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Lesson 5.1.1 5-7. x ≈ 7.50 and y ≈ 8.04 units; Use either sine or cosine to get the first leg, then any one of

the trig ratios or the Pythagorean Theorem to get the other. 5-8. a: False (a rhombus and square are counterexamples) b: True c: False (it does not mention that the lines must be parallel) 5-9. B 5-10. a: (4 cards less than 5)× (4 suits)

52 = 1652 . If Aces are not included, 1252 .

b: 1− 1652 =3652 . If Aces are included, 4052 .

c: P(red) + P(face) – P(red and face) = 2652 +

1252 −

652 = 32

52 5-11. area = 74 sq ft, perimeter = 47.66 ft 5-12. a: x = –3 b: m = 10 c: p = −4 or 23 d: x = 23

Lesson 5.1.2 5-17. a: sin 22° = x

17 b: x ≈ 6.37 , tan 49° = 7x , 6.09 c: cos 60° = x

6 , x = 3 5-18. ≈ 26.92 feet 5-19. a: G; an = 100( 110 )

n−1 = 103−n b: A; an = 0 − 50(n −1) = 50 − 50n 5-20. Region A is 14 of the circle. Since the spins are independent, the probability of A and A

is 14 ⋅14 =

116 . In 80 games, we expect A and A to occur 116 (80) = 5 times.

5-21. a: False (a 30°- 60°- 90° triangle is a counterexample) b: False (this is only true for rectangles and parallelograms) c: True 5-22. a: 6x2 − x − 2 b: 6x3 − x2 −12x − 5 c: −3xy + 3y2 + 8x − 8y d: x2 − 9y2 5-23. ΔABC ~ ΔEFD by SAS ~

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Lesson 5.1.3 5-29. a: x = ±5 b: All numbers c: x = 2 d: No solution 5-30. Using cos A = 5

13 , sin A = 1213 , or tan A = 12

5 , A ≈ 67.4º 5-31. ≈11.5 seconds 5-32. a: 136 b: 2036 5-33. Area ≈ 294.55 sq m, perimeter ≈ 78.21 m 5-34. a: It uses circular logic. b: Reverse the arrow between“Marcy likes chocolate” and “Marcy likes Whizzbangs.”

Also, remove the arrow connecting “Marcy likes chocolate” and “Whizzbangs are 100% chocolate.”

5-35. 9.38 minutes

Lesson 5.1.4 5-41. All of the triangles are similar. They are all equilateral triangles. 5-42. Since tan(33.7°) ≈ 2

3 , y ≈ 23 x + 7 .

5-43. a: an = 108 +12(n −1) = 96 +12n b: an = 2

5 (2)n−1 = 1

5 (2)n

c: an = 3741− 39(n −1) = 3780 − 39n d: an = 117(0.2)n−1 = 585(0.2)n 5-44. a: sinθ = b

a b: tanθ = ab c: cosθ = a

b 5-45. a: 2252 ; union b: 352 ; intersection c: 1− 22

52 =3052

5-46. a: cos 23° = 18

x or 0.921= 18x

b: Since 67° is complementary to 23°, then sin 67° = cos 23° . So sin 67° ≈ 0.921 .

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Lesson 5.2.1 5-52. a: A = 1 sq. m, P = 2 + 2 2 m

b: A = 25 32 ≈ 21.65 square ft, P = 15 + 5 3 ≈ 23.66 ft

5-53. a: y = 111º ; x = 53º b: y = 79º ; x = 47º

c: y = 83º ; x = 53º d: y = 3; x = 3 2 units 5-54. a: 4 2 units; Use the Pythagorean Theorem or that it is a 45°- 45°- 90° triangle. b: It is a trapezoid; 24 square units 5-55. 10.1% by using the Addition Rule. 5-56. a: Answers vary; sample responses: x < 3, x is even, etc. b: The length of each leg is 6 units. 5-57. a: an = 500 +1500(n −1) = 1500n −1000 b: an = 30(5)

n−1 = 6(5)n 5-58. a: Not similar b: Similar: Rotate ΔGHI, translate, then dilate. c: Similar: Reflect ΔMNP, translate, then dilate. d: Similar: Rotate ΔTUV, translate, then dilate.

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Lesson 5.2.2 5-64. a: 16 inches b: 4 yards and 4 2 yards

c: 24 feet d: 10 meters and 10 3 meters 5-65. a: m∠A = 35º, m∠B = 35º, m∠ACB = 110º, m∠D = 35º, m∠E = 35º, m∠DCE = 110º b: Answers vary. Once all the angles have been found, state that two pairs of

corresponding angles have equal measure, such as m∠A = m∠D and m∠B = m∠E to reach the conclusion that ΔABC ~ ΔDEC by AA ~ or AC = BC and DC = EC, so ACDC = BC

EC and m∠ACB = m∠DCE therefore ΔABC ~ ΔDEC SAS ~.

c: They are both correct. Since both triangles are isosceles, we cannot tell if one is the reflection or the rotation of the other (after dilation).

