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# Bell work 1 - PowerPoint PPT Presentation

Bell work 1. Find the measure of the inscribed angles , R, given that their common intercepted TU = 92º. T. •. TU = 92 º. R. •. •. U. Bell work 1 Answer. Angles R = ½ the intercepted arc TU since their intercepted Arc TU = 92º, then Angle R = 46º. T. •. TU = 92 º. R. •. •.

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Find the measure of the inscribed angles , R,

given that their common intercepted TU = 92º

T

TU = 92º

R

U

Angles R = ½ the intercepted arc TU since their

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

T

TU = 92º

R

U

A quadrilateral WXYZ is inscribed in circle P, if

∕_ X = 130º and ∕_ Y = 106º , Find the measures of

∕_ W = ? and ∕_ Z = ?

The Quadrilateral WXYZ is inscribed in the circle iff

/ X + / Z = 180º, and

/ W + / Y = 180º

X

Y

130º

106º

P

W

Z

From Theorem 10.11

∕_ W = 180º – 106º = 74º and

∕_ Z = 180º – 130º = 50º

The Quadrilteral WXYZ is inscribed in the circle iff

/ X + / Z = 180º, and

/ W + / Y = 180º

X

Y

130º

106º

P

W

Z

Unit 3 : Circles: 10.4 Other Angle Relationships in Circles

Objectives: Students will:

1. Use angles formed by tangents and chords to solve problems related to circles

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

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

its intercepted arc

A

m∕_ 1

= ½ m minor AC

P

m∕_ 2

= ½ m Major ABC

B

Angle 1

Angle 2

C

m

(p. 621) Theorem 10.12 Example 1

Find the measure of Angle 1 and Angle 2, if

the measure of the minor Arc AC is 130º

A

m minor AC = 130º

P

B

Angle 1

Angle 2

C

m

(p. 621) Theorem 10.12 Example 1 Answer

The measure of Angle 1 = 65º and

Angle 2 = 115º

A

m minor AC = 130º

P

B

65º = Angle 1

Angle 2 = 115º

C

m

(p. 621) Theorem 10.12 Example 2

Find the measure of Angle 1, if Angle 1 = 6xº, and

the measure of the minor Arc AC is (10x + 16)º

m minor AC = (10x + 16)º

A

Angle 1= 6xº

P

C

B

m

(p. 621) Theorem 10.12 Example 2 Answer

Angle 1 = 6xº = ½ Arc AC = ½ (10x + 16)º

6xº = ½ (10x + 16)º

6xº = 5x + 8

x = 8º thus,

Angle 1 = 48º

A

m minor AC = (10x + 16)º

P

Angle 1= 6xº

C

B

m

There are three places two lines can

intersect with respect to a circle.

Outside the cirlce

On the circle

In the circle

If two chords intersect in the interior of a circle , then the

measure of each angle is ½ the sum of the measures of

the arcs intercepted by the angle and its vertical angle.

D

Angle 1

m∕_ 1

= ½ (m AB + m CD)

C

P

Angle 2

A

m∕_ 2

B

(p. 622) Theorem 10.13Example

Find the value of x.

m CD = 16º

D

C

P

Angle 1

A

m AB = 40º

B

x = ½ (m AB + m CD) = ½ (40º + 16º)

x = ½ (56º)

x = 28 º

m CD = 16º

D

C

P

Angle 1

A

m AB = 40º

B

If a tangent and a secant, two tangents, or two secants

intersect in the exterior of a circle, then the measure of the

angle formed is ½ the difference of the intercepted arcs.

1 Tangent

and 1 Secant

2 Tangents

2 Secants

P

X

B

W

A

2

3

1

Q

Z

R

Y

C

m∕_ 1

= ½ (m BC – m AC)

m∕_ 2

= ½ (m PQR – m PR)

m∕_ 3

= ½ (m XY – m WZ)

B

(p. 622) Theorem 10.14Example 1

Find the value of x

P

Major Arc PQR

= 266º

Q

R

m∕_ x

= ½ (m PQR - mPR)

(p. 622) Theorem 10.14Example 1 Answer

m PR = (360º - m PQR) = (360º - 266º) = 94º

x = ½ (m PQR - m PR) = ½ (266º - 94º) = ½ (172º)

x = 86 º

P

Major Arc PQR

= 266º

Q

R

m∕_ x

= ½ (m PQR - mPR)

(p. 622) Theorem 10.14Example 2

Find the value of x, GF. The m EDG = 210º

The m angle EHG = 68º

E

F

Major Arc EDG

= 210º

D

68º

H

G

m∕_ EHG = 68º

= ½ (m EDG – m GF)

(p. 622) Theorem 10.14Example 2 Answer

m∕_ EHG = 68º = ½ (m EDG – m GF)

68º = ½ ( 210º - xº )

136º = 210º - xº

xº = 210º - 136º

xº = 74º

E

F

Major Arc EDG

= 210º

D

68º

H

G

m∕_ EHG = 68º

= ½ (m EDG – m GF)

PWS 10.4 A

P. 624 (8 -34) even