Bell work 1Bell work 1

T

R

•

•

TU = 92º

U

•

Find the measure of the inscribed angles , R, Find the measure of the inscribed angles , R,

given that their common intercepted TU = 92ºgiven that their common intercepted TU = 92º

Bell work 1 AnswerBell work 1 Answer

T

R

•

•

TU = 92º

U

•

Angles R = ½ the intercepted arc TU since their Angles R = ½ the intercepted arc TU since their

intercepted Arc TU = 92º, then Angle R = 46ºintercepted Arc TU = 92º, then Angle R = 46º

Bell work 2Bell work 2

A quadrilateral WXYZ is inscribed in circle P, if A quadrilateral WXYZ is inscribed in circle P, if ∕ ∕__ X = 130º and X = 130º and ∕_ ∕_ Y = 106º , Find the measures of Y = 106º , Find the measures of ∕ ∕__ W = ? and W = ? and ∕_ ∕_ Z = ? Z = ?

••

••••X

Y

P

••Z

The Quadrilateral WXYZ is inscribed in the circle iff

/ X + / Z = 180º, and

/ W + / Y = 180ºW

••

130º 106º

Bellwork 2 AnswerBellwork 2 Answer

From Theorem 10.11From Theorem 10.11 ∕ ∕__ W = 180º – 106º = 74º and W = 180º – 106º = 74º and ∕ ∕__ Z = 180º – 130º = 50º Z = 180º – 130º = 50º

••

••••X

Y

P

••Z

The Quadrilteral WXYZ is inscribed in the circle iff

/ X + / Z = 180º, and

/ W + / Y = 180ºW

••

130º 106º

Unit 3 : Circles: Unit 3 : Circles: 10.4 Other Angle Relationships in 10.4 Other Angle Relationships in

CirclesCircles

Objectives: Students will:Objectives: Students will:1.1. Use angles formed by tangents and Use angles formed by tangents and

chords to solve problems related to chords to solve problems related to circlescircles

2.2. Use angles formed by lines Use angles formed by lines intersecting on the interior or intersecting on the interior or exterior of a circle to solve exterior of a circle to solve problems related to circlesproblems related to circles

(p. 621) Theorem 10.12 (p. 621) Theorem 10.12

If a tangent and a chord intersect at a point on a circle, then If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is ½ the measure of the measure of each angle formed is ½ the measure of its intercepted arc its intercepted arc

•P

A

B

•

Angle 1

C

m ∕_ 1

= ½ m minor AC

•

m ∕_ 2

= ½ m Major ABC

Angle 2

•

m

(p. 621) Theorem 10.12 (p. 621) Theorem 10.12 Example 1Example 1

Find the measure of Angle 1 and Angle 2, if Find the measure of Angle 1 and Angle 2, if the measure of the minor Arc AC is 130the measure of the minor Arc AC is 130ºº

•P

A

B

•

Angle 1

C

m minor AC = 130ºº

•Angle 2

•

m

(p. 621) Theorem 10.12 (p. 621) Theorem 10.12 Example 1 AnswerExample 1 Answer

The measure of Angle 1 = 65The measure of Angle 1 = 65ºº and and

Angle 2 = 115Angle 2 = 115ºº

•P

A

B

•

65º = Angle 1

C

m minor AC = 130ºº

•

Angle 2 = 115º

•

m

(p. 621) Theorem 10.12 (p. 621) Theorem 10.12 Example 2Example 2

Find the measure of Angle 1, if Angle 1 = 6xFind the measure of Angle 1, if Angle 1 = 6xºº, and , and the measure of the minor Arc AC is (10x + 16)the measure of the minor Arc AC is (10x + 16)ºº

•P

A

B

•

Angle 1= 6xº

C

m minor AC = (10x + 16)ºº

•

•

m

(p. 621) Theorem 10.12 (p. 621) Theorem 10.12 Example 2 AnswerExample 2 Answer

Angle 1 = 6xAngle 1 = 6xºº = ½ = ½ Arc ACArc AC = ½ = ½ (10x + 16)(10x + 16)ºº6x6xº = ½ º = ½ (10x + 16)(10x + 16)ºº6x6xº = 5x + 8º = 5x + 8x = 8º thus,x = 8º thus,Angle 1 = 48ºAngle 1 = 48º

•P

A

B

•

Angle 1= 6xºC

m minor AC = (10x + 16)ºº

•

•

m

Intersections of lines with respect to Intersections of lines with respect to a circlea circle

There are three places two lines can There are three places two lines can

intersect with respect to a circle.intersect with respect to a circle.

