =
54
⎛ 40sin 72.782… ⎞
∠H = sin −1 ⎜
⎟
54
⎝
⎠
∠H = 45.035…
Use angle sum of a triangle to determine ∠P.
∠P = 180° − (45° + 73°)
∠P = 62°
The interior angles of the bike frame are 73°, 45°, and 62°.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 55 of 96
Section 2.4
Page 122
Question 18
d2 = 222 + 1022 − 2(22)(102) cos 74.63°
d2 = 9698.449…
d = 9698.449…
d = 98.480…
Barbora Spotakova threw the javelin
98.48 m, to the nearest hundredth of a
metre.
Section 2.4
Page 122
d
22 m
102 m
74.63°
Question 19
d2 = 16602 + 16602 − 2(1660)(1660) cos 55.5°
d2 = 2 389 621.947…
d = 2 389 621.947…
d = 1545.840…
The distance from Bermuda to San Juan is
1546 km, to the nearest kilometre.
Bermuda
1660 km
Miami
d
55.5°
1660 km
San Juan
Section 2.4
Page 123
Question 20
In ∆DTS,
∠DTS = 180° − 71° (supplementary angles)
∠DTS = 109°
Then, by angle sum of a triangle,
∠RDS = 180° − (61° + 109°)
∠RDS = 10°
Now use the sine law to determine DT.
D
S
T
F
DT
92
=
sin 61° sin10°
92sin 61°
DT =
sin10°
DT = 463.379…
Then, in ∆DFT:
h
sin 71° =
463.379…
h = (463.379…) sin 71°
h = 438.133…
The height of Della Falls is 438.1 m, to the nearest tenth of a metre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 56 of 96
Section 2.4
Page 123
Question 21
First use the cosine law to determine ∠A.
343.72 = 2002 + 3752 − 2(200)(375) cos A
150 000 cos A = 62 495.31
62 495.31
cos A =
150 000
⎛ 62 495.31 ⎞
∠A = cos −1 ⎜
⎟
⎝ 150 000 ⎠
∠A = 65.377…
∠A is 65°, to the nearest degree.
C
A
B
Next use the sine law to determine ∠B.
sin B sin 65.377…°
=
200
343.7
200sin 65.377…°
sin B =
343.7
⎛ 200sin 65.377…° ⎞
∠B = sin −1 ⎜
⎟
343.7
⎝
⎠
∠B = 31.937…
∠B is 32°, to the nearest degree.
Then use angle sum of a triangle to determine ∠C.
∠C = 180° − (65° + 32°)
∠C = 83°
The interior angles of the building are 65°, 32°, and 83°.
Section 2.4
Page 123
Question 22
Statement
c2 = (a – x)2 + h2
c2 = a2 – 2ax + x2 + h2
b2 = x2 + h2
c2 = a2 – 2ax + b2
x
cos C =
b
x = b cos C
2
c = a2 – 2ab cos C + b2
c2 = a2 + b2 – 2ab cos C
MHR • Pre-Calculus 11 Solutions Chapter 2
Reason
Use Pythagorean Theorem in
∆ABD.
Expand the square of a binomial.
Use Pythagorean Theorem in
∆ACD.
Substitute b2 for x2 + h2.
Use the cosine ratio in ∆ACD.
Multiply both sides by b.
Substitute b cos C for x in step 4.
Rearrange the terms.
Page 57 of 96
Section 2.4
Page 123
Question 23
From 4 p.m. until 6 p.m. is 2 h. In this time, the first
ships will have travelled 2(11.5) or 23 km. The
second ship will have travelled 2(13) or 26 km.
∠SPT = 180° − (38° + 47°)
∠SPT = 95°
d2 = 232 + 262 − 2(23)(26) cos 95°
d2 = 1309.238…
d = 1309.238…
d = 36.183…
At 6 p.m. the two ships are 36.2 km apart, to the
nearest tenth of a kilometre.
Section 2.4
Page 123
N
38°
d
port
47°
S
2(13) km
Page 123
2
C?
7 cm
16 cm
B
Question 25
At 1:30 the hour hand will be half way between
12:00 and 3:00, so the angle from 12:00 is 45°.
The angle between the two hands will be 135°.
2
ship 2
Question 24
It is not possible to draw a triangle with side lengths
7 cm, 8 cm, and 16 cm because the sum of the two
shorter sides is less than the longest side.
