ssrs barcode font Circles in Objective-C

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CHAPTER 6 Circles
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An inscribed angle is an angle whose vertex is on the circle and whose sides are chords. An angle inscribed in an arc has its vertex on the arc and its sides passing through the ends of the arc. Thus, j A in Fig. 6-30 is an inscribed angle whose sides are the chords AB and AC. Note that j A intercepts BC and is inscribed in BAC.
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Fig. 6-30
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6.3A Angle-Measurement Principles
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PRINCIPLE
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A central angle is measured by its intercepted arc. An inscribed angle is measured by one-half its intercepted arc.
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PRINCIPLE 2:
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A proof of this principle is given in 16.
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PRINCIPLE
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In the same or congruent circles, congruent inscribed angles have congruent intercepted arcs.
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Thus in Fig. 6-31, if j 1 > j 2, then BC > DE.
Fig. 6-31
PRINCIPLE 4:
In the same or congruent circles, inscribed angles having congruent intercepted arcs are congruent. (This is the converse of Principle 3.)
Thus in Fig. 6-31, if BC > DE, then j 1 > j 2.
PRINCIPLE 5:
Angles inscribed in the same or congruent arcs are congruent.
Thus in Fig. 6-32, if j C and j D are inscribed in ACB, then j C > j D.
Fig. 6-32
CHAPTER 6 Circles
6: An angle inscribed in a semicircle is a right angle.
PRINCIPLE
Thus in Fig. 6-33, since j C is inscribed in semicircle ACD,, mj C
Fig. 6-33
PRINCIPLE 7:
Opposite angles of an inscribed quadrilateral are supplementary.
Thus in Fig. 6-34, if ABCD is an inscribed quadrilateral, j A is the supplement of j C.
Fig. 6-34
PRINCIPLE
Parallel lines intercept congruent arcs on a circle.
Thus in Fig. 6-35, if AB i CD, then AC > BD. If tangent FG is parallel to CD, then PC > PD.
Fig. 6-35
PRINCIPLE
An angle formed by a tangent and a chord is measured by one-half its intercepted arc. An angle formed by two intersecting chords is measured by one-half the sum of the intercepted arcs. An angle formed by two secants intersecting outside a circle is measured by one-half the difference of the intercepted arcs. An angle formed by a tangent and a secant intersecting outside a circle is measured by onehalf the difference of the intercepted arcs. An angle formed by two tangents intersecting outside a circle is measured by one-half the difference of the intercepted arcs.
PRINCIPLE 10: PRINCIPLE 11: PRINCIPLE 12: PRINCIPLE 13:
Proofs of Principles 10 to 13 are given in 16.
Fig. 6-36
CHAPTER 6 Circles
6.3B Table of Angle-Measurement Principles
Note: To find the angle formed by a secant and a chord meeting on the circle, first find the measure of the inscribed angle adjacent to it and then subtract from 180 . Thus if secant AB meets chord CD at C on the circle in Fig. 6-36, to find mj y, first find the measure of inscribed j x. Obtain mj y by subtracting mj x from 180 .
CHAPTER 6 Circles
SOLVED PROBLEMS
6.10 Applying principles 1 and 2 (a) In Fig. 6-37(a), if mj y (b) In Fig. 6-37(b), if mj y (c) In Fig. 6-37(c), if mj x 46 , find mj x. 112 , find mj x. 75 , find m y .
Fig. 6-37
Solutions
(a) j y BC, so mBC 46 . Then j x
1 2 BC 1 2
(46 )
23 , so mj x
1 2 BC 1 2
23 . (68 ) 34 , so mj x 90 . 34 .
