Appendix A — Answer Key, Chapter 4: Sides and Angles of a Triangle

SOL G.TR.1 (a, b, c, d, e) · Covers textbook Chapter 4 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 108 across the chapter.

Conventions used in every answer below: in $\triangle ABC$, side $a$ is $\overline{BC}$, side $b$ is $\overline{AC}$, and side $c$ is $\overline{AB}$ — each lowercase letter names the side opposite its vertex. Ordering answers are written as chains, and a tie is written with an equals sign rather than forced into a strict order. The Triangle Inequality is strict: a sum equal to the third length is the degenerate case and is not a triangle.

The relationships used throughout:

Lesson 4.1 — The Angle Sum and the Exterior Angle

Guided practice

1. $54°$, $71°$, and $55°$. They sum to $180°$.
2. The Alternate Interior Angles Theorem, used twice — once for $\angle A$ and once for $\angle B$.
3. Only a parallel line makes the angles at $C$ alternate interior angles with the angles at $A$ and $B$. Any other line through $C$ would create angles with no relationship to the triangle's, and the proof would have nothing to substitute.
4. The exterior angle measures $118°$. Its remote interior angles are the $48°$ angle at $A$ and the $70°$ angle at $C$ — the two not adjacent to it.
5. $48° + 70° = 118°$, which matches the exterior angle.
6. The peak measures $44°$. The two rafters are congruent (tick marks), so the base angles are equal; $180° - 44° = 136°$, and $136° \div 2 = 68°$ each.

Independent practice

7. a) $180° - 42° - 63° = 75°$ b) $180° - 90° - 37° = 53°$ c) $180° - 58° - 58° = 64°$
8. $180° - 132° = 48°$ (linear pair).
9. $105° - 47° = 58°$.
10. $x + (2x + 10) + (3x - 4) = 180 \Rightarrow 6x + 6 = 180 \Rightarrow x = 29$. The angles are $29°$, $68°$, and $83°$; check $29 + 68 + 83 = 180$.
11. $5x + 6x + 7x = 180 \Rightarrow 18x = 180 \Rightarrow x = 10$. The angles are $50°$, $60°$, and $70°$ — all less than $90°$, so the triangle is acute.
12. $5y + 10 = 3y + (y + 26) \Rightarrow 5y + 10 = 4y + 26 \Rightarrow y = 16$. The exterior angle measures $5(16) + 10 = 90°$; the remote interior angles are $48°$ and $42°$, and $48 + 42 = 90$ checks.
13. $(180° - 52°) \div 2 = 64°$ at each base.
14. $(180° - 105°) \div 2 = 37.5°$ each.
15. Two angles of $90°$ already total $180°$, leaving $0°$ for the third — and an angle of $0°$ means the two sides lie on top of each other, so there is no triangle. A triangle has at most one angle of $90°$ or more.
16. An exterior angle equals the sum of the two remote interior angles — the two not adjacent to it. It does not involve $\angle B$, which is its linear-pair partner. (The correct relationship with $\angle B$ is that the two are supplementary.)
17. The exterior angle equals the sum of the two remote interior angles, and both of those are positive. A sum of two positive numbers is greater than either one, so the exterior angle exceeds each of them.
18. The other two must sum to $60°$, so both are acute and each is less than $60°$. In particular neither can be right or obtuse.

Exit ticket 4.1

19. $180° - 71° - 46° = 63°$.
20. Adjacent interior angle: $180° - 118° = 62°$. Sum of the two remote interior angles: $118°$ — that is exactly what the Exterior Angle Theorem says.
21. No. $88 + 94 = 182 > 180$, which leaves nothing for the third angle.
22. An exterior angle of a triangle equals the sum of the two interior angles that are not adjacent to it.

Lesson 4.2 — Ordering the Sides from the Angles

Guided practice

23. Largest angle: $\angle B = 85°$. The side opposite it is $b$.
24. Smallest angle: $\angle A = 35°$. The side opposite it is $a$.
25. $a < c < b$.
26. Because the third angle is not known, and it might be the largest — as it turns out to be. Ordering two angles orders only two sides, and the question asks for all three.
27. $m\angle F = 180° - 57° - 44° = 79°$. Angles rank $\angle E < \angle D < \angle F$.
28. $e < d < f$.