5-66. cos 52° = b

c ; tan 52° = ab ; cos 38° = a

c ; cos 38º= sin 52º 5-67. 14

27 =x40 , x ≈ 20.74 inches

5-68. a: explicit b: an = −3+ 4(n −1) = 4n − 7 c: a50 = 193 d: an = 3− 1

3 (n −1) = 313 −

13 n

5-69. $3(135360 )+ $5(

135360 )+ (−$6)(

90360 ) = $1.50 . It is not fair because the expected value is not 0.

5-70. a: 12 b: 0 c: 34 d: 1

Lesson 5.3.1 5-77. ≈ 61° 5-78. a: Impossible because a leg is longer than the hypotenuse. b: Impossible because the sum of the angles is more than 180°. 5-79. William is correct. 5-80. a: ′A (–3, –6), ′B (–5, – 4), ′C (0, – 4) b: ′′A (3, 3), ′′B (1, 1), ′′C (1, 6) 5-81. a: x = 16

5 b: No solution c: x = –11 or 3 d: x = 288 5-82. b is correct; if two sides of a triangle are congruent, the angles opposite them must be

equal.

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Lesson 5.3.2 5-89. They must have equal length. Since a side opposite a larger angle must be longer than a

side opposite a smaller angle, sides opposite equal angles must be the same length. 5-90. ≈ 10.6 mm 5-93. 72 square units 5-92. 12.6 5-93. See tree diagram at right (an

area model is not practical). P(three yogurts) = 12.5%. 100%−12.5% = 87.5% chance of not getting three yogurts.

5-94. a: x = 45

4 = 11.25

b: x = −10 or x = 10 c: x = 1.3 d: No solution 5-95. (−2, 4)

yogurt

green apple

red apple

yogurt

green apple

red apple

yogurt

green apple

red apple

yogurt

green apple

red apple yogurt

green apple

red apple

yogurt

green apple

red apple

yogurt

green apple

red apple

yogurt

green apple

red apple yogurt

green apple

red apple

yogurt

green apple

red apple

yogurt

green apple

red apple

yogurt

green apple

red apple yogurt

green apple

red apple

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Lesson 5.3.3 5-100. a: 29° b: cos 29° = y

42 , y ≈ 36.73 5-101. a: (–1, –2) b: (4, – 4) c: (3, 4) 5-102. tan−1( 34 ) ≈ 36.87° 5-103. sin−1 78 ≈ 61.0º 5-104. a: 2x2 + 6x b: 3x2 − 7x − 6 c: x = 1 or 7 d: y = –3 or 5 5-105. a: 112 b: 13

Lesson 5.3.4 5-111. a: The diagram should be a triangle with sides marked 116 ft. and 224 ft. and the angle

between them marked 58°. b: ≈ 190 feet, Law of Cosines 5-112. a: Corresponding angles have equal measure. b: The ratio of corresponding sides is constant, so corresponding sides are proportional. 5-113. y = (tan 25°)x + 4 or y ≈ 0.466x + 4 5-114. It must be longer than 5 and shorter than 23 units. 5-115. 31 terms 5-116. 3

12 (3)+712 (−1)+

212 (10) =

116 ≈ $1.83

The game is not fair because the expected value is not zero. 5-117. 7 years

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Lesson 5.3.5 5-126. The third side is 12.2 units long. The angle opposite the side of length 10 is

approximately 35.45°, while the angle opposite the side of length 17 is approximately 99.55°.

5-127. x ≈11.3 units; Methods include using the Pythagorean Theorem to set up the equation

x2 + x2 = 162 , using the 45°- 45°- 90° triangle shortcut to divide 16 by 2 , or to use sine or cosine to solve using a trigonometric ratio.

5-128. No, because to be a rectangle, the parallelogram needs to have 4 right angles.

Counterexample: A parallelogram without 4 right angles. 5-129. a: P ≈ 40.32 mm, A = 72 sq mm b: P = 30 feet, A = 36 square feet 5-130. A(2, 4), B(6, 2), C(4, 5) 5-131. The expected value per throw is 14 (2)+

14 (3)+

12 (5) =

154 = 3.75 , so her expected

winnings over 3 games are 3(3.75) = 11.25; yes, she should win enough tickets to get the panda bear.

5-132. y = 3

4 x + 4 5-133. a: m∠ABE = 80º, m∠EBC = 60º, m∠BCE = 40º, m∠ECD = 80º, m∠DEC = 40º,

m∠CEB = 80º, m∠BEA = 60º b: 360° 5-134. a: ≈ 8.64 cm b: PS = SR = 5.27 cm, so the perimeter is ≈ 25.5 cm 5-135. Area ≈ 21.86 sq. units, perimeter ≈ 24.59 units 5-136. a: Explicit t(n) = −2 + 3n ; Recursive t(0) = −2, t(n +1) = t(n)+ 3

b: Explicit t(n) = 6(12 )n ; Recursive t(0) = 6, t(n +1) = 1

2 t(n) c: t(n) = 24 − 7n

d: t(n) = 5(1.2)n e: t(4) = 1620 5-137. a: See diagram at right. b: x = 10 3

3 ≈ 5.77 5-138. a: 5 + 20 + 37 ≈15.55 units b: ≈ 31.11 c: (–2, 0)

Chain

Shed Bush

x x