(b) j y AB so mAB 112 . mBC m(ABC AB) 180 112 (c) j x
ADC, so mADC
150 . Then m y
68 . Then j x m(ADC
150 60
6.11 Applying principles 3 to 8 Find x and y in each part of Fig. 6-38.
Fig. 6-38
Solutions
(a) Since mj 1
mj 2, m x
50 . Since AD > CD, mj y
mj ABD
(b) j ABD and j x are inscribed in ABD; hence, mj x mj ABD 40 . ABCD is an inscribed quadrilateral; hence, mj y 180 mj B 95 . (c) Since j x is inscribed in a semicircle, mj x 90 . Since AC i DE, m y
CHAPTER 6 Circles
6.12 Applying principle 9 In each part of Fig. 6-39, CD is a tangent at P. (a) If m y
X y (b) If mX
220 in part (a), find mj x. 140 in part (b), find mj x. 75 in part (c), find mj x.
(c) If mj y
Fig. 6-39
Solutions
(a) / z
1 2 (220
110 . So mj x
(b) Since AB Since / x (c) / y
AP, mAB m y 1 40 , mj x 2 z
180 110
140 . Then m z 40 .
70 . 80 .
360 140 140 110 .
AP, so mAP
Since /x
150 . Then m z 55 , m/ x 55 .
360 100 150
6.13 Applying Principle 10 (a) If mj x 95 in Fig. 6-40(a), find m y . (b) If m y
X x (c) If mX
80 in Fig. 6-40(b), find mj x. 78 in Fig. 6-40(c), find mj y.
Fig. 6-40
Solutions
(a) /x (b) /z
1 2 (AC 1 2(
X); thus 95 y
AB )
1 2 (80
1 2 (70
m y ), so m y
X X x
120 . 180 mj z CD) 80 . 180 mj z 102 .
(c) Because BC i AD, mCD
120 )
100 . Then mj x
1 2(
78 . Also, / z
78 . Then mj y
CHAPTER 6 Circles
6.14 Applying principles 11 to 13 (a) If mj x (b) If mj x (c) If mj x
40 in Fig. 6-41(a), find m y .
X y 67 in Fig. 6-41(b), find mX. y 61 in Fig. 6-41(c), find mX.
Fig. 6-41
Solutions
(a) / x (b) / x (c) jx my
1 2 (BC 1 2 (BC
X), so 40 y
BE), so 67 360 200 66
1 2 (200 1 2
m y ) or m y
120 . 66 .
1 2 [(360
Then m y
(200 94 . 360
mBE) or m BE
1 2 (BFC
X and mBFC y
m y . Then 61
my)
180 m y . Thus
6.15 Using equations in two unknowns to find arcs In each part of Fig. 6-42, find x and y using equations in two unknowns.
Fig. 6-42
Solutions
(a) By Principle 10, 70 By Principle 11, 25
1 2 (m x 1 2 (m x
X X X X
my) my)
If we add these two equations, we get m x (b) Since m x
By Principle 13,
1 2 (m x
360 , (m x my)
my) 62
95 . If we subtract one from the other, we get m y 180
If we add these two equations, we find that m x my 118 .
242 . If we subtract one from the other, we get
CHAPTER 6 Circles
6.16 Measuring angles and arcs in general Find x and y in each part of Fig. 6-43.
Fig. 6-43
Solutions
(a) By Principle 2, 50 Then mPR By Principle 9, x By Principle 13, y
1 2 m PQ 1 2 m PR
or mPQ mQR 60 . mPQ)
100 . Also, by Principle 9, 70 120 .
1 2 (260
1 2 mQR
or mQR
360 mPQ
1 2 (mPRQ
100 )
80 . mPA 85 . Then mPC 360 mPA
(b) By Principle 1, mAB 80 . Also, by Principle 8, mBC mAB mBC 110 . By Principle 9, x By Principle 12, y
1 2 m PC 1 2 (mPCB
55 . mPA)
(195 85 )
6.17 Proving an angle measurement problem
Given:
To Prove:
Plan: First prove CD
Use this to show that jB > jC.
PROOF:
Statements 1. 2. 3. 4. BD CE DE DE BE CD jB > jC AC 1. 2. 3. 4.
Reasons Given Reflexive property If s are added to s, the sums are . In a circle, inscribed angles having equal intercepted arcs are >. 5. In a triangle, sides opposite > angles are equal in length.
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