8 cm
Try substituting into the cosine law:
A
162 = 72 + 82 − 2(7)(8) cos C
112 cos C = −143
143
cos C = −
112
⎛ 143 ⎞
∠C = cos −1 ⎜ −
⎟
⎝ 112 ⎠
∠C = error
An error message is returned, because cosine cannot be less than −1.
Section 2.4
ship 1
2(11.5) km
45°
2
d = 7.5 + 15.2 − 2(7.5)(15.2) cos 135°
d2 = 448.510
d = 448.510…
d = 21.178…
7.5 cm
15.2 cm
d
At 1:30 p.m. the distance between the tip of the
minute and hour hands is 21.2 cm, to the nearest
tenth of a centimetre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 58 of 96
Section 2.4
Page 123
Question 26
Use the Pythagorean Theorem to determine the length of each side of ∆ABC.
AB2 = (2 −(−4))2 + (8 − (−5))2
AB2 = 36 + 169
AB = 205
BC2 = (7 − 2)2 + (2 − 8)2
BC2 = 25 + 36
BC = 61
CA2 = (7 − (−4))2 + (2 − (−5))2
CA2 = 121 + 49
CA = 170
Use the cosine law to determine ∠ABC.
b2 = a2 + c2 − 2ac cos B
170 = 61 + 205 − 2( 61 )( 205 ) cos B
2( 61 )( 205 ) cos B = 266 − 170
96
cos B =
2 61 205
96
⎛
⎞
∠B = cos −1 ⎜
⎟
⎝ 2 61 205 ⎠
∠B = 64.580…
The measure of interior angle ∠ABC is 65°, to the nearest degree.
Now use the sine law to determine ∠BCA.
sin C sin 64.580…°
=
205
170
sin C =
205 sin 64.580…°
170
⎛ 205 sin 64.580…° ⎞
∠C = sin −1 ⎜
⎟
⎜
⎟
170
⎝
⎠
∠C = 82.665…
Then, ∠ACD = 180° − 83°
∠ACD = 97°, to the nearest degree.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 59 of 96
Section 2.4
Page 124
Question 27
First determine the height.
h
sin 40° =
45.9
45.9 km
h = 45.9sin 40° h.
h
h = 29.503…
40°
Now, use the formula for area of a triangle.
bh
A
B
A=
40.4 km
2
(40.4)(29.503…)
A=
2
A = 595.979…
The area of the region is 596 km2, to the nearest square kilometre.
C
Section 2.4
Page 124
Question 28
P
C
Q
R
Join the centre, C, to two vertices of the
triangle, Q and R. CR and CQ are radii.
Use the cosine law to determine ∠QPR.
22 = 32 + 42 − 2(3)(4) cos P
24 cos P = 21
21
cos P =
24
∠P = cos −1 ( 0.875 )
∠P = 28.955…
The angle at the centre is twice the angle subtended from the same arc.
So, ∠QCR = 2(∠QPR)
∠QCR = 2(29°)
∠QCR = 58°
1
∆CRQ is isosceles, so ∠CQR = ∠CRQ = (180° − 58°) or 61°.
2
Use the sine law in ∆CQR to determine r.
r
2
=
sin 61° sin 58°
2sin 61°
r=
sin 58°
r = 2.062…
The radius of the circle is 2.1 m, to the nearest tenth of a metre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 60 of 96
Section 2.4
Page 124
Question 29
x
cos θ R = − cos θ = −
x + y2
2
b = x2 + y 2
c=
(a + x)
2
+ y2
Prove c 2 = a 2 + b 2 − 2ab cos C:
Left Side =
(
( a + x) + y2
2
)
2
= ( a + x) + y2
2
= a 2 + 2ax + x 2 + y 2
Right Side = a 2 +
(
x2 + y 2
) − 2a (
2
⎛
x
x2 + y 2 ⎜ −
2
⎜
x + y2
⎝
)
⎞
⎟
⎟
⎠
= a 2 + x 2 + y 2 + 2ax
= a 2 + 2ax + x 2 + y 2
Left Side = Right Side
Therefore, the cosine law is true.
