1. 4. a. BAD and BCD; ABC and ADC; DCB and DAB; ADC and ABC
Inscribed Angles
GEOMETRY LESSON 11-3
12. a = 85; b = 47.5; c = 90
13. a = 50; b = 90; c = 90
14. p = 90; q = 122
15. w = 123
16. x = 65; y = 130
17. e = 65; f = 130
18. a = 26; b = 64; c = 42
6. 180
7. a = 218; b = 109
8. a = 54; b = 30; c = 96
9. a = 112; b = 120; c = 38
10. a = 101; b = 67; c = 84; d = 80
11. x = 36; y = 36
Pages Exercises
1. ACB; AB
2. RQS; RS
3. MPN; MN
4. a. BAD and BCD; ABC and ADC; DCB and DAB; ADC and ABC
b. ABC and BCD; obtuse
5. 58
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2. Inscribed Angles 19. a = 22; b = 78; c = 156
GEOMETRY LESSON 11-3
19. a = 22; b = 78; c = 156
20. a = 30; b = 60; c = 62; d = 124; e = 60
21. a. 96
b. 55
c. 77
d. 154
22. a. 40
b. 50
c. 40
d. 40
e. 65
23. a. 30
b. 78
c. 95
d. 105
e. 85
f. 75
24. a. 64
b. 116
c. 58
d. 90
e. 90
f. 58
PQR QRS since they are alt. int Since inscribed intercept arcs, mPR = mQS. s
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3. b. isosc. trapezoid; justifications may vary.
Inscribed Angles
GEOMETRY LESSON 11-3
26. a. Check students’ work.
b. isosc. trapezoid; justifications may vary.
Sample: arcs between two || chords are .
27. a. 77
b. 36
28. Rectangle; || chords are equidist. from center. 1 7
29. about 7.1 cm by 7.1 cm
30. about 4.3 cm for each side
31. about 7.1 cm legs, and a 10 cm base
32. Answers may vary. Sample:
a. If the cameras’ lenses open at = , then in the positions shown they share the same arc of the scene.
b. No; the distances from each position of the scene to each camera affect the look of the scene. s
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4. 37. a. CFE and FEG; ECD and EGD
Inscribed Angles
GEOMETRY LESSON 11-3
37. a. CFE and FEG; ECD and EGD
b. CED
c. EFG and EDG; FED and FGD
38.
33. false
34. true
35. true
ACB is a rt. because it is inscribed in semicircle ACB, and if a line is to a radius at its endpoint, it is tangent to the circle.
39. In the construction below, RS = x, ST = y, O is the mdpt. of RT, and QS RT.
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5. 40. 1. O with inscribed ABC (Given) 2. m ABO = mAP; m OBC = mPC
Inscribed Angles
GEOMETRY LESSON 11-3
40. 1.  O with inscribed ABC (Given)
2.  m ABO = mAP;
m OBC = mPC
(Inscribed Thm., Case I) 
3. m ABO + m OBC = m ABC ( Add. Post.) 
4.  mAP + mPC = m ABC
(Subst.) 
5.  (mAP + mPC) = m ABC
(Distr. Prop.) 
6.  mAC = m ABC (Arc Add. Post.) . 1 2
41. 1.  S with inscribed PQR (Given) 
2.  m PQT = mPT
(Inscribed Thm., Case I)
3.  m RQT = mRT
(Inscr. Thm., Case I) 
4.  mPR = mPT – mRT (Arc Add. Post.) 
5.  m PQR = m PQT – m RQT ( Add. Post.) 
6.  m PQR = mPT – mRT (Subst.)
7.  m PQR = mPR (Subst.)
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6. 42. 1. O, A intercepts BC, and D intercepts BC (Given)
Inscribed Angles
GEOMETRY LESSON 11-3
42. 1.  O, A intercepts BC, and D intercepts BC (Given) 
2.  m A = mBC and
m D = mBC (Inscr. Thm.)
3.  m A = m D (Subst.) 
4.  A D (Def. of )
43. 1.  O with inscribed CAB in a semicircle (Given) 
2. m CAB = mBDC
(Inscr. Thm.)
3.  mBDC = (Meas. of semicircle = 180.)
4.  m CAB = 90 (Subst.) 
CAB is a rt (Def. of rt. ) . 1 2
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7. 44. 1. quadrilateral ABCD inscribed in O (Given) 2. m A = mBCD and
Inscribed Angles
GEOMETRY LESSON 11-3
quadrilateral ABCD inscribed in O (Given) 
2.  m A = mBCD and
m C = mBAD (Inscr. Thm.) 
3. m A + m C =
mBCD + mBAD (Add. Prop.) 
4. mBCD + mBAD = (Entire circle = 360°.) 
5. m A + m C = (Subst. & Mult. Prop.)
6.  A and C are suppl. (Def. of suppl.)
44. (continued)
7. m B = mADC and
m D = mABC (Inscr. Thm.) 
8. m B + m D = mADC + mABC
(Add. Prop.) 
9. mADC + mABC = (Entire circle = 360.) 
10. m B + m D = (Subst. and Mult. Prop.) 
11.  B and D are suppl. (Def. of suppl.) . 1 2
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8. 45. 1. GH and tangent intersecting at H on E (Given)
Inscribed Angles
GEOMETRY LESSON 11-3
GH and tangent intersecting at H on E (Given) 
2. Construct diameter HD intersecting circle E at D. (Constr.) 
3.  DHI is a rt (Tangent and radius are .) 
4. DGH is a semicircle of measure (Def. of semicircle) 
5. m DHG + m GHI = m DHI ( Add. Post.)
6. mDG + mGFH = mDGH (Arc Add. Post.)    .
45. (continued)
7. 90 = m DHG + m GHI (Subst.) 
8. 180 = mDG + mGFH (Subst.) 
9. 90 = (mDG + mGFH) (Div. Prop.) 
10. m DHG + m GHI =
mDG + mGFH (Subst. and
Distr. Prop.) 
11. m DHG = mDG
(Inscr. Thm.) 
12. m GHI = mGFH (Subtr. Prop.) 1 2
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9. = 108; mXY = 112 = 2m XYZ; thus, m XYZ = 56.
Inscribed Angles
GEOMETRY LESSON 11-3
46. A
47. I
48. C
49. [2] a. mWY = 140; mWX = 2 • 54
= 108; mXY = = 2m XYZ; thus, m XYZ = 56.
b. 56
[1] incorrect answer OR incorrect explanation
50. [4] a. mADC = 140, so mABC = 220, and m D = 110.
b. m ACB = 90, because it is inscribed in a semicircle.
c. 180 = x. Thus, 70 = 4x and x = 17.5.
[3] appropriate methods, but with one computational error
[2] incorrect methods solved correctly OR correct methods solved incorrectly
[1] correct answers to a–c, without work shown
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10. 57. Both have rt. and the vertical are , so are ~ by AA ~ Post.
Inscribed Angles
GEOMETRY LESSON 11-3
in.2
cm2 m2
57. Both have rt. and the vertical are , so are ~ by AA ~ Post.
58. a. 960 ft
b. about 0.18 mi s
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