10.4 Other Angle Relationships in Circles

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# 10.4 Other Angle Relationships in Circles - PowerPoint PPT Presentation

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.

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### 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.

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.

n

x

m1= ½m

Angle x = ½ n

m2= 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:

m1= ½

m1= ½ (150°)

m1= 75°

150°

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°

130°

Angle x = ½ n

Ex. 2: Finding an Angle Measure
• In the diagram below,

is tangent to the circle. Find mCBD

• Solution:

mCBD = ½ m

5x = ½(9x + 20)

10x = 9x +20

x = 20

 mCBD = 5(20°) = 100°

(9x + 20)°

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

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.

m1 = ½ m + m

m2 = ½ m + m

Theorem 10.13

n

x

• 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.

f

Angle x = ½ ( f + n)

m1 = 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.

x

n

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.

x

n

f

Angle x = ½ ( f - n )

m2 = 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.

x

3

n

f

Angle x = ½ ( f - n )

m3 = 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

Angle x = ½ (f +n)

174°

Apply Theorem 10.13

Substitute values

Simplify

mGHF = ½ m( - m )

Ex. 4: Using Theorem 10.14

Angle x = ½ (f – n)

200°

• Find the value of x

Solution:

72° = ½ (200° - x°)

144 = 200 - x°

- 56 = -x

56 = x

72°

Apply Theorem 10.14

Substitute values.

Multiply each side by 2.

Subtract 200 from both sides.

Divide by -1 to eliminate negatives.

mGHF = ½ m( - m )

Ex. 4: Using Theorem 10.14

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

92°

• Find the value of x

Solution:

= ½ (268 - 92)

= ½ (176)

= 88

Apply Theorem 10.14

Substitute values.

Subtract

Multiply

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.
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 mCBA  87.9°. So, mCBD  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°.

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.