Section 2.4
Page 124
Question 30
In right triangle ABR:
∠RBA = 35°
(alternate angles)
70
sin 35° =
RB
70
RB =
sin 35°
RB = 122.041…
R
A
B
C
In ∆RBC:
∠CRB = 35° − 18° = 17°
∠RCB = 18°
(alternate angles)
Use the sine law to determine BC.
BC
122.041…
=
sin17°
sin18°
122.041…sin17°
BC =
sin18°
BC = 115.467…
The car has travelled 115.5 m, to the nearest tenth of a metre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 61 of 96
Section 2.4
Page 124
Question 31
a) c2 = 12.22 + 8.92 − 2(12.2)(8.9) cos 90°
c2 = 228.05
b) c2 = 12.22 + 8.92
c2 = 228.05
A
c
8.9 cm
C
12.2 cm
B
c) The cosine law is similar to the Pythagorean Theorem, but it has the extra term
−2ab cos C. The equations are the same when ∠C = 90°.
d) The two formulas are the same in a right triangle because cos 90° = 0, so the extra
term has value 0.
Section 2.4
Page 124
Question 32
Concept Summary for Solving a Triangle
Begin by Using the
Method of
Given
Right triangle
A
Two angles and any side
B
Three sides
C
Three angles
D
Two sides and the included
C
angle
Two sides and the angle
B
opposite one of them
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 62 of 96
Section 2.4
Page 125
Question 33
Step 1
a)-c) Construct the figure, actual size, on
a full sheet of paper.
Step 2
a) Use the cosine law to determine ∠A.
42 = 62 + 82 − 2(6)(8) cos A
96 cos A = 84
84
cos A =
96
∠A = cos −1 (0.875)
∠A = 28.955…
Use the sine law to determine ∠C.
sin C sin 28.955…
=
6
4
6sin 28.955…
sin C =
4
−1
∠C = sin (1.5sin 28.955…)
∠C = 46.567…
Use the angle sum of a triangle to determine ∠B.
∠B = 180° − (29° + 47°)
∠B = 104°
b) There are four angles meeting at each vertex of ∆ABC. Two of the angles are corners
of squares, so they are each 90°. At any vertex, A, B, or C, the other two angles must
have a sum of 360° − 2(90°) or 180°.
∠GBF = 180° − ∠ABC
∠HCI = 180° − ∠BCA
∠DAE = 180° − ∠BAC
∠GBF = 180° − 104°
∠HCI = 180° − 47°
∠DAE = 180° − 29°
∠GBF = 76°
∠HCI = 133°
∠DAE = 151°
c) In ∆BGF, BG = BC = 4 cm, BF = BA = 6 cm.
Use the cosine law to determine GF.
GF2 = 42 + 62 − 2(4)(6)cos 76°
GF2 = 40.387…
GF = 40.387…
GF = 6.355…
The length of GF is 6.4 cm, to the nearest tenth of a centimetre.
In ∆ADE, AD = AB = 6 cm, AE = AC = 8 cm.
Use the cosine law to determine GF.
DE2 = 62 + 82 − 2(6)(8)cos 151°
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 63 of 96
DE2 = 183.963…
DE = 183.963…
DE = 13.563…
The length of DE is 13.6 cm, to the nearest tenth of a centimetre.
In ∆CHI, CH = CB = 4 cm, CI = AC = 8 cm.
Use the cosine law to determine HI.
HI2 = 42 + 82 − 2(4)(8)cos 133°
HI2 = 123.647…
HI = 123.647…
HI = 11.119…
The length of HI is 11.1 cm, to the nearest tenth of a centimetre.
Step 3
a) and b)
In ∆ABC, draw an altitude from B to AC. Let its height be h1.
h
sin A = 1
AB
4 cm
h1 = 6 sin 29°
B
6 cm
h1
h1 = 2.908…
29°
Now, use the formula for area of a
8 cm
C
triangle.
bh
A=
2
(8)(2.908…)
A=
2
A = 11.635…
The altitude of ∆ABC from B to AC is 2.9 cm, to the nearest tenth of a centimetre and the
area of ∆ABC is 11.6 cm2, to the nearest tenth of a square centimetre.
