CirclesSection 9.5Inscribed Angles

Warm Up

Circles Crossword puzzle Some answers are 2 words, do not

leave blanks

Homework Answers p. 347

1. 82. 53. 9√24. 555. 806. 457. 248. 129. 10√5

Central Angle

vertex at the center of a circle.angle equals arc

O

B

A

If AB = 100° then AOB = 100°

Inscribed Angle

An angle whose vertex is on the edge of a circle.

C is a point on circle O ACB is an inscribed angle

AB is the intercepted arc of inscribed angle ACB

O

B

C

A

Theorem 9.7The measure of an inscribed angle

is equal to HALF the intercepted arc

O

B

C

If AB = 100° then ACB = 50°

A

Example 1 Find x, y, z

55°80°

90°

z°

y°x°

x= ½ (80°) = 40°y= 2(55°) = 110°z= 360°-(90°+80°+110°)= 80°

Example 2 Find 1B

C

A

1

3. BAC = 280°, BC = 360°- 280°= 80° 1 = ½ (80°) = 40°

1. BC = 70°, 1 = ½ (70°) = 35°

2. BC = 84°, 1 = ½ (84°) = 42°

Example 3 Find BC

BC

A

1

1. 1 = 40°, BC= 2(40°) = 80°

2. 1 = 36° ,BC = 2(36°) = 72°

Theorem 9.7 – Corollary 1If two inscribed angles intersect the same arc, then the angles are congruent.

1 = 2

BC

A

12

1 intersects arc BC

2 intersects arc BC

Theorem 9.7 – Corollary 2An angle inscribed in a semicircle is a right angle.

B

O

A

R1

arc AB is a semicircle

1 intersects arc AB

1 = ½ (AB) = 90°

O is the center of the circle

Theorem 9.7 – Corollary 3If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

O is the center of the circleB

O

A

Z1

X2

3

4

BXAZ is a quadrilateral inscribed in circle O

1 and 2 are opposite angles 1 + 2 = 180°

Example 6 Find x, y, z

x°

160°

z°

y°

68°

a°

x= ½ (160°) = 80°

a°+ 80° + 68° = 180°a° = 32°z = 2(32°) = 64°

Y intersects arcs 160° + 64°y= ½(160° + 64°) = 112°

80°64°

Example 7 Find x, y, zO is the center

x°

O

z°

y°

68°

x and y are inscribed in the same semicirclex = y = ½ (180°) = 90°

2 chords are marked congruent, therefore the 2 arcs are congruent.The 2 arcs are inscribed in a semicircleTherefore z = ½ (180°) = 90°

Example 8 Find x, y, zO is the center

z°= 2(43°) = 86°

x°

110°

z°

y°

43° OBA

CAB is a semicircle = 180°

CB = 180°- 110° = 70°y is a cental angle = arcy = 70°

x = ½ (CB) = ½(70°) = 35°

Example 9 Find x, y, z

x°

60°

z°

y°

85°

x° and 85° are opposite anglesTherefore they are supplementary x= 180°- 85° = 95°

y intersects arcs z° + 60°y= ½(60° + z°) 70°= ½ (60° + z°)140° = (60° + z°) z° = 80°

70°

110°

A quadrilateral is inscribed in the circle

y° and 110° are opposite anglesTherefore they are supplementaryy= 180°- 110° = 70°

Theorem 9.8The measure of an angle formed by a chord and a tangent is equal to

HALF its intercepted arc

If CB = 120° then FCB = 60°

vertex on the edge of the circle

O

B

C

A

G

F

ACB = _______GCB= _______

Examples 10-12

CBD = ½(240°) = 120°

BD = 360°- 240° = 120°O

B

C

A

D

240°

ABD = ½(120°) = 60°

Examples 13-15

PRT = ½(100°) = 50°

O s

R

Q

100°

170°

P

T

PRQ = ½(170°) = 85°

QRS = ½(QR) QR = 360°- (170°+100°) = 90°QRS = ½(90°) = 45°

Examples 16-18AC is tangent to circle Z at B

BE is a diameter, therefore ABC =CBE= 90°

ZB

C

AF

75°

D

E

EBD = 90°- 75° = 15°DE = 2(15°) = 30°DB = 2(75°) = 150°

Examples 19-21AC is tangent to circle Z at B

BED = ½ (150°) = 75°

ZB

C

AF

75°

D

E

30°

150°

15°

BDE is inscribed in a semicircle

BDE = 90°

BFE is a semicircle BFE = 180°

Cool Down

Complete exercises 1-6 on bottom of notesheet

Check your answers with Mrs. Baumher

Begin HW p. 354 1-9, 19-21