Appendix A — Answer Key, Chapter 13: Circles: Central and Inscribed Angles

SOL G.PC.3 (b, c) · Covers textbook Chapter 13 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 126 across the chapter.

Conventions used in every answer below. Arcs are measured in degrees only — arc length is Chapter 14. A minor arc takes two letters and a major arc takes three. Arcs add, and the arcs around a circle total $360°$. Congruent arcs require one circle (or congruent circles): equal measure is not equal size.

Everything else in this chapter is the second row applied again — same arc, semicircle, inscribed quadrilateral. Before using either rule, find the vertex. That single decision is where nearly every wrong answer below would come from.

Lesson 13.1 — Circles and Arcs

Guided practice

1. A radius.
2. It passes through the centre.
3. A diameter is twice a radius: $d = 2r$.
4. The minor arc $\widehat{AB}$, measuring $110°$.
5. The major arc $\widehat{AXB}$, measuring $250°$. It needs three letters because two letters always names the minor arc, so a third point is needed to say which way round you are going.
6. $50°$, $110°$, $100°$, $100°$ — totalling $360°$.

Independent practice

7. A diameter.
8. $2(9) =$ $18$.
9. $26 \div 2 =$ $13$.
10. Minor — it is less than $180°$.
11. $360 - 74 =$ $286°$.
12. $360 - 128 =$ $232°$.
13. $360 - (70 + 140 + 60) = 360 - 270 =$ $90°$.
14. $360 - (45 + 135 + 90) = 360 - 270 =$ $90°$.
15. $360 - (100 + 60 + 85) = 360 - 245 =$ $115°$.
16. $65 + 85 =$ $150°$.
17. $40 + 75 + 95 =$ $210°$.
18. $180°$. An arc is a semicircle exactly when its endpoints are the ends of a diameter.
19. $360 \div 8 =$ $45°$.
20. Two letters always names the minor arc, and a minor arc is less than $180°$ — so "$\widehat{AB} = 250°$" contradicts its own notation. The fix: name a point $X$ on the long way round and write $\widehat{AXB} = 250°$.
21. Because measure is a fraction of a full turn, not a distance. Doubling a circle's radius doubles the length of every arc and changes no arc's measure.
22. Because together they cover the whole circle exactly once, with no overlap and no gap — and one full turn is $360°$.

Exit ticket 13.1

23. $360 - 96 =$ $264°$.
24. $360 - (110 + 85 + 75) = 360 - 270 =$ $90°$.

Lesson 13.2 — Central Angles

Guided practice

25. The angle is $110°$ and the arc is $110°$ — a central angle equals its arc.
26. Because arc measure is defined as the central angle that opens onto the arc. There is nothing prior to prove it from; the equality is what "arc measure" means.
27. Both central angles are $75°$; both arcs are $75°$.
28. In one circle: congruent central angles $\Leftrightarrow$ congruent arcs $\Leftrightarrow$ congruent chords.
29. Because across circles of different sizes, equal central angles still give equal arc measures — but the arcs and chords are different sizes, and congruence is about size. The chain is about one circle at a time.
30. $\dfrac{360°}{n}$.

