Download So, the measure of arc TS is 144.

So, the measure of arc TS is 144.
ANSWER: 144
10-6 Secants, Tangents, and Angle Measures
Find each measure. Assume that segments that
appear to be tangent are tangent.
3. 1. SOLUTION: SOLUTION: ANSWER: 110
ANSWER: 73
4. 2. SOLUTION: SOLUTION: So, the measure of arc TS is 144.
ANSWER: 144
3. ANSWER: 31
5. SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
Page 1
So, the measure of arc QTS is 248.
So, the measure of arc LP is 150.
10-6ANSWER: Secants, Tangents, and Angle Measures
31
ANSWER: 150
7. STUNTS A ramp is attached to the first of several
barrels that have been strapped together for a circus
motorcycle stunt as shown. What is the measure of
the angle the ramp makes with the ground?
5. SOLUTION: SOLUTION: Let x be the measure of the angle the ramp makes
with the ground which is formed by two intersecting
tangents to the circle formed by the barrel. One arc
has a measure of 165. The other arc is the major arc
with the same endpoints, so its measure is 360 – 165
or 195.
So, the measure of arc QTS is 248.
ANSWER: 248
6. Therefore, the measure of the angle the ramp makes
with the ground is 15.
SOLUTION: ANSWER: 15
Find each measure. Assume that segments that
appear to be tangent are tangent.
So, the measure of arc LP is 150.
8. ANSWER: 150
7. STUNTS A ramp is attached to the first of several
barrels that have been strapped together for a circus
motorcycle stunt as shown. What is the measure of
the angle the ramp makes with the ground?
SOLUTION: ANSWER: 82
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Let x be the measure of the angle the ramp makes
with the ground which is formed by two intersecting
9. Page 2
Therefore, the measure of the angle the ramp makes
with the ground is 15.
10-6ANSWER: Secants, Tangents, and Angle Measures
15
Find each measure. Assume that segments that
appear to be tangent are tangent.
ANSWER: 71.5
10. 8. SOLUTION: SOLUTION: ∠JMK and ∠HMJ form a linear pair.
ANSWER: 82
9. ANSWER: 102
11. SOLUTION: SOLUTION: Therefore, the measure of arc RQ is 28.
ANSWER: 71.5
10. ANSWER: 28
12. SOLUTION: SOLUTION: ∠JMK and ∠HMJ form a linear pair.
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ANSWER: Page 3
Therefore, the measure of arc RQ is 28.
So, the measure of arc PM is 144.
10-6ANSWER: Secants, Tangents, and Angle Measures
28
12. ANSWER: 144
14. SOLUTION: Arc BD and arc BCD are a minor and major arc that
share the same endpoints.
SOLUTION: ANSWER: 97
ANSWER: 103
13. 15. SOLUTION: SOLUTION: So, the measure of arc PM is 144.
By Theorem 10.13,
ANSWER: 144
.
14. ANSWER: 125
SOLUTION: Arc BD and arc BCD are a minor and major arc that
share the same endpoints.
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ANSWER: 103
16. SOLUTION: By Theorem 10.13,
Page 4
.
10-6ANSWER: Secants, Tangents, and Angle Measures
125
ANSWER: 196
17. SPORTS The multi-sport field shown includes a
16. softball field and a soccer field. If
find each measure.
,
SOLUTION: By Theorem 10.13,
Solve for
.
a.
b.
.
We know that
Substitute.
SOLUTION: .
a. By Theorem 10.13,
.
Substitute.
ANSWER: 196
Simplify.
17. SPORTS The multi-sport field shown includes a
softball field and a soccer field. If
find each measure.
b. By Theorem 10.14,
,
.
Substitute.
Simplify.
ANSWER: a. 100
b. 20
a.
b.
SOLUTION: CCSS STRUCTURE Find each measure.
a. By Theorem 10.13,
.
18. Substitute.
Simplify.
b. By Theorem 10.14,
SOLUTION: .
Substitute.
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By Theorem 10.14,
.
Page 5
Substitute.
ANSWER: 100
10-6a.Secants,
Tangents, and Angle Measures
b. 20
ANSWER: 81
CCSS STRUCTURE Find each measure.
