12.2 HW Solutions 3.14 4.2 5. 7 6.50 7.8 8.10 9.a) CE b) DE c) angle CEB d) angle DEA 10.The Center...
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12.2 HW Solutions 3.14 4.2 5. 7 6.50 7.8 8.10 9.a) CE b) DE c) angle CEB d) angle DEA 10.The Center of the circle 11.6 12.5.4 13.8.9 14.12.5 15.9.9 16.20.8 17.108 18.90 19.123.9 or 124 23. a) PL b) PM c) All radii of a circle are congruent d) Triangle LPN e) SAS f) CPCTC 24. a) All radii of a circle are congruent b) AB ≅ CD c) Given d) SSS e) angle AEB ≅ CED f) Thm 12.1
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Transcript of 12.2 HW Solutions 3.14 4.2 5. 7 6.50 7.8 8.10 9.a) CE b) DE c) angle CEB d) angle DEA 10.The Center...
12.2 HW Solutions3. 144. 25. 76. 507. 88. 109. a) CEb) DEc) angle CEBd) angle DEA10. The Center of the circle11. 612. 5.413. 8.914. 12.515. 9.916. 20.817. 10818. 9019. 123.9 or 124
23. a) PLb) PMc) All radii of a circle are congruentd) Triangle LPNe) SASf) CPCTC24. a) All radii of a circle are congruentb) AB ≅ CDc) Givend) SSSe) angle AEB ≅ CEDf) Thm 12.1
Inscribed Angles
Section 12.3
Three high-school soccer players practice kicking goals from the points shown in the diagram. All three points are along an arc of a circle. Player A says she is in the best position because the angle of her kicks towards the goal is wider than the angle of the other players’ kicks. Do you agree?
Player B
Player A
Player C
Vocabulary
• Inscribed Angle– An angle whose vertex is on the circle and whose
sides are chords of the circle
• Intercepted Arc– The portion of the circle intercepted by an
inscribed angle
• Theorem 12-9: Inscribed Angle Theorem– The measure of an inscribed angle is half the
measure of its intercepted arc
mACBm2
1 B
C
A
60°? ° ?°45 °
Three Cases for 12-9
Q R F
1) The center is on a side of the angle
2) The center is inside of the angle
3) The center is outside of the angle
1. BC is a diameter2. mAC = mAOC3. mAC = mA + mB4. mA = mB5. mAC =2mB6. ½mAC = mB
GivenDefinition of AC Tri. Ext. Ang. Thm.Isosceles Tri. ThmSubstitutionDivision Prop. Of Equal.
Given:
Prove:
Circle O, with inscribed B and diameter BC
mB = ½mAC
B
O
C
A
Find a and b
P
Q
R
S
T60˚
b˚
a˚
30˚
= 120°
= 75°
Corollaries: Inscribed Angle Theorem
1) Two inscribed angles that intercept the same arc are congruent
70˚
x
Corollaries: Inscribed Angle Theorem
2) An angle inscribed in a semicircle is a right angle( A semicircle is 180, ½ of that is 90)
x° = 90°
Corollaries: Inscribed Angle Theorem
3) The opposite angles of a quadrilateral inscribed in a circle are supplementary
(opposite angles intercept the entire circle, ½ of 360 is 180)
70˚
x
Corollaries: Inscribed Angle Theorem
1) Two inscribed angles that intercept the same arc are congruent
2) An angle inscribed in a semicircle is a right angle( A semicircle is 180, ½ of that is 90)
3) The opposite angles of a quadrilateral inscribed in a circle are supplementary
(opposite angles intercept the entire circle, ½ of 360 is 180)
Find the measure of angles 1 and 2
70˚
1
40˚
38˚
2
70˚
= 90°
= 38°
• Theorem 12-10– The measure of an angle formed by a tangent and
a chord is half the measure of the intercepted arc.
mBDCCm2
1
C
O
B
D
C
O
B
D
GEOGEBRA
Find x and y
K
L
QJ
35˚
y˚
x˚90°
= 55°
110°
70°
= 35°
Homework
• Section: 12-3• Pages: 681-683• Questions: 4-22, 36, 38
1. Circle O, inscribed ∠ABC2. Draw diameter BP3. mABP = ½AP4. mCBP = ½PC5. mABC = mABP + mABP 6. mABC = ½AP + ½PC7. mABC = ½(AP+PC)8. mABC = ½AC
GivenDef. of DiameterThm. 12-9Thm. 12-9Ang. Add. Post.SubstitutionDistributing (factoring)Arc Addition Post.
Given:
Prove: mABC = ½mAC
B
O
C
A
PCircle O, with inscribed ∠ABC
Post-Homework Review
• CHAPTER REVIEW• Page: 707• Questions: 6-14
• Section: 12-3• Pages: 681-683• Questions: 25