Independent practice

31. $75°$.
32. $128°$.
33. $64°$.
34. $95°$.
35. $360 - 110 =$ $250°$.
36. $360 - 145 =$ $215°$.
37. $360 \div 6 =$ $60°$.
38. $360 \div 10 =$ $36°$.
39. $360 \div 9 =$ $40°$.
40. $68°$ — congruent central angles cut congruent arcs.
41. $4y + 5y + 3y = 360 \Rightarrow 12y = 360 \Rightarrow y =$ $30$, so the arcs are $120°$, $150°$, $90°$. (Check: they total $360$.)
42. $2x + 3x + 4x + 6x = 360 \Rightarrow 15x = 360 \Rightarrow x =$ $24$, so the arcs are $48°$, $72°$, $96°$, $144°$. (Check: they total $360$.)
43. $\widehat{AC} = 50 + 110 = 160°$, and a central angle equals its arc, so $m\angle AOC =$ $160°$.
44. $360 \div 8 =$ $45°$ between neighbouring cars, and a car travels a $45°$ arc between two stops.
45. $\dfrac{54}{360} = \dfrac{3}{20}$ — the category is $\tfrac{3}{20}$ of the whole.
46. Equal measure is not congruence. Both arcs are $70°$ of their own circle, but the merry-go-round's arc is enormously longer. Congruent arcs require congruent circles.
47. Because one full turn around the centre is $360°$, and a central angle is part of a single turn. An angle of exactly $180°$ cuts off a semicircle, and its two sides together form a diameter.
48. Their arcs are congruent and their central angles are congruent — the three-way chain runs in every direction, so any one of the three gives the other two.

Exit ticket 13.2

49. $\widehat{AB} =$ $84°$, and the major arc is $360 - 84 =$ $276°$.
50. $360 \div 5 =$ $72°$.

Lesson 13.3 — Inscribed Angles

Guided practice

51. The arc is $110°$ and the inscribed angle is $55°$.
52. The central angle on the same arc is $110°$, so central : inscribed is $2 : 1$.
53. The vertex is on the circle, and both sides are chords.
54. Because $\overline{OV}$ and $\overline{OP}$ are both radii of the same circle, so they are congruent — which is the definition of an isosceles triangle.
55. The Exterior Angle Theorem — an exterior angle of a triangle equals the sum of the two remote interior angles.
56. Because it is a central angle, and a central angle equals its arc.

Independent practice

57. $\tfrac12(110) =$ $55°$.
58. $\tfrac12(86) =$ $43°$.
59. $\tfrac12(74) =$ $37°$.
60. $\tfrac12(208) =$ $104°$.
61. $\tfrac12(96) =$ $48°$.
62. $2(34) =$ $68°$.
63. $2(63) =$ $126°$.
64. $2(27) =$ $54°$.
65. $2(48) =$ $96°$.
66. $2(70) =$ $140°$.
67. Central $128°$; inscribed $\tfrac12(128) =$ $64°$.
68. Central $250°$; inscribed $\tfrac12(250) =$ $125°$.
69. Arc $2(55) =$ $110°$; central angle $110°$.
70. $3x + 5 = \tfrac12(8x - 10) = 4x - 5$, so $x =$ $10$. The angle is $35°$ and the arc is $70°$. (Check: $\tfrac12(70) = 35$. ✓)
71. $2x = \tfrac12(5x - 30)$, so $4x = 5x - 30$ and $x =$ $30$. The angle is $60°$ and the arc is $120°$. (Check: $\tfrac12(120) = 60$. ✓)
72. $\tfrac12(74) =$ $37°$.
73. They halved when they should have doubled. Inscribed $=$ half the arc, so the arc is twice the angle: $2(40) =$ $80°$.
74. They used the central rule. A vertex on the circle makes the angle inscribed, so it is half the arc, not equal to it.
75. Because the angle is half its arc, and an arc is at most the whole circle. Half of $360°$ is $180°$ — and an arc of exactly $360°$ would put both sides of the angle on the same ray, which is not an angle at all. So every genuine inscribed angle is less than $180°$.
76. Draw the diameter through the vertex. It splits any inscribed angle into one or two angles of the diameter case, and it splits the arc the same way. Adding the two halved results (or subtracting, when the centre falls outside the angle) gives the general statement, because half of a sum is the sum of the halves.

Exit ticket 13.3

77. $\tfrac12(96) =$ $48°$.
78. $2(63) =$ $126°$.