19. 18. SOLUTION: By Theorem 10.14,
SOLUTION: .
By Theorem 10.14,
Substitute.
Substitute.
Simplify.
Simplify.
ANSWER: 81
ANSWER: 74
.
20. m(arc JNM)
19. SOLUTION: By Theorem 10.14,
.
SOLUTION: Substitute.
Simplify.
So, the measure of arc JNM is 205.
ANSWER: 205
ANSWER: 74
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20. m(arc JNM)
21. Page 6
So, the measure of arc JNM is 205.
10-6ANSWER: Secants, Tangents, and Angle Measures
205
21. ANSWER: 30
23. SOLUTION: By Theorem 10.14,
SOLUTION: By Theorem 10.14,
.
Substitute.
.
Substitute.
Solve for
.
Simplify.
ANSWER: 22
ANSWER: 185
24. JEWELRY In the circular necklace shown, A and B
are tangent points. If x = 260, what is y?
22. SOLUTION: By Theorem 10.14,
.
Substitute.
SOLUTION: Simplify.
By Theorem 10.14,
.
Substitute.
Simplify.
ANSWER: 30
23. ANSWER: 80
25. SPACE A satellite orbits above Earth’s equator.
Find x, the measure of the planet’s arc, that is visible
to the satellite.
Page 7
SOLUTION: eSolutions
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By Theorem 10.14,
.
ANSWER: 9
10-6ANSWER: Secants, Tangents, and Angle Measures
80
25. SPACE A satellite orbits above Earth’s equator.
Find x, the measure of the planet’s arc, that is visible
to the satellite.
27. SOLUTION: By Theorem 10.14,
SOLUTION: The measure of the visible arc is x and the measure
of the arc that is not visible is 360 – x. Use Theorem
10.14 to find the value of x.
Solve for
Therefore, the measure of the planet’s arc that is
visible to the satellite is 168.
.
.
ANSWER: 20
ANSWER: 168
ALGEBRA Find the value of x.
28. SOLUTION: 26. By Theorem 10.14,
SOLUTION: By Theorem 10.14,
Solve for
Solve for
.
.
.
.
ANSWER: 9
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27. SOLUTION: ANSWER: 19
29. PHOTOGRAPHY A photographer frames a
carousel in his camera shot as shown so that the lines
of sight form tangents to the carousel.
a. If the camera’s viewing angle is
, what is the
arc measure of the carousel that appears in the shot?
b. If you want to capture an arc measure of
in the photograph, what viewing angle should be used?
Page 8
ANSWER: a. 145
b. 30
10-6ANSWER: Secants, Tangents, and Angle Measures
19
29. PHOTOGRAPHY A photographer frames a
carousel in his camera shot as shown so that the lines
of sight form tangents to the carousel.
a. If the camera’s viewing angle is
, what is the
arc measure of the carousel that appears in the shot?
b. If you want to capture an arc measure of
in the photograph, what viewing angle should be used?
CCSS ARGUMENTS For each case of
Theorem 10.14, write a two-column proof.
30. Case 1
Given: secants
and Prove:
SOLUTION: Statements (Reasons)
1.
and are secants to the circle. (Given)
2.
SOLUTION: a. Let x be the measure of the carousel that appears
in the shot.
By Theorem 10.14,
Solve for
.
.
b. Let x be the measure of the camera’s viewing
angle.
By Theorem 10.14,
Solve for
.
(The ,
measure of an inscribed
the measure of its
intercepted arc.)
3.
Theorem)
(Exterior
4.
(Substitution)
5.
(Subtraction Prop.)
6.
(Distributive Prop.)
ANSWER: Statements (Reasons)
1.
and are secants to the circle. (Given)
2.
(The ,
.
ANSWER: a. 145
b. 30
CCSS ARGUMENTS For each case of
Theorem 10.14, write a two-column proof.
30. Case 1
Given: secants
and measure of an inscribed
the measure of its
intercepted arc.)
3.
Theorem)
(Exterior
4.
(Substitution)
5.
(Subtraction Prop.)
6.
(Distributive Prop.)
31. Case 2
Given: tangent
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Prove:
and secant Page 9
Prove:
(Subtraction Prop.)
5.
(Distributive Prop.)
