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10.4 Other Angle Relationships in CirclesPowerPoint Presentation

10.4 Other Angle Relationships in Circles

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10.4 Other Angle Relationships in Circles. Geometry Mrs. Spitz Spring 2005. 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. Assignment: pp. 624-625 #2-35.

10.4 Other Angle Relationships in Circles

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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.
- Assignment: pp. 624-625 #2-35

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.

mADB = ½m

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.

m1= ½m

m2= ½m

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°

150°

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°

130°

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°

(9x + 20)°

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

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.

m1 = ½ m + m

m2 = ½ m + m

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.

m1 = ½ m( - m )

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.

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.

m2 = ½ m( - m )

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.

3

m3 = ½ m( - m )

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

106°

- Find the value of x
- Solution:
x° = ½ (m +m

x° = ½ (106° + 174°)

x = 140

x°

174°

Apply Theorem 10.13

Substitute values

Simplify

mGHF = ½ m( - m )

Ex. 4: Using Theorem 10.14200°

- Find the value of x
Solution:

72° = ½ (200° - x°)

144 = 200 - x°

- 56 = -x

56 = x

x°

72°

Apply Theorem 10.14

Substitute values.

Multiply each side by 2.

Subtract 200 from both sides.

Divide by -1 to eliminate negatives.

mGHF = ½ m( - m )

Ex. 4: Using Theorem 10.14Because and make a whole circle, m =360°-92°=268°

x°

92°

- Find the value of x
Solution:

= ½ (268 - 92)

= ½ (176)

= 88

Apply Theorem 10.14

Substitute values.

Subtract

Multiply

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.

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.

Ex. 5: Describing the View from Mount Rainier

- 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

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°.

Reminder:

- Quiz after section 10.5
- Deficiencies go out the week of April 25-28; so all work is due no later than the end of this week.