In ∆BFG, draw an altitude from F to BG. Let its height be h2.
h
sin B = 2
BF
6.4 cm
h2 = 6 sin 76°
h2 = 5.821…
Now, use the formula for area of a
triangle.
bh
A=
2
(4)(5.821…)
A=
2
A = 11.643…
MHR • Pre-Calculus 11 Solutions Chapter 2
G
F
h2
6 cm
4 cm
76°
B
Page 64 of 96
A
An altitude of ∆BFG is 5.8 cm, to the nearest tenth of a centimetre and the area of ∆BFG
is 11.6 cm2, to the nearest tenth of a square centimetre.
In ∆ADE, draw an altitude from D to EA extended. Let its height be h3.
h
h3
D
sin A = 3
DA
6 cm
h3 = 6 sin 29°
29°
h3 = 2.908…
151°
A
Now, use the formula for area of a
triangle.
13.6 cm
bh
A=
8 cm
2
(8)(2.908…)
A=
2
E
A = 11.635…
An altitude of ∆DAE is 2.9 cm, to the nearest tenth of a centimetre and the area of ∆ADE
is 11.6 cm2, to the nearest tenth of a square centimetre.
In ∆CHI, draw an altitude from H to IC extended. Let its height be h4.
h
h4
sin C = 4
H
CH
4 cm
h4 = 4 sin 47°
h4 = 2.925…
Now, use the formula for area of a
triangle.
bh
A=
2
(8)(2.925…)
A=
2
A = 11.701…
47°
133°
C
11.1 cm
8 cm
I
An altitude of ∆CHI is 2.9 cm, to the nearest tenth of a centimetre and the area of ∆BFG
is 11.7 cm2, to the nearest tenth of a square centimetre.
Step 4
Within rounding errors, the areas of the triangles are the same. This is because the area of
∆ABC can be found by using an altitude from any vertex to its opposite side. For
example, using the altitude from vertex B, the height of ∆ABC could be either 6 sin 29°
or 4 sin 47° with base 8. Notice each of these possible combinations was used to calculate
the areas of ∆ADE and ∆CHI, respectively. If an altitude from vertex A in ∆ABC is used,
the height would be 6 sin 76° with base 4, which is the same as the area calculation for
∆BFG. This would hold true for any ∆ABC.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 65 of 96
Chapter 2 Review
Chapter 2 Review
Page 126
Question 1
a) E An angle in standard position is an angle whose vertex is at the origin and whose
arms are the x-axis and the terminal arm.
b) D The reference angle is the acute angle formed by the terminal arm and the x-axis.
c) B An exact value is is not an approximation and may involve a radical.
d) A The sine law is a formula that relates the lengths of the side of a triangle to the sine
values of its angles.
e) F The cosine law is a formula that relates the lengths of the side of a triangle to the
cosine value of one of its angles.
f) C A terminal arm is the final position of the rotating arm of an angle is standard
position.
g) G An ambiguous case is a situation that is open to two or more interpretations.
Chapter 2 Review
Page 126
Question 2
a)
quadrant III,
reference angle is 200° − 180° = 20°
b)
quadrant II,
reference angle is 180° − 130° = 50°
c)
quadrant I,
reference angle is 20°
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 66 of 96
d)
quadrant IV,
reference angle is 360° − 330° = 30°
Chapter 2 Review
Page 126
Question 3
The 30° angle is not a reference angle because it is measured from the vertical. In this
situation, the reference angle would be 60°.
Chapter 2 Review
Page 126
Question 4
Consider a reference angle of 35° in each
quadrant as shown.
II
In quadrant II, the angle in standard position is
180° − 35° = 145°.
35°
35°
In quadrant III, the angle in standard position is
180° + 35° = 215°
III
I
35°
35°
IV
In quadrant IV, the the angle in standard position is
360° − 35° = 325º
The angles, in standard position, 0° ≤ θ
35°, 145°, 215° and 325°.