10-66.Secants, Tangents, and Angle
Measures
31. Case 2
Given: tangent
5.
(Subtraction Prop.)
6.
(Distributive Prop.)
32. Case 3
Given: tangent
and secant and Prove:
Prove:
SOLUTION: Statements (Reasons)
1.
is a tangent to the circle and
the circle. (Given)
SOLUTION: Statements (Reasons)
1.
and are tangents to the circle. (Given)
2.
is a secant to
( The ,
the measure of its meas. of an inscribed
intercepted arc.)
3.
Theorem)
(Exterior
2.
of an secant-tangent
(The meas.
the measure of its
intercepted arc.)
3.
Theorem)
(Exterior 4.
(Substitution)
4.
(Substitution)
5.
(Subtraction Prop.)
5.
(Subtraction Prop.)
6.
(Distributive Prop.)
6.
(Distributive Prop.)
ANSWER: Statements (Reasons)
1.
is a tangent to the circle and
the circle. (Given)
2.
is a secant to
( The ,
meas. of an inscribed
the measure of its intercepted arc.)
3.
Theorem)
(Exterior
4.
(Substitution)
5.
(Subtraction Prop.)
6.
(Distributive Prop.)
32. Case 3
Given: tangent
2.
of an secant-tangent
(The meas.
the measure of its
intercepted arc.)
3.
Theorem)
(Exterior 4.
(Substitution)
5.
(Subtraction Prop.)
6.
(Distributive Prop.)
33. PROOF Write a paragraph proof of Theorem 10.13.
and eSolutions Manual - Powered by Cognero
Prove:
ANSWER: Statements (Reasons)
1.
and are tangents to the circle. (Given)
Page 10
6.
(Distributive Prop.)
measure of 180. By substitution, m∠CAB = m∠CAF
+ 90 and m(arc CDA) = m(arc CF) + 180. Since
∠CAF is inscribed, m∠CAF = m(arc CF) and by
10-6 Secants, Tangents, and Angle Measures
33. PROOF Write a paragraph proof of Theorem 10.13.
a. Given:
is a tangent of
.
is a secant of .
∠CAE is acute.
Prove: m∠CAE = m(arc CA)
b. Prove that if ∠CAB is obtuse, m∠CAB = m
(arc CDA)
SOLUTION: a. Proof: By Theorem 10.10,
. So, ∠FAE
is a right ∠ with measure of 90 and arc FCA is a
semicircle with measure of 180. Since ∠CAE is
acute, C is in the interior of ∠FAE, so by the Angle
and Arc Addition Postulates, m∠FAE = m∠FAC +
m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA).
By substitution, 90 = m∠FAC + m∠CAE and 180 =
substitution, m∠CAB = m(arc CF) + 90. Using the
Division and Subtraction Properties on the Arc
Addition equation yields m(arc CDA) – m(arc
CF) = 90. By substituting for 90, m∠CAB = m(arc
CF) + m(arc CDA) – m(arc CF). Then by
subtraction, m∠CAB = m(arc CDA).
ANSWER: a. Proof: By Theorem 10.10,
. So, ∠FAE
is a right ∠ with measure of 90 and arc FCA is a
semicircle with measure of 180. Since ∠CAE is
acute, C is in the interior of ∠FAE, so by the Angle
and Arc Addition Postulates, m∠FAE = m∠FAC +
m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA).
By substitution, 90 = m∠FAC + m∠CAE and 180 =
m(arc FC) + m(arc CA). So, 90 = m(arc FC) +
(arc CA) by Division Prop., and m∠FAC + m∠CAE
= m(arc FC) + m(arc CA) by substitution.
m∠FAC = m(arc FC) since ∠FAC is inscribed, so
m(arc FC) + m(arc CA). So, 90 = m(arc FC) +
substitution yields m(arc FC) + m∠CAE = m
(arc CA) by Division Prop., and m∠FAC + m∠CAE
(arc FC) + m(arc CA). By Subtraction Prop.,
= m(arc FC) + m(arc CA) by substitution.
m∠CAE = m(arc CA).