Lesson 13.4 — Three Consequences

Guided practice

79. The arc is $140°$ and both angles are $70°$. They are equal because both are half the same arc.
80. Nothing changes — every vertex on the major arc gives $70°$.
81. Arc $180°$, angle $90°$, from $\tfrac12(180) = 90$.
82. $\overline{MN}$ must be a diameter — that is what makes the arc a semicircle.
83. $105 + 75 =$ $180°$ and $100 + 80 =$ $180°$.
84. Because adjacent angles stand on arcs that overlap rather than on the two arcs that together make the whole circle. Their halves therefore add to nothing fixed — here $105 + 100 = 205$.

Independent practice

85. $34°$ — both are half the same arc.
86. $52°$, and the arc is $2(52) =$ $104°$.
87. $90°$ — inscribed in a semicircle.
88. The angle at $V$ is $90°$, so $180 - 90 - 35 =$ $55°$.
89. $180 - 90 - 52 =$ $38°$.
90. $180 - 112 =$ $68°$ — $P$ and $R$ are opposite.
91. $180 - 95 =$ $85°$.
92. $m\angle R = 180 - 105 =$ $75°$ and $m\angle S = 180 - 100 =$ $80°$.
93. $(2x + 10) + (3x + 20) = 180 \Rightarrow 5x = 150 \Rightarrow x =$ $30$, giving $70°$ and $110°$. (Check: $70 + 110 = 180$. ✓)
94. Each angle is half the sum of the two arcs not touching it: $\tfrac12(90 + 110) = 100°$, $\tfrac12(110 + 90) = 100°$, $\tfrac12(90 + 70) = 80°$, $\tfrac12(70 + 90) = 80°$. So $100°$, $100°$, $80°$, $80°$ — and $100 + 80 = 180$ both ways. ✓
95. $\tfrac12(140 + 90) = 115°$, $\tfrac12(90 + 90) = 90°$, $\tfrac12(90 + 40) = 65°$, $\tfrac12(40 + 140) = 90°$. So $115°$, $90°$, $65°$, $90°$ — and $115 + 65 = 180$, $90 + 90 = 180$. ✓
96. The arc is $2(90) =$ $180°$, which is a semicircle — so the chord joining its endpoints is a diameter.
97. Draw any chord through the centre — that is a diameter. Join each of its ends to any third point on the rim. The angle at that third point is $90°$, because it is inscribed in a semicircle and $\tfrac12(180) = 90$. No protractor is needed.
98. Yes, that is possible. Opposite angles of an inscribed quadrilateral must be supplementary, and $88 + 92 = 180$. ✓
99. Proximity is irrelevant. Two inscribed angles are congruent when they stand on the same arc, because both are half the same number. Two inscribed angles standing on different arcs are generally not equal, however close their vertices happen to be.
100. Only opposite angles are supplementary. Adjacent angles stand on overlapping arcs and have no fixed sum.
101. All three are one line of arithmetic from inscribed $=$ half the arc. Same arc: both angles are half the same number, so they are equal. Semicircle: the arc is $180°$, and half of that is $90°$. Inscribed quadrilateral: opposite angles stand on the two arcs that together make $360°$, so their halves total $180°$.
102. One side is a diameter, so it cuts off a semicircle of $180°$. The third vertex lies on the circle, so the angle there is inscribed and equals $\tfrac12(180) = 90°$. The right angle is at the vertex opposite the diameter.

Exit ticket 13.4

103. $90°$ — inscribed in a semicircle.
104. $180 - 43 =$ $137°$.

Lesson 13.5 — Putting It Together

Guided practice

105. $110°$ and $55°$. The only thing that differs is where the vertex sits — the arc and its endpoints are identical.
106. At the centre → the angle equals the arc. On the circle → the angle is half the arc.
107. Central and inscribed. The rest are the halving rule restated for a particular arc, so a student who forgets them can rebuild each in one line.
108. The inscribed rule, with the arc equal to $180°$.
109. $360 \div 8 =$ $45°$.
110. Draw a diameter, then join both of its ends to any third point on the rim. That angle is inscribed in a semicircle, so it is exactly $90°$.

