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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.
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°.
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° < θ < 360°, that have a reference angle of 45° are 135°, 225°, and 315°.
b) 180° − 60° = 120°, 180° + 60° = 240°, 360° − 60° = 300° The three other angles in standard position, 0° < θ < 360°, that have a reference angle of 60° are 120°, 240°, and 300°.
c) 180° − 30° = 150°, 180° + 30° = 210°, 360° − 30° = 330° The three other angles in standard position, 0° < θ < 360°, that have a reference angle of 30° are 150°, 210°, and 330°.
d) 180° − 75° = 105°, 180° + 75° = 255°, 360° − 75° = 285° The three other angles in standard position, 0° < θ < 360°, that have a reference angle of 75° are 105°, 255°, and 285°.
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
cos 30 50
x
x
x
By symmetry, the horizontal distance that the tip of the wiper travels in one swipe will be 2 x , 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
1
1
1
sin 60 10
v
v
v
2
2
2
sin 30 10
10 5
v
v
v
Then, v 1 (^) − v 2 = 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°.
50 cm
x
30°
150°
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 Question 16
a) The terminal arm of the blue angle falls at the end of the 12th day of 20 on the second
ring. Since there are 360° in the circle, the blue angle measures 12 360 20
⎜⎝ ⎟⎠ or 216°.
b) If the angle was in quadrant II, it would be the 8th day of 20. The angle would
measure 8 360 20
⎜⎝ ⎟⎠ or 144°.
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.
b) S50°W is equivalent to 180° + 40° or 220° in standard position.
220°
50°
N
W
S
E
S50°W
70°
20°
N
W
S
E
N20°E
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 y (^) s represent the height of the seat above the centre of rotation.
sin 72 9 9sin 72 8.559...
s
s s
y
y y
So, the height of Carl’s seat above the ground is 11 + 8.559… or approximately 19.56 m.
b) i) If the speed is 4 rev/min, then in 5 s Carl has moved 4 5 60
⎜⎝ ⎟⎠ or
of a revolution.
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
sin 9 9sin 1.871...
s
s s
y
y y
In this case, Carls seat is 11 − 1.871… or approximately 9.13 m above the ground.
Section 2.1 Page 86 Question 21
a) sin θ = CD OC sin θ = CD, since the radius 1
is 1.
So, option B is correct.
b) tan θ = BA OA sin θ = BA, since the radius 1
is 1.
So, option D is correct.
Section 2.1 Page 86 Question 22
Using the Pythagorean Theorem, x^2 + y^2 = r^2
Section 2.1 Page 86 Question 23
a) θ 20° 40° 60° 80° sin θ 0.3420 0.6428 0.8660 0. sin (180° − θ) 0.3420 0.6428 0.8660 0. sin (180° + θ) −0.3420 −0.6428 −0.8660 −0. sin (360° − θ) −0.3420 −0.6428 −0.8660 −0.
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
(^2) cos θ sin θ 16
d =^ V. (^2) cos sin 16 3025 3
d
d
d
⎝ ⎠⎝^ ⎠
The exact distance that Daria hit the ball with this driver was^3025 16
ft.
θ
y
x
P( x , y ) r
Section 2.2 Page 96 Question 3
a) Use the Pythagorean Theorem to determine the hypotenuse: r = 5.
sin θ , cos θ , tan θ
sin θ 4 , cos θ 3 , t^4 5 5 3
an θ
y x y r r x
b) r^2 = x^2 + y^2 r^2 = (−12) 2 + (−5)^2 r^2 = 144 + 25 r^2 = 169 r = 13 sin θ , cos θ , tan θ
sin θ , cos θ , tan θ 5 1
y x y r r x − −
c) r^2 = x^2 + y^2 r^2 = (8) 2 + (–15) 2 r^2 = 64 + 225 r^2 = 289 r = 17 sin θ , cos θ , tan θ
sin θ 15 , cos θ^8 17
, tan θ 17 8
y x y r r x
d) r^2 = x^2 + y^2 r^2 = (1) 2 + (−1) 2 r^2 = 2 r = 2 sin θ , cos θ , tan θ
sin θ or 2 , cos θ or 2 , tan θ 1 2
y x y r r x
= −^ − = =− = −
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.
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. r^2 = x^2 + y^2 r^2 = (−5) 2 + (12) 2 r^2 = 25 + 144 r^2 = 169 r = 13
sin θ
sin θ 12 13
y r
cos θ
cos θ 5 5 13
or 13
x r −
t 2
an θ
tan θ or^12 5 5
y x
−
b) r^2 = x^2 + y^2 r^2 = (5) 2 + (−3) 2 r^2 = 25 + 9 r^2 = 34 r = 34
s
in
θ
sin θ or 3 34 34
y r
= −^ −^5
c
os
θ
cos θ or 5 34 34
x r
tan θ
tan 3 3 5
θ or 5
y x
c) r^2 = x^2 + y^2 r^2 = (6) 2 + (3) 2 r^2 = 36 + 9 r^2 = 45 r = 45
s
in
θ
sin θ 1 or 5 5 5
y r
co
s θ
cos θ 2 or 2 5 5 5
x r
tan θ
ta 3 1 6
n θ 2
y x
d) r^2 = x^2 + y^2 r^2 = (−24) 2 + (−10)^2 r^2 = 576 + 100 r^2 = 676 r = 26
Then, sin θ
sin 5 3
θ
y r
and tan θ
tan 5 5 2
θ 2
y x
b) sin θ 3 5
y r
= = , so y = 3 and r = 5 for an angle in quadrant I.