m∠FAC = m(arc FC) since ∠FAC is inscribed, so
substitution yields m(arc FC) + m∠CAE = m
b. Proof: Using the Angle and Arc Addition
Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar
CDA) = m(arc CF) + m(arc FDA). Since
and
is a diameter, ∠FAB is a right angle with a
measure of 90 and arc FDA is a semicircle with a
measure of 180. By substitution, m∠CAB = m∠CAF
+ 90 and m(arc CDA) = m(arc CF) + 180. Since
(arc FC) + m(arc CA). By Subtraction Prop.,
m∠CAE = m(arc CA).
b. Proof: Using the Angle and Arc Addition
Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar
CDA) = m(arc CF) + m(arc FDA). Since
and
is a diameter, ∠FAB is a right angle with a
measure of 90 and arc FDA is a semicircle with a
measure of 180. By substitution, m∠CAB = m∠CAF
+ 90 and m(arc CDA) = m(arc CF) + 180. Since
∠CAF is inscribed, m∠CAF = m(arc CF) and by
substitution, m∠CAB = m(arc CF) + 90. Using the
Division and Subtraction Properties on the Arc
Addition equation yields m(arc CDA) – m(arc
∠CAF is inscribed, m∠CAF = m(arc CF) and by
CF) = 90. By substituting for 90, m∠CAB = m(arc
substitution, m∠CAB = m(arc CF) + 90. Using the
Division and Subtraction Properties on the Arc
CF) + m(arc CDA) – m(arc CF). By
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Addition
equation
yields
m(arc
CDA) – m(arc
subtraction, m∠CAB = m(arc CDA).
Page 11
CF) = 90. By substituting for 90, m∠CAB = m(arc
34. OPTICAL ILLUSION The design shown is an
Division and Subtraction Properties on the Arc
Addition equation yields m(arc CDA) – m(arc
10-6CF) = 90. By substituting for 90, m
Secants, Tangents, and Angle Measures
∠CAB = m(arc
CF) + m(arc CDA) – m(arc CF). By
subtraction, m∠CAB = m(arc CDA).
34. OPTICAL ILLUSION The design shown is an
example of optical wallpaper. is a diameter of If m∠A = 26 and
SOLUTION: a. Redraw the circle with points A, B, and C in the
same place but place point D closer to C each time.
Then draw chords
and
.
= 67, what is Refer to the image on page 748.
SOLUTION: First, find the measure of arc BD.
b.
Since
is a diameter, arc BDE is a semicircle and
has a measure of 180.
c. As the measure of
ANSWER: 98
35. MULTIPLE REPRESENTATIONS In this
problem, you will explore the relationship between
Theorems 10.12 and 10.6.
a. GEOMETRIC Copy the figure shown. Then
draw three successive figures in which the position
of point D moves closer to point C, but points A, B,
and C remain fixed.
gets closer to 0, the measure of x approaches half of
becomes an inscribed angle.
d.
ANSWER: a.
b. TABULAR Estimate the measure of
for each successive circle, recording the measures of
b.
and in a table. Then calculate and record the value of x for each circle.
c. VERBAL Describe the relationship between
and the value of x as
approaches zero. What type of angle does ∠AEB become when
d. ANALYTICAL Write an algebraic proof to
show the relationship between Theorems 10.12 and
10.6 described in part c.
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SOLUTION: a. Redraw the circle with points A, B, and C in the
c. As the measure of
gets closer to 0, the measure of x approaches half of
becomes an inscribed angle.
∠AEB
d.
36. WRITING IN MATH Explain how to find the
measure of an angle formed by a secant and a
tangent that intersect outside a circle.
SOLUTION: Find the difference of the two intercepted arcs and
divide by 2.
Page 12
ANSWER: Find the difference of the two intercepted arcs and
measure of x approaches half of
becomes an inscribed angle.
∠AEB
10-6d.Secants, Tangents, and Angle Measures
36. WRITING IN MATH Explain how to find the
measure of an angle formed by a secant and a
tangent that intersect outside a circle.
ANSWER: 15
38. REASONS Isosceles
What can you conclude about 37. CHALLENGE The circles below are concentric.
What is x?
and Explain.
SOLUTION: Find the difference of the two intercepted arcs and
divide by 2.