Chapter 2 Review
Page 126
a)
sin 225° =
225°
−1
x −1
2
=
or −
r
2
2
y −1
=1
tan 225° = =
x −1
2
b)
y
=
r
x
cos 120° = =
r
y
tan 120° = =
x
sin 120° =
2
3
60°
y −1
2
=
or −
r
2
2
cos 225° =
45°
−1
Question 5
120°
−1
MHR • Pre-Calculus 11 Solutions Chapter 2
3
2
−1
2
3
=− 3
−1
Page 67 of 96
c)
y −1
=
r
2
x
3
cos 120° = =
r
2
y −1
3
or −
tan 120° = =
x
3
3
sin 330° =
3
30°
330°
−1
2
d)
sin 135° =
2
1
135°
45°
−1
Chapter 2 Review
Page 126
y
1
2
or
=
r
2
2
x −1
2
=
or −
r
2
2
y1
= −1
tan 135° = =
x −1
cos 135° =
Question 6
b) r2 = x2 + y2
r2 = (−3)2 + 62
r2 = 45
r = 45
a)
r =3 5
c) sin θ =
y
6
2
25
=
=
or
r 35
5
5
x
−3
−1
5
=
=
or −
r 35
5
5
y6
tan θ = =
= −2
x −3
cos θ =
Chapter 2 Review
Page 126
⎛ −1 ⎞
d) θ = cos−1 ⎜
⎟
⎝ 5⎠
θ = 116.565…
θ = 117°, to the nearest degree.
Question 7
P(2, −5) is in quadrant IV.
Points with the same reference angle, in the
other quadrants, are (2, 5), (−2, 5), and
(−2, −5).
P(2, −5)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 68 of 96
Chapter 2 Review
y
r
1
sin 90° = = 1
1
a) sin θ =
y
r
0
sin 180° = = 0
3
b) sin θ =
Chapter 2 Review
Page 127
Question 8
x
r
0
cos 90° =
1
y
x
1
tan 90° = , which is undefined
0
cos θ =
tan θ =
x
r
−3
= −1
cos 180° =
3
cos θ =
Page 127
y
x
0
=0
tan 180° =
−3
tan θ =
Question 9
a) Since both sine and cosine are negative, ∠θ is in quadrant III.
x
−3
θ
5
x2 + y 2 = r 2
x + (−3)2 = 52
x2 = 25 – 9
x2 = 16
x = −4
x −4
Then, cos θ = =
r
5
y −3 3
=
tan θ = =
x −4 4
2
b) Since cosine is positive and tangent negative, ∠θ is in quadrant IV.
x2 + y 2 = r 2
2
1 + y2 = 32
1
y2 = 9 – 1
y2 = 8
θ
y
y = − 8 or − 2 2
3
y −2 2
=
r
3
y −2 2
= −2 2
tan θ = =
1
x
Then, sin θ =
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 69 of 96
c) Since both tangent and sine are positive, ∠θ is in quadrant I.
x2 + y2 = r2
2
5 + 122 = r2
25 + 144 = r2
r
r2 = 169
12
r = 169
θ
r = 13
5
y 12
Then, sin θ = =
r 13
x5
cos θ = =
r 13
Chapter 2 Review
Page 127
Question 10
a) Given tan θ = −1.1918.
Since tan θ is negative, ∠θ is in quadrant II or IV.
tan−1 (1.1918) = 50.001…
So, the reference angle is 50°, to the nearest degree.
In quadrant II, ∠θ = 180° − 50° = 130°.
In quadrant IV, ∠θ = 360° − 50° = 310°.
b) Given sin θ = −0.3420.
Since sin θ is negative, ∠θ is in quadrant III or IV.
sin−1 (0.3420) = 19.998…
So, the reference angle is 20°, to the nearest degree.
In quadrant III, ∠θ = 180° + 20° = 200°.
In quadrant IV, ∠θ = 360° − 20° = 340°.
c) Given cos θ = 0.3420.
Since cos θ is positive, ∠θ is in quadrant I or IV.
cos−1 (0.3420) = 70.001…
So, the reference angle is 70°, to the nearest degree.
In quadrant I, ∠θ = 70°.
In quadrant IV, ∠θ = 360° − 70° = 290°.
Chapter 2 Review
Page 127
Question 11
a)
Yes. One side and its corresponding angle are given, and
you know the measures of all the angles, so you can
determine the other two sides using the sine law.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 70 of 96
b)
You could use the sine law to determine ∠θ, by using
sin θ sin 90°
=
. However, since this is a right triangle it
20
32
is more efficient to use the sine ratio to determine ∠θ
and the Pythagorean Theorem to determine the unknown
side length.
No. You are not given the right pairs of sides and
angles to use the sine law.
c)
Chapter 2 Review
Page 127
a)
Question 12
∠C = 180° − (35° + 88°)
∠C = 57°
Then, using the sine law:
44
c
=
sin 57° sin 88°
44sin 57°
c=
sin 88°
c = 36.923…
c = 36.9 mm, to the nearest tenth of a millimetre.
sin B sin 42°
=
22
17
22sin 42°
sin B =
17
⎛ 22sin 42° ⎞
∠B = sin −1 ⎜
⎟
17
⎝
⎠
∠B = 59.989…
∠B = 60°, to the nearest degree.
