Download presentation
Presentation is loading. Please wait.
Published byWinfred Perkins Modified over 5 years ago
1
Warm-Up A circle and an angle are drawn in the same plane. Find all possible ways in which the circle and angle intersect at two points.
2
12.5 Other Angle Relationships in Circles
Geometry Mr. Calise
3
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.
4
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
5
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
6
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°
7
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°
8
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
9
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
10
Inside the circle
11
Outside the circle
12
Summary When an angle has its vertex on the circle the angle is equal to ½ the measure of its intercepted arc. When the angle has a vertex inside the circle the angle is equal to ½ the sum of its two intercepted arcs. When the angle has a vertex outside of the circle then the angle is equal to ½ the difference of the two intercepted arcs.
13
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 m2 = ½ m m
14
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. m1 = ½ m( m )
15
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 )
16
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 )
17
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
18
Ex. 4: Using Theorem 10.14 Find the value of x Solution:
200° Find the value of x Solution: 72° = ½ (200° - x°) 144 = x° - 56 = -x 56 = x x° mGHF = ½ m( m ) 72° Apply Theorem 10.14 Substitute values. Multiply each side by 2. Subtract 200 from both sides. Divide by -1 to eliminate negatives.
19
Ex. 4: Using Theorem 10.14 Find the value of x Solution:
Because and make a whole circle, m =360°-92°=268° x° 92° Find the value of x Solution: = ½ ( ) = ½ (176) = 88 mGHF = ½ m( m ) Apply Theorem 10.14 Substitute values. Subtract Multiply
20
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.
21
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.
22
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
23
From the peak, you can see an arc about 4°.
175.8 ½[(360 – x) – x] 175.8 ½(360 – 2x) 175.8 180 – x x 4.2 Apply Theorem Simplify. Distributive Property. Solve for x. From the peak, you can see an arc about 4°.
24
Homework Page 617 #’s 1 – 6, 16, 17
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.