ANSWER: Find the difference of the two intercepted arcs and
divide by 2.
is inscribed in SOLUTION: Sample answer:
m∠BAC = m∠BCA
because the triangle is isosceles. Since ∠BAC and
∠BCA are inscribed angles, by theorem 10.6, m
(arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC.
So, m(arc AB) = m(arc BC).
ANSWER: Sample answer:
m∠BAC = m∠BCA
because the triangle is isosceles. Since ∠BAC and
∠BCA are inscribed angles, by theorem 10.6, m
(arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC.
SOLUTION: So, m(arc AB) = m(arc BC).
Use the arcs intercepted on the smaller circle to find
m∠A.
39. CCSS ARGUMENTS In the figure,
is a diameter and
is a tangent.
a. Describe the range of possible values for m∠G.
Explain.
b. If m∠G = 34, find the measures of minor arcs HJ
and KH. Explain.
Use the arcs intercepted on the larger circle and
m∠A to find the value of x.
SOLUTION: a.
;
for all values except when
.
at G, then
b.
Because a diameter is involved the intercepted arcs measure (180 – x) and
ANSWER: 15
38. REASONS Isosceles
is inscribed in What can you conclude about Explain.
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and x degrees. Hence solving
leads to the answer.
ANSWER: a.
;
at G, then
b.
for all values except when
Page 13
.
Because a diameter is because the triangle is isosceles. Since ∠BAC and
∠BCA are inscribed angles, by theorem 10.6, m
(arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC.
m(arcTangents,
AB) = m(arc
10-6So,
Secants,
andBC).
Angle Measures
b.
Because a diameter is involved the intercepted arcs measure (180 – x) and
leads to x degrees. Hence solving
the answer.
39. CCSS ARGUMENTS In the figure,
is a diameter and
is a tangent.
a. Describe the range of possible values for m∠G.
Explain.
b. If m∠G = 34, find the measures of minor arcs HJ
and KH. Explain.
SOLUTION: a.
;
for all values except when
.
at G, then
b.
Because a diameter is involved the intercepted arcs measure (180 – x) and
leads to x degrees. Hence solving
the answer.
ANSWER: a.
;
at G, then
40. OPEN ENDED Draw a circle and two tangents
that intersect outside the circle. Use a protractor to
measure the angle that is formed. Find the measures
of the minor and major arcs formed. Explain your
reasoning.
SOLUTION: Sample answer:
By Theorem 10.13,
So, 50° = Therefore, x (minor arc) = 130,
and y (major arc) = 360° – 130° or 230°.
ANSWER: Sample answer:
for all values except when
.
b.
Because a diameter is involved the intercepted arcs measure (180 – x) and
leads to x degrees. Hence solving
By Theorem 10.13,
the answer.
So, 50 = Therefore, x (minor arc) = 130,
40. OPEN ENDED Draw a circle and two tangents
that intersect outside the circle. Use a protractor to
measure the angle that is formed. Find the measures
of the minor and major arcs formed. Explain your
reasoning.
SOLUTION: Sample answer:
and y (major arc) = 360 – 130 or 230.
41. WRITING IN MATH A circle is inscribed within
If m∠P = 50 and m∠Q = 60, describe how
to find the measures of the three minor arcs formed
by the points of tangency.
SOLUTION: By Theorem 10.13,
So, 50° = Sample answer: Use Theorem 10.14 to find each
minor arc. Therefore, x (minor arc) = 130,
and y (major arc) = 360° – 130° or 230°.
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ANSWER: Sample answer:
Page 14
By Theorem 10.13,
So, 50 = Therefore, x (minor arc) = 130,
10-6 Secants, Tangents, and Angle Measures
and y (major arc) = 360 – 130 or 230.
41. WRITING IN MATH A circle is inscribed within
If m∠P = 50 and m∠Q = 60, describe how
to find the measures of the three minor arcs formed
by the points of tangency.
ANSWER: Sample answer: Using Theorem 10.14, 60° = 1/2
((360 – x) – x) or 120°; repeat for 50° to get 130°. The third arc can be found by adding 50° and 60° and subtracting from 360 to get 110°
42. What is the value of x if
and SOLUTION: A 23°
B 31°
C 64°
D 128°
Sample answer: Use Theorem 10.14 to find each
minor arc. SOLUTION: By Theorem 10.14,
.
