Other Angle Relationships in Circles

Objectives/Assignment

• Use angles formed by tangents and chords to solve problems in geometry.

• Use angles formed by lines that intersect a circle to solve problems.

Using Tangents and Chords

• 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

Tangent to a Chord Conjecture

• 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.

21

A

B

C

m1= ½m

m2= ½m

ABBCA

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

Ex. 2: 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

B

C

A

Ex. 3: 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

5x°

D

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.

Chords intersecting Inside the circle

• 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

1

A

B

C

Tangent and Secant Exterior Intersections

• 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

Tangent and Secant Exterior Intersections

• 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.

PQRm2 = ½ m( - m ) PR

2

R

P

Q

Tangent and Secant Exterior Intersections

• 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.

XYm3 = ½ m( - m ) WZ

Z

W

Y

X

3

Ex. 4: 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°

Ex. 5: Tangent & Secant Intersections

• Find the value of x

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°

P

N

M

L

• Find the value of x

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

Ex. 6: Tangent & Secant Intersections

Ex. 7: 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

Ex. 7: 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. 7: 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°.