Download 10. 4 Other Angle Relationships in Circles - Geometry 10-12

10.3: P.616 #3-6, 9-16
40o
3.
(20× 2)
4. 180o (90× 2)
5. 210o (105×2)
6. x=115o (230÷2)
9. 64o (32×2)
10. 156o (78×2)
11. 228o (114×2)
12. 55o (110÷2)
13. 109o (218÷2)
14. 90o (180÷2)
15. 47o
16. x=90o, y=50o
L
M
x° 40
°
y°
K
Inscribed Angles & Their Arcs: DO NOT COPY!
A
Inscribed Angle:
an angle whose vertex is on
a circle and whose sides
contain the chords of a
circle. (<CAD&<CBD)
E
Intercepted Arc:
arc that lies in the interior of
an inscribed angle
If two inscribed angles of a
circle intercept the same arc,
then the angles are
congruent.
(m<AEB = mAB, measure of
the central angle = measure
of the intercepted arc)
C
D
m<CAD = ½(CD)
m<CBD = ½(CD)
<CAD  <CBD
B
10.4:
Secants, Tangents and Chords in Circles
5/6/2017
GOALS
Use angles formed
by tangents and
chords to solve
problems in
geometry
Use angles formed
by lines that
intersect a circle to
solve problems
There are 3 theorems you will use to solve for missing angles or arcs in a
circle. You choose which theorem to use based on where 2 lines intersect.
In other words, determine whether the intersection is:
#1
On the circle:
1 chord and 1 tangent
C
B
#2
Inside the circle:
2 chords
A
C
C
4
D
3
A
1
1
2
2
The angle is half of
its intercepted arc
B
D
A
m<1 or m<2 = (mAB + mCD)
M<1 = mACB
2 secants or tangents
B
2
1
#3
Outside the circle:
m<3 or m<4 = (mBC + mAD)
2
The angles are the average of
their intercepted arcs
m<1 = (CD – AB)
2
The angle is half of the
difference between the
intercepted arcs
Example #1: Tangent & Chord
Line m is tangent to the circle.
A) Solve for <VRT
1) m<VRT =
V
R
m
102°
m RST
2
2) m<VRT = (204°)
2
3) m<VRT = 102°
•S
T
204°
Example #2: 2 Chords
Solve for the value of x.
S
40°
1) m<x ° = (mST + mRU)
T
2
2) m<x° = (40° + 120°)
2
100°
y°
3) m<x° = (160) = 80°
X°
80°
2
U
R
120°
4) m<y ° = 180° - 80°
5) m<y° = 100°
Example #3: 1 secant & 1 tangent
Solve for the value of x.
56°
P
x° 102°
50°
S
Solve for x:
R
202°
T
1) x = (mRTS - mPS)
2
2) x = (202° - 102°)
2
3) x = (100°)
2
4) x = 50°
Example #4: 2 tangents
Solve for the value of x.
mRTS: 360°- 80°= 280°
T 280°
R
100°
x°
P
80°
S
1) x° = (mRTS - mRS)
2
2) x° = ( 280° - 80°)
2
3) x° = (200°)
2
4) x° = 100°
Assignment
P.624
#2-4, 8, 9, 17, 18, 20-22
Draw circles…
Show work…
SUMMARY and ?’s
Take a few minutes to write your summary.
Make sure that you are writing 3 complete
sentences. Also, don’t forget to come up
with the questions for your notes as well…
10.4, P.624: # 2-4, 8, 9, 17, 18, 20-22
2. 210o (105×2)
3. 60o ((55+65)/2)
4. 65o ((190-60)/2)
8. 110o (220/2)
9. 280o (140×2)
17. 112.5o ((130+95)/2)
18. 50o ((25+75)/2)
20. 27o ((105-51)/2)
21. 26o ((122-70)/2)
22. 45o ((142-52)/2)