Substitute.
Simplify.
The sum of the angles of a triangle is 180, so m∠R =
180 – (50 + 60) or 70.
Therefore, the measures of the three minor arcs are 130, 120, and 110.
ANSWER: Sample answer: Using Theorem 10.14, 60° = 1/2
((360 – x) – x) or 120°; repeat for 50° to get 130°. The third arc can be found by adding 50° and 60° and subtracting from 360 to get 110°
42. What is the value of x if
and So, the correct choice is C.
ANSWER: C
43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on
circle C, and
is a diameter. What are the coordinates of C?
F (2, 10)
G (10, –6)
H (5, –3)
J (1, 5)
SOLUTION: A 23°
B 31°
C 64°
D 128°
Here, C is the midpoint of
Use the midpoint formula,
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.
, to find the coordinates of
C.
.
Page 15
Substitute.
So, the correct choice is J.
So, the correct choice is C.
10-6ANSWER: Secants, Tangents, and Angle Measures
C
43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on
circle C, and
is a diameter. What are the coordinates of C?
F (2, 10)
G (10, –6)
H (5, –3)
J (1, 5)
.
ANSWER: 35
45. SAT/ACT If the circumference of the circle below
is 16π, what is the total area of the shaded regions?
2
A 64π units
B 32π units2
SOLUTION: Here, C is the midpoint of
Use the midpoint formula,
,
Since
2
C 12π units
D 8π units2
.
2
, to find the coordinates of
D 2π units
C.
SOLUTION: First, use the circumference to find the radius of the
circle.
So, the correct choice is J.
ANSWER: J
The shaded regions of the circle comprise half of the
circle, so its area is half the area of the circle.
44. SHORT RESPONSE If m∠AED = 95 and
what is m∠BAC?
The area of the shaded regions is 32π.
SOLUTION: So, the correct choice is B.
Since
.
ANSWER: B
We know that vertical angles are congruent.
So,
.
Since
,
.
We know that the sum of the measures of all interior
angles of a triangle is 180.
Substitute.
Since
Find x. Assume that segments that appear to be
tangent are tangent.
46. ,
.
ANSWER: 35
SOLUTION: By theorem 10.10,
. So,
triangle.
Use the Pythagorean Theorem.
is a right Substitute.
45. SAT/ACT If the circumference of the circle below
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is 16π,
what
is the total
area of the shaded regions?
Page 16
The area of the shaded regions is 32π.
So, the correct choice is B.
10-6ANSWER: Secants, Tangents, and Angle Measures
B
ANSWER: 8
Find x. Assume that segments that appear to be
tangent are tangent.
48. SOLUTION: Use Theorem 10.10 and the Pythagorean Theorem.
46. SOLUTION: By theorem 10.10,
. So,
triangle.
Use the Pythagorean Theorem.
is a right Solve for x.
Substitute.
ANSWER: ANSWER: 3
49. PROOF Write a two-column proof.
Given:
is a semicircle; Prove:
47. SOLUTION: If two segments from the same exterior point are
tangent to a circle, then they are congruent.
SOLUTION: Given:
is a semicircle.
Prove:
ANSWER: 8
48. SOLUTION: Use Theorem 10.10 and the Pythagorean Theorem.
Solve for x.
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ANSWER: Proof:
Statements (Reasons)
1. MHT is a semicircle;
(Given)
2.
is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.)
3.
is a right angle (Def. of ⊥ lines)
4.
(All rt. angles are .)
5. (Reflexive Prop.)
6.
(AA Sim.)
7.
(Def. of Page 17
intercepts a semicircle, the angle is a right angle.)
3.
is a right angle (Def. of ⊥ lines)
.)
(All rt. angles are 10-64.Secants, Tangents,
and Angle Measures
5. (Reflexive Prop.)
6.
(AA Sim.)
7.
(Def. of ANSWER: Given:
is a semicircle.
4.
5. 6.
7.
(All rt. angles are .)
(Reflexive Prop.)
(AA Sim.)
(Def. of 50. REMODELING The diagram at the right shows
the floor plan of Trent’s kitchen. Each square on the
diagram represents a 3-foot by 3-foot area. While
remodeling his kitchen, Trent moved his refrigerator
from square A to square B. Describe one possible
combination of transformations that could be used to
make this move.