Independent practice

29. a) $a < b < c$ b) $f < e < d$ c) $r < p = q$
30. a) $m\angle C = 60°$; angles $\angle A(55°) < \angle C(60°) < \angle B(65°)$, so $a < c < b$. b) $m\angle Z = 118°$; angles $\angle X(28°) < \angle Y(34°) < \angle Z(118°)$, so $x < y < z$. c) $m\angle L = 38°$; angles $\angle L(38°) < \angle K(52°) < \angle J(90°)$, so $l < k < j$.
31. $m\angle T = 180° - 47° - 47° = 86°$, the largest angle, so the longest side is the one opposite it: $\overline{RS}$.
32. $m\angle B = 180° - 25° - 115° = 40°$. The smallest angle is $\angle A$ at $25°$, so the shortest side is $\overline{BC}$.
33. The longest side is the one opposite the $89°$ angle. The triangle is not a right triangle — $89° \ne 90°$, so it is acute.
34. The exterior angle at $F$ is $130°$, so $m\angle F = 180° - 130° = 50°$, and the two remote interior angles satisfy $62° + m\angle E = 130°$, giving $m\angle E = 68°$. Check: $62 + 68 + 50 = 180$. Angles rank $\angle F < \angle D < \angle E$, so sides rank $f < d < e$.
35. The edge opposite the $95°$ angle.
36. The shortest side is the one opposite the $36°$ angle. The smallest angle always faces the shortest side.
37. The student ordered from two angles without finding the third. Here $m\angle C = 180° - 60° - 70° = 50°$, so $\angle C$ is the smallest, not $\angle A$, and the full ordering is $c < a < b$. It is true that $a < b$, but "$b$ is the longest" needed the third angle to confirm.
38. $\overline{AB}$ is adjacent to $\angle A$, not opposite it. The side opposite $\angle A$ is $\overline{BC}$. Knowing $m\angle A = 80°$ tells you about $\overline{BC}$, and nothing yet about which side is longest — the other two angles are still unknown.
39. Hold two sides of an angle fixed in length and open the angle wider. The two far endpoints move apart, so the segment joining them — the side opposite the angle — has to be longer. Closing the angle brings them together and shortens that side.
40. All three sides are equal: $a = b = c$. Equal angles face equal sides, so the triangle is equilateral as well as equiangular.

Exit ticket 4.2

41. $a < b < c$.
42. $m\angle R = 180° - 38° - 52° = 90°$. The longest side is the one opposite $\angle R$: side $r$, which is $\overline{PQ}$.
43. $\overline{XZ}$.
44. In any triangle, the longest side is opposite the largest angle and the shortest side is opposite the smallest angle.

Lesson 4.3 — Ordering the Angles from the Sides

Guided practice

45. $a = 9$, $b = 6$, $c = 5$.
46. Sides rank $c < b < a$, so angles rank $\angle C < \angle B < \angle A$.
47. $\angle A$ is largest, and it is opposite side $a$, the longest.
48. The two angles opposite the two congruent $7$-unit sides. Equal sides face equal angles.
49. Each is $60°$. All three sides are equal, so all three angles are equal, and three equal angles summing to $180°$ are $60°$ each.
50. Because the two $7$-unit sides are equal, so the angles opposite them are equal, not one larger than the other. The correct answer keeps the tie: $\angle C < \angle A = \angle B$.

Independent practice

51. a) $\angle A < \angle C < \angle B$ b) $\angle C < \angle A = \angle B$ c) $\angle A = \angle B = \angle C$
52. Side opposite $\angle A$ is $\overline{BC} = 8$; opposite $\angle B$ is $\overline{AC} = 10$; opposite $\angle C$ is $\overline{AB} = 13$. So $\angle A < \angle B < \angle C$.
53. Side opposite $\angle X$ is $\overline{YZ} = 12$; opposite $\angle Y$ is $\overline{XZ} = 12$; opposite $\angle Z$ is $\overline{XY} = 5$. So $\angle Z < \angle X = \angle Y$ — a tie between $\angle X$ and $\angle Y$.
54. The angle opposite the $29$-unit side.
55. Side opposite $\angle P$ is $\overline{QR} = 24$, the shortest, so $\angle P$ is the smallest.
56. Angles rank: opposite $9$ $<$ opposite $12$ $<$ opposite $15$. Since $9^2 + 12^2 = 81 + 144 = 225 = 15^2$, the triangle is a right triangle, with the right angle opposite the $15$.
57. All three sides equal, so $\angle D = \angle E = \angle F$, and each measures $60°$.
58. The widest opening is at the corner opposite the $30$-inch side. That side is across from it.
59. The angle opposite the $7$-foot side.
60. Two errors. First, the ordering is inverted: side $a = 8$ is the longest, so $\angle A$ is the largest, not the smallest. Second, $b = c = 5$, so $\angle B = \angle C$ — a tie, not a strict order. The correct answer is $\angle B = \angle C < \angle A$.
61. All three sides are equal, so by the ordering rule no angle can be larger than another — all three angles are equal. Three equal angles summing to $180°$ must each be $60°$.
62. The third side is not determined. Two sides of $7$ and $10$ can close with a third side anywhere from just over $3$ to just under $17$, and each choice produces a different set of angles with a different ordering. Without the third side you do not know which of the two given sides' opposite angles is larger relative to the third angle.