Use the Pythagorean Theorem to determine x. r^2 = x^2 + y^2 52 = x^2 + 3^2 25 = x^2 + 9 x = 4
Then, cos θ
cos 4 5
θ
x r
and tan θ
tan 3 4
θ
y x
c) tan θ 4 5
y x
= = − , so x = 5 and y = −4 for an angle in quadrant IV.
Use the Pythagorean Theorem to determine r. r^2 = x^2 + y^2 r^2 = (5) 2 + (−4) 2 r^2 = 25 + 16 r = 41
Then,
4
s
in
θ
sin θ or 4 41 41
y r
and
5
c
os
θ
cos θ or 5 41 41
x r
d) sin θ 1 3
y r
= = − , so y = −1 and r = 3 for an angle in quadrant III.
Use the Pythagorean Theorem to determine x. r^2 = x^2 + y^2 32 = x^2 + (−1) 2 9 = x^2 + 1 x = − 8
Then, cos θ
cos θ 2 2 3
or
x r
and
1
tan θ
tan θ or 1 or^2 8 2 2 4
y x − −
e) tan θ 1 1
y x
= =^ −
, so x = −1 and y = −1 for an angle in quadrant III.
Use the Pythagorean Theorem to determine r. r^2 = x^2 + y^2 r^2 = (–1) 2 + (–1) 2 r = 2
Then, sin θ
sin 1 2 2
θ or 2
y r
and cos θ
cos 1 2 2
θ or 2
x r
Section 2.2 Page 97 Question 9
a) The diagram shows the two possible positions of θ, 0° ≤ θ < 360°, for which
cos θ =^1 2
cos θ 1 2
= is part of 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 = 360° − 60° or 300°.
b) The diagram shows the two possible positions of θ, 0° ≤ θ < 360°, for which
cos θ = 1 2
For cos θ = 1 2
− , refer to the 45°-45°-90°
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°.
− 1
y^2 θ
1
(^2) y
θ
Section 2.2 Page 97 Question 10
θ sin θ cos θ tan θ 0° 0 1 0 90° 1 0 undefined 180° 0 − 1 0 270° − 1 0 undefined 360° 0 1 0
Section 2.2 Page 97 Question 11
a) Given P(−8, 6), use x = −8, y = 6, and the Pythagorean Theorem to determine r. r^2 = x^2 + y^2 r^2 = (−8) 2 + 6^2 r^2 = 64 + 36 r^2 = 100 r = 10 sin θ
sin θ 3 5
or
y r
cos θ
cos θ 8 4 10
or 5
x r
tan θ
tan θ or 3 4
y x
b) Given P(5, −12), use x = 5, y = −12, and the Pythagorean Theorem to determine r. r^2 = x^2 + y^2 r^2 = 5^2 + (−12) 2 r^2 = 25 + 144 r^2 = 169 r = 13 sin θ
sin θ 12 13
y r
cos θ
cos θ 5 13
x r
tan θ
ta 12 5
n θ
y x
Section 2.2 Page 97 Question 12
a)
b)
1
tan θ 4 9 θ tan 4 9 θ 24
R
R
R
−
= ⎛^ ⎞
c) θ = 180° − 24° θ = 156°, to the nearest degree
Section 2.2 Page 97 Question 13
a) (^) b)
1
tan θ 24 7 θ tan 24 7 θ 74
R
R
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. r^2 = x^2 + y^2 r^2 = 2^2 + 4^2 r^2 = 20 r = 20
Then,
4 20
sin θ
sin θ = 4 or 2 5 2 5 5
y r
b) Given Q(4, 8), then x = 4 and y = 8. Then, r^2 = x^2 + y^2 r^2 = 4^2 + 8^2 r^2 = 80 r = 80
sin θ
sin θ = 8 or 2 5 4 5 5
y r
c) Given R(8, 16), then x = 8 and y = 16. Then, r^2 = x^2 + y^2 r^2 = 8^2 + 16^2
Section 2.2 Page 97 Question 16
cos θ x r
= , so given cos θ =^1 5
, x = 1, and r = 5.
tan θ y x
= , so given tan θ = 2 6 , y = 2 6.
Then, sin θ
sin θ 2 5
y r
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.
Then, sin θ
sin 0 0 1
or 0
y r
cos θ
cos 0 1 1
or 1
x r
tan θ
tan 0 1
y x
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.
Then, sin θ
sin 90 1 1 1
or
y r
cos θ
cos 9 0 1
0 or 0
x r
tan θ
tan 90 or undefined
y x
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
triangle both have a value of 3 2
e) sin 270° = cos 180° is a true statement, because both have a value of −1.
Section 2.2 Page 98 Question 19
θ sin θ cos θ tan θ 0° 0 1 0 30° 1 2
(^1) or 3 3 3
45° 1 or 2 2 2
(^1) or 2 2 2
90° 1 0 undefined 120° 3 2
2 −^3
135° 1 or^2 2 2
− 1 or 2 2 2
2 −^
(^1) or 3 3 3
210° −^1
2 −^
(^1) or 3 3 3
225° − 1 or 2 2 2
− − 1 or 2 2 2
270° − 1 0 undefined 300° (^) − 3 2
2 −^3
315° − 1 or 2 2 2
− 1 or 2 2 2
330° −^
2 −^
(^1) or 3 3 3
Section 2.2 Page 98 Question 20
a) The measure for ∠A is 45°, for ∠B is 135°, for ∠C is 225°, and for ∠D is 315°.