Then, ∠A = 180° − (42° + 60°)
∠A = 78°
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 71 of 96
Chapter 2 Review
Page 127
Question 13
∠R = 180° − (63.5° + 51.2°)
∠R = 65.3°
Then, using the sine law:
6.3
6.3
p
q
=
=
sin 63.5° sin 65.3°
sin 51.2° sin 65.3°
6.3sin 63.5°
6.3sin 51.2°
p=
q=
sin 65.3°
sin 65.3°
p = 6.205…
q = 5.404…
The unknown sides are RQ = 6.2 cm and PR = 5.4 cm, to the nearest tenth of a
centimetre.
Chapter 2 Review
Page 127
Question 14
M
4.1°
h
8.7°
J
21 km
K
L
In ∆JKM,
∠JMK = 8.7° − 4.1° (exterior angle of a triangle)
∠JMK = 4.6°
Then, using the sine law:
MK
21
=
sin 4.1° sin 4.6°
21sin 4.1°
MK =
sin 4.6°
MK = 18.721…
In ∆KLM, h represents the height of the mountain.
h
sin 8.7° =
18.721…
h = 18.721…sin 8.7°
h = 2.831…
The height of the mountain is approximately 2.8 km, to the nearest tenth of a kilometre.
This is 2800 m.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 72 of 96
Chapter 2 Review
Page 127
Question 15
a) Ship B is closer to Sarah.
BS
68
AS
68
Check AS:
=
=
sin 47° sin 84°
sin 49° sin 84°
68sin 47°
68sin 49°
BS =
AS =
sin 84°
sin 84°
BS = 50.005…
AS = 51.602…
Sarah is 50.0 km from ship B.
x
h
b) Let h represent the perpendicular distance from Sarah to the line from ship A to ship
B. Let x represent the distance from the foot of the perpendicular to B.
Use the tangent ratio in each right triangle:
h
h
tan 47° =
tan 49° =
68 − x
x
h = tan 47°(68 − x)
h = x tan 49°
So,
x tan 49° = tan 47°(68 − x)
x(tan 49° + tan 47°) = 68 tan 47°
68 tan 47°
x=
tan 49° + tan 47°
x = 32.806…
32.806…
68 − 32.806…
Then, cos 49° =
and
cos 47° =
AS
BS
68 − 32.806…
32.806…
AS =
BS =
cos 47°
cos 49°
AS = 51.602…
BS = 50.005…
This verifies the answer in part a), that ship B is closer to Sarah.
Chapter 2 Review
Page 127
Question 16
There is one solution if a = b sin A, because this creates a
right triangle. If a ≥ b then only one triangle can be drawn.
If b sin A
one with an acute angle at B, and one with an obtuse angle
at B. If a
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 73 of 96
Chapter 2 Review
Page 128
Question 17
a) On the first leg, the plane goes for 1 h at
720 km/h, so AB = 720 km.
On the second part, the plane travels for 30 min
before reaching the original eastward path. So,
BC = 360 km.
B
A
D
C
b) Let D be at the foot of the perpendicular from B to the East-West line AC.
Then, in ∆ABD:
BD
sin 20° =
720
BD = 720(sin 20°)
Now in ∆BCD:
720(sin 20°)
cos ∠CBD =
360
⎛ 720(sin 20°) ⎞
∠CBD = cos −1 ⎜
⎟
360
⎝
⎠
∠CBD = 46.839…
The heading of the second part is S47°E, to the nearest degree.
c) In ∆BCD, ∠C = 180° − (46.839…° + 90°) = 43.161…°
In ∆ABC, ∠B = 180° − (20° + 43.161…°)
∠B = 116.839…°
Use the sine law to determine AC.
AC
360
=
sin116.839…° sin 20°
360sin116.839…°
AC =
sin 20°
AC = 939.185…
The jet resumes its original path 939.2 km east of the point where it changed course.
