10.5 Other Angles10.5 Other Angles

In a CircleIn a Circle

Vertex on the circleVertex on the circle

When the vertex of the angle is When the vertex of the angle is on the circle, the measure of the on the circle, the measure of the angle is half the intercepted arc.angle is half the intercepted arc.

Formed by two chordsFormed by two chords Formed by chord and tangentFormed by chord and tangent

A

B

DC

m CD on AB = 53.73

mCBD = 26.86

E

F

H

G

m GF on EF = 130.06

mGFH = 65.02

Two Chords Chord/Tangent

What are the missing measures?

Angles inside the Angles inside the CircleCircle If two chords intersect inside the circle, then If two chords intersect inside the circle, then

the angle is the average of the intercepted the angle is the average of the intercepted arcs of the vertical angles. (half the sum)arcs of the vertical angles. (half the sum)

1

80 °

30°

What is the measure of Angle 1?

½( 80 + 30) = 55 °

Angles Outside the Angles Outside the CircleCircle If a tangent/ tangent, secant/secant or If a tangent/ tangent, secant/secant or

tangent/secant intersect outside a circle tangent/secant intersect outside a circle the angle formed is half the difference of the angle formed is half the difference of the intercepted arcs.the intercepted arcs.

Tangent/tangentSecant/secant Tangent/secant

aa a

b bb

Angle = ½ (a-b)

GUIDED PRACTICE for Example 1

Find the indicated measure.

SOLUTION

= 12 (210o) = 105om 1 = 2 (98o) = 196om TSR

SOLUTION

GUIDED PRACTICE

Find the value of the variable.

SOLUTION

The chords AC and CD intersect inside the circle.

Use Theorem 10.12.

Substitute.

Simplify.

= 12

(yo + 95o) 78o

= y 61

78° (mAB + mCD)= 12

156 = y +95

GUIDED PRACTICE

Find the value of the variable.

SOLUTION

The tangent JF and the secant JG intersect outside the circle.

Use Theorem 10.13.

Substitute.

Simplify.

= 12

(ao – 44o) 30o

= 104 a

m FJG (mFG – mKH)= 12

60 = a - 44

GUIDED PRACTICE

Find the value of the variable.

SOLUTION

Use Theorem 10.13.

Substitute.73.7o 12

[(xo) –(360 –x)o]

Solve for x.

xo 253.7

= 12

m TQR (mTUR – mTR)

CP

Congruent triangles (HL)Trig using 3-4-5

147.4 = x – 360 + x

507.4 = 2x

Review of all angles with circles• Central angle = the intercepted arc.

• Vertex on circle = half the intercepted arc.

(chords sharing common endpoint or

a chord and a tangent intersecting on

circle.)

• Vertex inside circle = half the sum

• Vertex outside circle = half the difference

Central =On –half arc

Inside –half sumOutside---half difference

GeometryGeometry

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Sophomore MathSophomore Math

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