Independent practice

111. Central — $128°$.
112. Inscribed — $\tfrac12(128) = 64°$.
113. Inscribed — $\tfrac12(180) = 90°$.
114. Central — $210°$.
115. Inscribed — $\tfrac12(74) = 37°$.
116. $\widehat{AC} = 50 + 110 = 160°$, so the inscribed angle is $\tfrac12(160) =$ $80°$.
117. The arc is $2(63) = 126°$, so the central angle is $126°$.
118. The arc is $96°$, so the inscribed angle is $48°$.
119. The fourth arc is $360 - 245 = 115°$, so the inscribed angle standing on it is $\tfrac12(115) =$ $57.5°$. (An arc need not give a whole-number half.)
120. Central $360 \div 12 =$ $30°$; inscribed $\tfrac12(30) =$ $15°$.
121. The centre sensor's angle is the central angle, $146°$. The rim sensor's is inscribed, $\tfrac12(146) =$ $73°$.
122. They confused the two rules — halving belongs to the inscribed angle. A central angle equals its arc, so there is no halving to do.
123. Look at the vertex. At the centre, with two radii as sides → central, and the angle equals the arc. On the circle, with two chords as sides → inscribed, and the angle is half.
124. Because the inscribed angle is exactly half the central angle on the same arc, and a number equals its own half only when it is zero — which is not an angle. So on any real arc the two measures always differ.

Exit ticket 13.5

125. Inscribed, so $\tfrac12(208) =$ $104°$.
126. "Where is the vertex — at the centre, or on the circle?"

Chapter 13 Review — answers

Review 1 (G.PC.3b).

• $360 - (60 + 120 + 80) =$ $100°$. The fact used is that the arcs around a circle total $360°$.
• The four central angles are $60°$, $120°$, $80°$, $100°$ — a central angle equals its arc.
• Across both: $60 + 120 =$ $180°$. Careful: what is left is $360 - 180 = 180°$ as well, so neither piece is a major arc — both are semicircles. A major arc has to be more than $180°$, and here the circle has been split exactly in half.
• Doubling the radius changes no arc measure and no angle — all of them are fractions of a full turn. What doubles is every arc's length and every chord's length, which is Chapter 14's subject.

Review 2 (G.PC.3c). $\overline{AB}$ is a diameter, $\widehat{AC} = 74°$.

• $m\angle ACB =$ $90°$, because $\overline{AB}$ is a diameter, so $\widehat{AB}$ is a semicircle of $180°$, and the inscribed angle standing on it is $\tfrac12(180) = 90$.
• $\widehat{CB} = 180 - 74 =$ $106°$ — the two arcs make up the semicircle.
• $m\angle ABC = \tfrac12 \widehat{AC} = \tfrac12(74) =$ $37°$, by the inscribed angle rule.
• The third angle is $m\angle BAC = \tfrac12 \widehat{CB} = \tfrac12(106) = 53°$, and $90 + 37 + 53 =$ $180°$. ✓ The inscribed rule was used at all three steps — on the $180°$ arc, on $\widehat{AC}$, and on $\widehat{CB}$. The central rule was not needed at all.

Review 3 (G.PC.3 b, c). Arcs $50°$, $110°$, $100°$, $100°$.

• They total $50 + 110 + 100 + 100 =$ $360°$, so they are possible. ✓
• Each angle is half the sum of the two arcs not touching it: $m\angle P = \tfrac12(110 + 100) =$ $105°$, $m\angle Q = \tfrac12(100 + 100) =$ $100°$, $m\angle R = \tfrac12(100 + 50) =$ $75°$, $m\angle S = \tfrac12(50 + 110) =$ $80°$.
• $105 + 75 =$ $180°$ and $100 + 80 =$ $180°$. ✓ Both pairs of opposite angles are supplementary.
• $m\angle QOR = \widehat{QR} =$ $110°$, because $O$ is the centre. $m\angle QPR = \tfrac12(110) = 55°$ — half as much, because $P$ is on the circle rather than at the centre. Same arc, same endpoints, different vertex.