Chapter 2 Review
Page 128
Question 18
a) Given a = 7, b = 2 and c = 4, the triangle cannot exist because the two shorter sides
are less than the longest side i.e., the three sides will not meet to make a triangle.
b) Given ∠A = 85°, b = 10, ∠C = 98°, the triangle cannot exist because the two given
angles sum to more than 180°.
c) Given a = 12, b = 20 and c = 8, the triangle cannot exist because the two shorter sides
equal the longest side i.e., the three sides will not meet to make a triangle.
d) ∠A = 65°, ∠B = 52°, ∠C = 35°, the triangle cannot exist because the sum of the three
angles is less than 180°.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 74 of 96
Chapter 2 Review
Page 128
Question 19
a) You can use the sine law. First you need to
find the measure of ∠X using angle sum of a
x
24
=
.
triangle. Then you can solve
sin 38° sin Y
b) You can use the cosine law directly:
y2 = 42 + 82 − 2(4)(8) cos 88°
Chapter 2 Review
Page 128
Question 20
a) a2 = 262 + 362 − 2(26)(36) cos 53°
a2 = 845.402…
a = 845.402…
a = 29.075…
a = 29.1 cm, to the nearest tenth of a centimetre.
b)
28.42 = 23.62 + 33.22 − 2(23.6)(33.2) cos θ
1567.04 cos θ = 852.64
852.64
cos θ =
1567.04
⎛ 852.64 ⎞
θ = cos −1 ⎜
⎟
⎝ 1567.04 ⎠
θ = 57.036…
∠B = 57°, to the nearest degree.
Chapter 2 Review
Page 128
Question 21
BG2 = 2402 + 3752 −2(240)(375) cos 20°
BG2 = 29 080.328…
BG = 29 080.328…
BG = 170.529…
The distance of the ball from the centre
of the hole is 170.5 yd, to the nearest
tenth of a yard.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 75 of 96
Chapter 2 Review
a)
Page 128
Question 22
9.62 = 10.82 + 18.42 − 2(10.8)(18.4) cos A
397.44 cos A = 363.04
363.04
cos A =
397.44
⎛ 363.04 ⎞
∠A = cos −1 ⎜
⎟
⎝ 397.44 ⎠
∠A = 24°
b)
AB2 = 102 + 92 − 2(10)(9) cos 48°
AB2 = 60.556…
AB = 60.556…
AB = 7.781…
The length of AB is 7.8 cm, to the
nearest tenth of a centimetre.
C
10 cm
A
48º
9 cm
B
Use the cosine law to determine b.
b2 = 82 + 152 −2(8)(15) cos 24°
b2 = 69.749…
b = 69.749…
b = 8.351…
Next, use the sine law to determine ∠A.
sin A sin 24°
=
8
8.351…
8sin 24°
sin A =
8.351…
⎛ 8sin 24° ⎞
∠A = sin −1 ⎜
⎟
⎝ 8.351… ⎠
∠A = 22.930…
Now, use angle sum of a triangle to determine ∠C.
∠C = 180° − (24° + 23°)
∠C = 133°
In ∆ABC, AC = 8.4 m, to the nearest tenth of a metre, ∠A = 23° and ∠C = 133°, both to
the nearest degree.
c)
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 76 of 96
Chapter 2 Review
Page 128
Question 23
a)
d
b) Determine the distance travelled by each boat in 4 h.
4(48) = 192
4(53.6) = 214.4
Use the cosine law to find their distance apart, d.
d2 = 1922 + 214.42 − 2(192)(214.4) cos 54°
d2 = 34 439.235…
d = 34 439.235…
d = 185.578…
After 4 h, the boats are 185.6 km apart, to the nearest tenth of a kilometre.
Chapter 2 Review
Page 128
Question 24
a)
b) Let s represent the length of the shorter diagonal.
s2 = 42 + 62 −2(4)(6) cos 58°
s2 = 26.563…
s = 26.563…
s = 5.154…
Let d represent the length of the longer diagonal.
d2 = 42 + 62 −2(4)(6)cos 122°
d2 = 77.436…
d = 77.436…
d = 8.799…
The lengths of the two diagonals are 5.2 cm and 8.8 cm, to the nearest tenth of a
centimetre.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 77 of 96
Chapter 2 Practice Test
Chapter 2 Practice Test
Page 129
Question 1
Angle
Reference Angle
A 125°
180° − 125° = 55°
B 155°
180° − 155° = 25°
C 205°
205° − 180° = 25°
D 335°
360° − 335° = 25°
Angle A has a different reference angle than the others.