Exit ticket 4.3

63. Sides rank $b(4) < a(7) < c(9)$, so angles rank $\angle B < \angle A < \angle C$.
64. The two angles opposite the $6$-unit sides are equal, and both are smaller than the angle opposite the $10$. Writing the vertices so that $c = 10$: $\angle A = \angle B < \angle C$.
65. $\overline{AC}$ is opposite $\angle B$, so $\angle B$ is the largest.
66. In any triangle the largest angle is opposite the longest side. It is the converse of the rule in Lesson 4.2 — same relationship, read in the other direction — and unlike most converses, this one is also true.

Lesson 4.4 — Does a Triangle Exist?

Guided practice

67. Lengths $4$, $5$, $6$. The comparison shown is $4 + 5 = 9 > 6$.
68. The two shorter pieces total only $7$, and the gap they have to close is $9$. Swinging them from the ends of the $9$ leaves their far ends $2$ units apart, so the arcs cannot meet.
69. Three lengths make a triangle only when every pair of them adds to more than the remaining one.
70. If the two shortest add to more than the longest, then the longest plus either other length certainly exceeds the remaining one — the longest alone is already at least as big as it. So the other two comparisons cannot fail once the first one passes.
71. $3 + 5 = 8$, which is exactly the long side. The two short pieces lie flat along the long one, so the figure is a segment: no height, no area, no angles.
72. Because the equal case is the degenerate one. Allowing "greater than or equal to" would count flattened figures as triangles, and they are not.

Independent practice

73. a) $5 + 7 = 12 > 11$ ✓ yes b) sorted $3$, $4$, $8$: $3 + 4 = 7 < 8$ ✗ no c) $10 + 10 = 20 > 19$ ✓ yes d) $6 + 6 = 12$, equal to $12$ ✗ no — the degenerate case
74. a) $2.5 + 3.5 = 6 > 5$ ✓ yes b) $14 + 9 = 23 > 22$ ✓ yes c) $1 + 1 = 2 > 1$ ✓ yes
75. The range is $9 - 4 = 5$ to $9 + 4 = 13$, so $5 < x < 13$. Only $12$ qualifies. ($4$ and $5$ are too small — $5$ is the excluded endpoint — and $13$ is the excluded upper endpoint.)
76. The range is $0 < x < 22$, so $1$ and $21$ qualify. $22$ is the excluded endpoint and $23$ is too long.
77. The range is $15 - 8 = 7 < x < 23$, so the smallest whole number is $8$.
78. The largest whole number is $22$.
79. Both comparisons the student checked involve the longest side ($10$) and therefore cannot fail. The comparison that decides it is $6 + 8 = 14 > 10$ ✓, so a triangle does exist.
80. Sorted: $18$, $25$, $46$. $18 + 25 = 43 < 46$ ✗ — no, the boards cannot form a triangle.
81. $12 + 16 = 28$, exactly equal to the long rope, so it is the degenerate case and will not work. The $28$-foot rope must be shortened to anything less than $28$ feet — cutting even a small amount off, say to $27$ feet, is enough ($12 + 16 = 28 > 27$).
82. $4 + 13 > 9$ is a comparison involving the longest side, so it cannot fail and decides nothing. The comparison that matters is $4 + 9 = 13$, which is equal to $13$, not greater. So no triangle exists — this is the degenerate case.
83. The two congruent sides total $s + s = 2s$. For a triangle, that total must be greater than the third side. If the third side is $2s$, the total merely equals it (degenerate); if it is more than $2s$, the total falls short. Either way, no triangle.
84. The two shorter sides swing from the ends of the longest and meet at exactly one point — a point that lies on the longest side. The "triangle" has zero height and zero area, and its three vertices are collinear, so there are no angles to measure. "Degenerate" names a figure that has collapsed out of the category it was heading for.