Prove:
Proof:
Statements (Reasons)
1. MHT is a semicircle;
(Given)
2.
is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.)
3.
is a right angle (Def. of ⊥ lines)
4.
(All rt. angles are .)
5. (Reflexive Prop.)
6.
(AA Sim.)
7.
(Def. of 50. REMODELING The diagram at the right shows
the floor plan of Trent’s kitchen. Each square on the
diagram represents a 3-foot by 3-foot area. While
remodeling his kitchen, Trent moved his refrigerator
from square A to square B. Describe one possible
combination of transformations that could be used to
make this move.
SOLUTION: The move is a translation 15 feet out from the wall
and 21 feet to the left, then a rotation of 90° counterclockwise.
ANSWER: The move is a translation 15 feet out from the wall
and 21 feet to the left, then a rotation of 90° counterclockwise.
SOLUTION: The move is a translation 15 feet out from the wall
and 21 feet to the left, then a rotation of 90° counterclockwise.
ANSWER: The move is a translation 15 feet out from the wall
and 21 feet to the left, then a rotation of 90° counterclockwise.
COORDINATE GEOMETRY Find the
measure of each angle to the nearest tenth of a
degree by using the Distance Formula and an
inverse trigonometric ratio.
51. ∠C in triangle BCD with vertices B(–1, –5), C(–6, –
5), and D(–1, 2)
SOLUTION: Use the Distance Formula to find the length of each
side.
Since
, triangle BCD is a right
triangle.
So, CD is the length of the hypotenuse and BC is the
length of the leg adjacent to
Write an equation using the cosine ratio.
If
Use a calculator.
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COORDINATE GEOMETRY Find the
measure of each angle to the nearest tenth of a
degree by using the Distance Formula and an
Page 18
ANSWER: ANSWER: The move is a translation 15 feet out from the wall
10-6and 21 feet to the left, then a rotation of 90° Secants, Tangents, and Angle Measures
counterclockwise.
COORDINATE GEOMETRY Find the
measure of each angle to the nearest tenth of a
degree by using the Distance Formula and an
inverse trigonometric ratio.
51. ∠C in triangle BCD with vertices B(–1, –5), C(–6, –
5), and D(–1, 2)
ANSWER: 54.5
52. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, –
2), and Z(7, –2)
SOLUTION: Use the Distance Formula to find the length of each
side.
SOLUTION: Use the Distance Formula to find the length of each
side.
In right triangle XYZ, ZX is the length of the
hypotenuse and YZ is the length of the leg opposite
Write an equation using the sine ratio.
Since
, triangle BCD is a right
triangle.
So, CD is the length of the hypotenuse and BC is the
length of the leg adjacent to
Write an equation using the cosine ratio.
If
Use a calculator.
ANSWER: 54.5
If
Use a calculator.
ANSWER: 51.3
Solve each equation.
2
53. x + 13x = –36
SOLUTION: 52. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, –
2), and Z(7, –2)
SOLUTION: Use the Distance Formula to find the length of each
side.
Therefore, the solution is –9, –4.
ANSWER: –4, –9
2
54. x – 6x = –9
SOLUTION: In right triangle XYZ, ZX is the length of the
hypotenuse and YZ is the length of the leg opposite
Write an equation using the sine ratio.
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If
Therefore, the solution is 3.
ANSWER: 3
2
55. 3x + 15x = 0
Page 19
Therefore, the solution is –9, –4.
Therefore, the solution is –6.
ANSWER: 10-6–4,
Secants,
Tangents, and Angle Measures
–9
ANSWER: –6
2
54. x – 6x = –9
58. SOLUTION: SOLUTION: Therefore, the solution is 3.
ANSWER: 3
2
55. 3x + 15x = 0
Therefore, the solution is
.
ANSWER: SOLUTION: Therefore, the solution is –5, 0.
ANSWER: 0, –5
2
56. 28 = x + 3x
SOLUTION: Therefore, the solution is –7, 4.
ANSWER: –7, 4
2
57. x + 12x + 36 = 0
SOLUTION: Therefore, the solution is –6.
ANSWER: –6
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SOLUTION: Page 20