Chapter 2 Practice Test
Page 129
Question 2
Angle
Reference Angle
A 35°
35°
B 125°
180° − 125° = 55°
C 235°
235° − 180° = 55°
D 305°
360° − 305° = 55°
Angle A does not have a reference angle of 55°.
Chapter 2 Practice Test
Page 129
Question 3
cos θ =
2
1
−3
2
C is the exact value of cos 150°.
150°
cos 150° =
30°
−3
Chapter 2 Practice Test
x
r
Page 129
Question 4
From the diagram,
sin θ sin 28°
=
34
70
The equation B can be used to determine the
measure of angle θ.
Chapter 2 Practice Test
Page 129
Question 5
A Since three sides are given, only one triangle can be drawn.
B Since ∠D is given as an obtuse angle only one such triangle can be drawn.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 78 of 96
C Check h = 27 sin 35° = 15.486…. Since h
21, two possible triangles can be drawn.
C
27 m
For the triangle in C, you must consider the
ambiguous case.
21 m
35°
B
A
D Here the three angles and one side are known, so only one such triangle can be drawn.
Chapter 2 Practice Test
Page 129
Question 6
Since cos θ is positive and tan θ is negative, the angle is in quadrant IV.
x
cos θ =
r
2
1
2
=
10 r
b
r = 2 10
Then, using the Pythagorean Theorem:
r2 = x2 + b2
r
( 2 10 )
2
= 22 + b 2
40 − 4 = b2
b2 = 36
b = −6
Chapter 2 Practice Test
Page 129
Question 7
a)
T
b) Consider the vertices of the triangle to be O, R, and T.
∠R = 79° − 57° = 22°
Use the sine law to determine a second angle inside ∆ORT.
MHR • Pre-Calculus 11 Solutions Chapter 2
Page 79 of 96
sin O sin 22°
=
1.9
1.1
1.9sin 22°
sin O =
1.1
⎛ 1.9sin 22° ⎞
∠O = sin −1 ⎜
⎟
1.1
⎝
⎠
∠O = 40.319…
∠O = 40°, to the nearest degree.
Then, ∠T = 180° − (22° + 40°)
∠T = 118°
Now, use the cosine law to determine the length RO.
t2 = 1.12 + 1.92 − 2(1.1)(1.9) cos 118°
t2 = 6.782…
t = 2.604…
The distance between Ross Bay and Oak Bay is 2.6 km, to the nearest tenth of a
kilometre.
Chapter 2 Practice Test
Page 129
Question 8
a) h = 16 sin 30° = 8
Since 8
case. Two distinct triangles can be
drawn.
C
16
h
10
30°
A
B′
B
sin B sin 30°
=
16
10
16sin 30°
sin B =
10
⎛ 16sin 30° ⎞
∠B = sin −1 ⎜
⎟
10
⎝
⎠
∠B = 53.130…
So, ∠B = 53° or ∠B = 180° − 53° = 127°.
b)
Case 1: ∠B = 53°.
∠C = 180° − (30° + 53°)
∠C = 97°
Use the sine law to determine c.
MHR • Pre-Calculus 11 Solutions Chapter 2
Case 2: ∠B = 127°
∠C = 180° − (30° + 127°)
∠C = 23°
Page 80 of 96
c
16
c
16
=
=
sin 97° sin 53°
sin 23° sin 53°
16sin 97°
16sin 23°
c=
c=
sin 53°
sin 53°
c = 19.884…
c = 7.827…
The unknown measures in ∆ABC are ∠B = 53°, ∠C = 97°, and AB = 19.9
or ∠B = 127°, ∠C = 23°, and AB = 7.8, where angles are given to the nearest degree and
lengths to the nearest tenth.
Chapter 2 Practice Test
Page 129
Question 9
62 = 202 + 202 − 2(20)(20) cos R
800 cos R = 764
764
cos R =
800
⎛ 764 ⎞
∠R = cos −1 ⎜
⎟
⎝ 800 ⎠
∠R = 17.253…
Rudy must fire the puck within an angle of 17°, to the nearest degree.
Chapter 2 Practice Test
Page 129
Question 10
a) h = 12 sin 56° = 9.948…, and 9.948…