Exit ticket 4.4

85. $6 + 7 = 13 < 14$ ✗ — no.
86. $9 + 12 = 21 > 20$ ✓ — yes.
87. $5 + 5 = 10$, exactly equal to the third length ✗ — no. This is the degenerate boundary case: the two $5$s lie flat along the $10$.
88. Sort the three lengths and check whether the two shorter ones sum to more than the longest. The other two comparisons each involve the longest length, which is already at least as big as either of the others, so they can never fail.

Lesson 4.5 — The Range for a Third Side, and Problems in Context

Guided practice

89. Two sides of $7$ and $10$. The range is $3 < x < 17$.
90. Because a third side of exactly $3$ or exactly $17$ produces the degenerate case — the three points would be collinear and the triangle would flatten. Neither endpoint is allowed, so both circles are open.
91. $4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16$ — $13$ values.
92. $m\angle C = 180° - 62° - 74° = 44°$.
93. The anchor line, which is the side opposite $\angle C$. $\angle C$ is the smallest angle, and the smallest angle faces the shortest side.
94. $m\angle C$ would become $180° - 70° - 74° = 36°$. $\angle C$ would still be the smallest angle, so the shortest side would still be the anchor line.

Independent practice

95. a) $4 < x < 16$ b) $0 < x < 8$ c) $7 < x < 37$
96. a) $3 < x < 19$; whole numbers $4$ through $18$, which is $15$ values. b) $8 < x < 18$; whole numbers $9$ through $17$, which is $9$ values.
97. Range $7 < x < 25$, so the smallest whole number is $8$ and the largest is $24$.
98. $19 - 12 = 7$ and $19 + 12 = 31$, so $7 < x < 31$. Whole numbers $8$ through $30$: $23$ values.
99. The longest edge is the one opposite the $84°$ angle. The two edges opposite the $48°$ angles are equal to each other.
100. Range $14 - 9 = 5 < x < 23$, so the possible whole-inch crosspieces are $6, 7, 8, \ldots, 22$.
101. $5.2 - 3.5 = 1.7$ and $5.2 + 3.5 = 8.7$, so the third leg must satisfy $1.7 < x < 8.7$ km.
102. $m\angle C = 180° - 96° - 42° = 42°$. Angles rank $\angle B = \angle C < \angle A$, so the sides rank $b = c < a$ — as segments, $\overline{AC} = \overline{AB} < \overline{BC}$. Shortest to longest: $\overline{AC}$ and $\overline{AB}$ tied, then $\overline{BC}$.
103. The student wrote the interval between the two given sides instead of between their difference and their sum. The correct range is $9 - 7 = 2 < x < 16 = 9 + 7$.
104. Picture the longer side lying flat, with the shorter one swinging from one of its ends and the unknown third side swinging from the other. The shorter side can at best reach back toward the far end, closing the gap by its own length — so the unknown side has at least the leftover distance to cover, which is the difference of the two known lengths. Anything shorter than that difference cannot bridge the gap, and exactly the difference lays the pieces flat.

Exit ticket 4.5

105. $11 - 6 = 5$ and $11 + 6 = 17$, so $5 < x < 17$.
106. Whole numbers $6$ through $16$: $11$ values.
107. Angles rank $27° < 63° < 90°$, so the sides rank: shortest opposite the $27°$ angle, then the side opposite $63°$, then the side opposite the $90°$ angle (the hypotenuse) as the longest.
108. (i) the third-side range, $\lvert p - q \rvert < x < p + q$; (ii) the ordering rule, angles → sides — the longest side is opposite the largest angle; (iii) the Triangle Inequality — the two shorter lengths must sum to more than the longest.

Chapter 4 Review

Review 1 (G.TR.1 a, e). $m\angle A = 41°$, exterior angle at $C$ is $118°$.

