Download Reteach 12.5

Name ________________________________________ Date __________________ Class__________________
LESSON
12-5
Reteach
Angle Relationships in Circles
JJJG
Tangent BC and
JJJG
secant BA
intersect at B.
If a tangent and a secant
(or chord) intersect on a circle at
the point of tangency, then the
measure of the angle formed is
half the measure of its
intercepted arc.
1
p
m∠ABC = m AB
2
If two secants or chords intersect in the
interior of a circle, then the measure of
the angle formed is half the sum of the
measures of its intercepted arcs.
Chords AB and CD
intersect at E.
m∠1 =
1
2
p + mBC
p
(mAD
)
Find each measure.
p
2. mLM
1. m∠FGH
_________________________________________
3. m∠JML
________________________________________
4. m∠STR
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
12-38
Holt McDougal Geometry
Name ________________________________________ Date __________________ Class__________________
Reteach
LESSON
12-5
Angle Relationships in Circles continued
If two segments intersect in the exterior of a circle, then the measure of the angle formed is
half the difference of the measures of its intercepted arcs.
A Tangent and a Secant
m∠1 =
p − mBD
p
(mAD
)
2
1
Two Tangents
m ∠2 =
q − mEG
p
(mEHG
)
2
1
Two Secants
m∠3 =
1
2
p − mKM
q
(mJN
)
Find the value of x.
q + mPR
p = 360°, mPVR
q + 142° = 360°,
Since mPVR
q = 218°.
and mPVR
x° =
=
1
2
1
2
q − mPR
p
(mPVR
)
( 218° − 142° )
x° = 38°
x = 38
Find the value of x.
5.
6.
_________________________________________
7.
________________________________________
8.
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
12-39
Holt McDougal Geometry
Reading Strategies
congruent arcs have congruent chords,
p ≅ TU
p . ∠RUS,
RU ≅ ST . It is given that RS
∠URT, ∠TSU, and ∠STR are all
p
inscribed angles that intercept either RS
1. half
2. m∠A + m∠C = 180; m∠B + m∠D = 180
3. 31°
4. 60°
5. 124°
6. 60°
7. 56°
8. 120°
p . Therefore all four angles have the
or TU
same measure and are congruent. By
SAS, UQRU and UQST are congruent
triangles. Furthermore, the base angles
are all the same, so they are isosceles
triangles. So RQ, UQ, and SQ (and TQ )
are congruent by CPCTC and the
Isosceles Triangle Theorem. Congruent
segments have equal lengths, so Q is
equidistant from points R, U, and S (and
T) that lie on the circle. Therefore Q is the
center of the circle.
12-5 ANGLE RELATIONSHIPS IN CIRCLES
Practice A
1. B
2. C
3. A
4. 45°
5. 150°
6. 55°
7. 116°
8. 82°
9. 40
11. 96°
10. 67
3. Possible answer: Draw chord KM.
q = 180°. Because a
Assume that m KM
q would also
KLM
circle contains
JJJG360°, m
JJJG
equal 180°. JK and JM intersect outside
1
q −
the circle, thus m∠KJM = (m KLM
2
q ) = 0°. A triangle cannot contain a
m KM
0° angle, so UJKM does not exist, and
q = 180° cannot be true. Assume
m KM
JJJG
q > 180°. Tangent JM and
that m KM
JJJJG
chord KM intersect at the point of
1 q
tangency, M. So m∠JMK = m KM
,
2
which means m∠JMK > 90°. Similar
reasoning shows that m∠JKM is also
greater than 90°. A triangle cannot
contain two obtuse angles, so UJKM
q > 180° cannot
does not exist, and m KM
q < 180°.
be true. Therefore m KM
12. 134°
13. 38°
Practice B
1. 64°; 96°
2. 119°; 42°
3. 130°
4. 99°
5. 64
6. 47
7. 8
8. 45
9. 60
10. 66.5°; 115°
11. 84°; 192°
Practice C
1. Possible answer: It is given that
AB ≅ EB. So UABE is an isosceles
triangle, and ∠BAC ≅ ∠BEA. ∠BEA is an
1 p
inscribed angle, so m∠BEA = m BC
.
2
1 p
By substitution, m∠BAC = m BC
. AD
2
and AE are secants that intersect in the
exterior of the circle. So m∠BAC =
1
p − m BC
p ). Substitution leads to
(m DE
2
1 p
1
p − m BC
p ). This
m BC = (m DE
2
2
p = 2m BC
p.
simplifies to m DE
4. 120°
5. 80°
Reteach
1. 108°
2. 128°
3. 61°
4. 103°
5. 33
6. 52
7. 23
8. 38
2. Possible answer: Draw chords RU and
q ≅ ST
p. Because
ST . It is given that RU
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A53
Holt McDougal Geometry