•

On the circle

•

In the circle Outside the cirlce

•

(p. 622) Theorem 10.13 (p. 622) Theorem 10.13

If two chords intersect in the If two chords intersect in the interiorinterior of a circle , then the of a circle , then the measure of each angle is ½ the measure of each angle is ½ the sumsum of the measures of of the measures of the arcs intercepted by the angle and its vertical angle.the arcs intercepted by the angle and its vertical angle.

•P

D

C•

Angle 1

B

m ∕_ 1

= ½ (m AB + m CD)

• m ∕_ 2

= ½ (mBC + mAD)

Angle 2

•

•A

m CD = 16º

(p. 622) Theorem 10.13(p. 622) Theorem 10.13Example Example

Find the value of x.Find the value of x.

•P

D

C•

Angle 1

Bm AB = 40º

•

•

•A

xº

m CD = 16º

(p. 622) Theorem 10.13(p. 622) Theorem 10.13Example Answer Example Answer

x = ½ (m AB + m CD) = ½ (40x = ½ (m AB + m CD) = ½ (40ºº + 16 + 16ºº))x = ½ (56º) x = ½ (56º) x = 28 ºx = 28 º

•P

D

C•

Angle 1

Bm AB = 40º

•

•

•A

xº

32

1

(p. 622) Theorem 10.14 (p. 622) Theorem 10.14 If a tangent and a secant, two tangents, or two secants If a tangent and a secant, two tangents, or two secants intersect in the intersect in the exteriorexterior of a circle, then the measure of the of a circle, then the measure of the angle formed is ½ the angle formed is ½ the difference difference of the intercepted arcs.of the intercepted arcs.

Bm ∕_ 1

= ½ (m BC – m AC)

•m ∕_ 2

= ½ (m PQR – m PR)

1 Tangentand 1 Secant

2 Tangents 2 Secants

•

•

••B

A

C

•

•

•

m ∕_ 3

= ½ (m XY – m WZ)

P

QR

••

••

••

W

X

Y

Z

xº

(p. 622) Theorem 10.14(p. 622) Theorem 10.14Example 1Example 1

Find the value of xFind the value of x

m ∕_ x

= ½ (m PQR - mPR)

Major Arc PQR= 266º

•

•

•

P

Q

R

•

xº

(p. 622) Theorem 10.14(p. 622) Theorem 10.14Example 1 AnswerExample 1 Answer

m PR m PR = (360= (360º - m PQR) = º - m PQR) = (360(360º - 266º)º - 266º) = 94º = 94ºx = ½ (m PQR - m PR) = ½ (x = ½ (m PQR - m PR) = ½ (266º266º - - 94º94º) = ½ (172) = ½ (172º)º)x = 86 ºx = 86 º

m ∕_ x

= ½ (m PQR - mPR)

Major Arc PQR= 266º

•

•

•

P

Q

R

•

(p. 622) Theorem 10.14(p. 622) Theorem 10.14Example 2 Example 2

Find the value of x, GFFind the value of x, GF.. The The mm EDG = 210 EDG = 210ºº

The The m m angle EHG = 68angle EHG = 68ºº

m ∕_ EHG = 68º

= ½ (m EDG – m GF)

•

•

•

•

E

H

G

xº

F

68ºD

Major Arc EDG= 210º

•

(p. 622) Theorem 10.14(p. 622) Theorem 10.14Example 2 Answer Example 2 Answer

m ∕_ EHG = 68º = ½ (m EDG – m GF) 68º = ½ ( 210º - xº ) 136º = 210º - xºxº = 210º - 136ºxº = 74º

m ∕_ EHG = 68º

= ½ (m EDG – m GF)

•

•

•

•

E

H

G

xº

F

68ºD

Major Arc EDG= 210º

•

Home workHome work

PWS 10.4 APWS 10.4 A

P. 624 (8 -34) evenP. 624 (8 -34) even

JournalJournal

Write two things about “the Write two things about “the

intersections of chords, secants, and/or intersections of chords, secants, and/or

tangents” related to circles from this lesson.tangents” related to circles from this lesson.