10.4 other angle relationships in circles

10.4 Other Angle Relationships in Circles

Learning Target

• I can use theorems about tangents, chords and secants to solve unknown measure of arcs and angles .

Review on inscribe angles

• You know that measure of an angle inscribed in a circle is half the measure of its intercepted arc. This is true even if one side of the angle is tangent to the circle.

C

B

A

D

ABmADB = ½m

n

x

Angle x = ½ n

Theorem 10.12

• If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one half the measure of its intercepted arc. 2

1

A

B

C

m1= ½m

m2= ½m

ABBCA

n

x

Angle x = ½ n

Ex. 1: Finding Angle and Arc Measures

• Line m is tangent to the circle. Find the measure of the red angle or arc.

• Solution:

m1= ½

m1= ½ (150°)

m1= 75°

m

1

B

A

150°AB

Angle x = ½ n

Ex. 1: Finding Angle and Arc Measures

• Line m is tangent to the circle. Find the measure of the red angle or arc.

• Solution:

m = 2(130°)

m = 260°

RSP

130°

RSP

RP

S

Angle x = ½ n

Ex. 2: Finding an Angle Measure

• In the diagram below,

is tangent to the circle. Find mCBD

• Solution:

mCBD = ½ m

5x = ½(9x + 20)

10x = 9x +20

x = 20

mCBD = 5(20°) = 100°

DAB

(9x + 20)°

BC

B

C

A

5x°

D

Angle x = ½ n

Lines Intersecting Inside or Outside a Circle

• If two lines intersect a circle, there are three (3) places where the lines can intersect.

on the circle

Inside the circle

Outside the circle

Lines Intersecting• You know how to find angle and arc

measures when lines intersect on the circle.

• You can use the following theorems to find the measures when the lines intersect

INSIDE or OUTSIDE the circle.

Theorem 10.13

• If two chords intersect in the interior of a circle, then the measure of each angle is one half the sum of the measures of the arcs intercepted by the angle and its vertical angle.

2

1

B

A C

D

CDm1 = ½ m + m

AB

BCm2 = ½ m + m

AD

n

x

f

Angle x = ½ ( f + n)

Theorem 10.14

• If a tangent and a secant, two tangents or two secants intercept in the EXTERIOR of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs.

BCm1 = ½ m( - m )

AC

1

A

B

C

nx

f

Angle x = ½ ( f - n )

Theorem 10.14


PQRm2 = ½ m( - m ) PR

2

R

P

Q

Angle x = ½ ( f - n )

fnx

Theorem 10.14


XYm3 = ½ m( - m ) WZ

Z

W

Y

X

3

fnx

Angle x = ½ ( f - n )

Ex. 3: Finding the Measure of an Angle Formed by Two Chords

• Find the value of x

• Solution:

x° = ½ (m +m

x° = ½ (106° + 174°)

x = 140

PS

RQ

R

S

P

Q

Apply Theorem 10.13

Substitute values

Simplify

174°

106°

x°Angle x = ½ (f +n)

Ex. 4: Using Theorem 10.14


Solution:

72° = ½ (200° - x°)

144 = 200 - x°

- 56 = -x

56 = x

Substitute values.

Subtract 200 from both sides.

Multiply each side by 2.

EDGmGHF = ½ m( - m )

GF Apply Theorem 10.14

Divide by -1 to eliminate negatives.

F

GH

E

D

200°

x°

72°

Angle x = ½ (f – n)

P

N

M

L

Ex. 4: Using Theorem 10.14


Solution:

= ½ (268 - 92)

= ½ (176)

= 88

Substitute values.

Multiply

Subtract

MLNmGHF = ½ m( - m )

MN Apply Theorem 10.14

x°92°

Because and make a whole circle, m =360°-92°=268°

MN

MLN

MLN

Angle x = ½ ( f – n )

Ex. 5: Describing the View from Mount Rainier

• You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level. Find the measure of the arc that represents the part of Earth you can see.

CD

• and are tangent to the Earth. You can solve right ∆BCA to see that mCBA 87.9°. So, mCBD 175.8°. Let m = x° using Trig Ratios

Ex. 5: Describing the View from Mount Rainier

BCCD

BD

CD

175.8 ½[(360 – x) – x]

175.8 ½(360 – 2x)

175.8 180 – x

x 4.2

Apply Theorem 10.14.

Simplify.

Distributive Property.

Solve for x.

From the peak, you can see an arc about 4°.

Reminders:

• Pair-Share:

Work on page. 624-625 #2-35

• Refer to the summary sheet “Angles Related to Circles” to identify what formula to use.

• Quiz on Friday about Trigonometry and Circles.

10.4 other angle relationships in circles

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