• $m\angle C = 180° - 118° = 62°$ — the exterior angle and $\angle C$ form a linear pair. Then $m\angle B = 118° - 41° = 77°$ by the Exterior Angle Theorem (the exterior angle at $C$ equals the sum of the remote interior angles $\angle A$ and $\angle B$). Check: $41 + 77 + 62 = 180$.
• Angles: $\angle A(41°) < \angle C(62°) < \angle B(77°)$. Sides: $a < c < b$.
• The longest side is $b$, which is $\overline{AC}$.
• The two pieces given were one interior angle and one exterior angle — that is only enough to place two of the three interior angles until the third is computed. Ordering all three sides requires all three angles, and $\angle B$, once found at $77°$, turned out to be the largest. Ordering from $\angle A$ and $\angle C$ alone would have named $\angle C$ the largest and $c$ the longest side, which is wrong.

Review 2 (G.TR.1 b, c, d). Two sides measure $13$ and $6$.

• $13 - 6 = 7$ and $13 + 6 = 19$, so $7 < x < 19$. On a number line, open circles at $7$ and at $19$ with the segment between them shaded.
• Whole numbers $8$ through $18$: $8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18$ — $11$ values.
• $x = 13$ makes the triangle isosceles, with sides $13$, $13$, and $6$. (There is no way to match the $6$: a third side of $6$ would need $6 + 6 = 12 > 13$, which is false, and $6$ is below the range anyway.) The two angles opposite the $13$s are equal, and the angle opposite the $6$ is the smallest: writing the vertices so that $c = 6$, the ordering is $\angle C < \angle A = \angle B$.
• For $x = 8$ the sides are $13$, $8$, $6$. Angles rank: the angle opposite $6$ is smallest, then the angle opposite $8$, then the angle opposite $13$ is largest.

Review 3 (G.TR.1 e). A triangular brace with corner angles $53°$ and $89°$.

• The third angle is $180° - 53° - 89° = 38°$.
• Shortest to longest: the side opposite the $38°$ angle, then the side opposite $53°$, then the side opposite $89°$.
• Boards of $4$, $9$, and $14$ ft: sorted, $4 + 9 = 13 < 14$ ✗ — no, those three lengths cannot form any triangle.
• Keeping $4$ ft and $9$ ft, the third board must satisfy $9 - 4 = 5 < x < 13 = 9 + 4$. The whole-number options are $6, 7, 8, 9, 10, 11, 12$ ft.

Workbook-only items

Page 2, fill in the blanks. The three interior angles of a triangle sum to $180°$. A triangle has at most one angle of $90°$ or more.

Page 3, proof frame. 1. Draw the line through $C$ parallel to $\overline{AB}$ — through a point not on a line, exactly one parallel line exists. 2. and 3. Alternate interior angles are congruent. 4. Definition of a straight angle. 5. Substitution.

Page 4, fill in the blanks. An exterior angle equals the sum of its two remote interior angles. The exterior angle and its adjacent interior angle form a linear pair, so they are supplementary.

Page 6, fill in the blank. The longest side is opposite the largest angle; the shortest side is opposite the smallest angle.

Page 6, opposite-side frame. In $\triangle ABC$: side $a$ is $\overline{BC}$, side $b$ is $\overline{AC}$, side $c$ is $\overline{AB}$.

Page 9, fill in the blank. Before ordering, find the third angle using the angle sum.

Page 11, tie frame. Isosceles: two congruent sides face two congruent angles, so the ordering carries an equals sign. Equilateral: all sides equal, so all angles equal $60°$.

Page 13, fill in the blank. Sort the three lengths and check that the two shorter ones sum to more than the longest.

Page 14, degenerate frame. When the two shorter lengths sum to exactly the longest, the figure flattens: no height, no area, no angles. That is why the inequality is strict.

Page 16, range frame. $\lvert p - q \rvert < x < p + q$ — greater than the difference, less than the sum. Both endpoints are open.

Page 18, tool-choice table. Two angles → angle sum. Exterior angle and one remote interior → Exterior Angle Theorem. Angles, which side is longest → ordering angles to sides. Sides, which angle is largest → ordering sides to angles. Three lengths, will it work → Triangle Inequality. Two lengths, what could the third be → third-side range.

Pages 7, 12, and 17, blank frames. Any assigned triangle or number-line graph. Expected conventions: mark congruent sides with matching ticks and congruent angles with matching arcs; write an ordering as a chain with an equals sign wherever there is a tie; graph a third-side range with open circles at both endpoints.