trig 3e ism chapter 1 final · the key to solving this problem is setting up the correct...

1

CHAPTER 1 Section 1.1 Solutions --------------------------------------------------------------------------------

1. Solve for x: 12 360

x=

360 2 , so that 180x x= = .


x=

360 4 , so that 90x x= = .


x− =

360 3 , so that 120x x= − = − . (Note: The angle has a negative measure since it is a clockwise rotation.)


x− =

720 2(360 ) 3 , so that 240x x= = − = − . (Note: The angle has a negative measure since it is a clockwise rotation.)


x=

1800 5(360 ) 6 , so that 300x x= = = .

6. Solve for x: 712 360

x=

2520 7(360 ) 12 , so that 210x x= = = .


x− =



x− =


9. a) complement: 90 18 72− =

b) supplement: 180 18 162− =

10. a) complement: 90 39 51− =

b) supplement: 180 39 141− = 11. a) complement: 90 42 48− =

b) supplement: 180 42 138− =

12. a) complement: 90 57 33− =

b) supplement: 180 57 123− = 13. a) complement: 90 89 1− =

b) supplement: 180 89 91− =

14. a) complement: 90 75 15− =

b) supplement: 180 75 105− =

15. Since the angles with measures ( )4x and ( )6x are assumed to be complementary,

we know that ( ) ( )4 6 90x x+ = . Simplifying this yields ( )10 90x = , so that 9x = .

So, the two angles have measures 36 and 54 .

Chapter 1

2

16. Since the angles with measures ( )3x and ( )15x are assumed to be supplementary,

we know that ( ) ( )3 15 180x x+ = . Simplifying this yields ( )18 180x = , so that

10x = . So, the two angles have measures 30 and 150 .

17. Since the angles with measures ( )8x and ( )4x are assumed to be supplementary, we

know that ( ) ( )8 4 180x x+ = . Simplifying this yields ( )12 180x = , so that 15x = .

So, the two angles have measures 60 and 120 .

18. Since the angles with measures ( )3 15x + and ( )10 10x + are assumed to be

complementary, we know that ( ) ( )3 15 10 10 90x x+ + + = . Simplifying this yields

( )13 25 90x + = , so that ( )13 65x = and thus, 5x = . So, the two angles have measures

30 and 60 . 19. Since 180α β γ+ + = , we know that

150

117 33 180 and so, 30γ γ=

+ + = = .

20. Since 180α β γ+ + = , we know that

155

110 45 180 and so, 25γ γ=

+ + = = .

21. Since 180α β γ+ + = , we know that ( ) ( )

6

4 180 and so, 30β

β β β β=

+ + = = .

Thus, 4 120 and 30α β γ β= = = = .

22. Since 180α β γ+ + = , we know that ( ) ( )

5

3 180 and so, 36β

β β β β=

+ + = = .

Thus, 3 108 and 36α β γ β= = = = .

23. ( )180 53.3 23.6 103.1α = − + = 24. ( )180 105.6 13.2 61.2β = − + = 25. Since this is a right triangle, we know from the Pythagorean Theorem that

2 2 2a b c+ = . Using the given information, this becomes 2 2 24 3 ,c+ = which simplifies to 2 25c = , so we conclude that 5c = .

26. Since this is a right triangle, we know from the Pythagorean Theorem that 2 2 2a b c+ = . Using the given information, this becomes 2 2 23 3 ,c+ = which simplifies to 2 18c = , so we conclude that 18 3 2c = = .

27. Since this is a right triangle, we know from the Pythagorean Theorem that 2 2 2a b c+ = . Using the given information, this becomes 2 2 26 10 ,b+ = which simplifies

to 2 236 100 and then to, 64b b+ = = , so we conclude that 8b = .

Section 1.1

3

28. Since this is a right triangle, we know from the Pythagorean Theorem that 2 2 2a b c+ = . Using the given information, this becomes 2 2 27 12 ,a + = which simplifies

to 2 95a = , so we conclude that 95a = . 29. Since this is a right triangle, we know from the Pythagorean Theorem that

2 2 2a b c+ = . Using the given information, this becomes 2 2 28 5 ,c+ = which simplifies to 2 89c = , so we conclude that 89c = .

30. Since this is a right triangle, we know from the Pythagorean Theorem that 2 2 2a b c+ = . Using the given information, this becomes 2 2 26 5 ,c+ = which simplifies to 2 61c = , so we conclude that 61c = .

31. Since this is a right triangle, we know from the Pythagorean Theorem that 2 2 2a b c+ = . Using the given information, this becomes 2 2 27 11 ,b+ = which simplifies

to 2 72b = , so we conclude that 72 6 2b = = . 32. Since this is a right triangle, we know from the Pythagorean Theorem that

2 2 2a b c+ = . Using the given information, this becomes 2 2 25 9 ,a + = which simplifies to 2 56a = , so we conclude that 56 2 14a = = .

33. Since this is a right triangle, we know from the Pythagorean Theorem that 2 2 2a b c+ = . Using the given information, this becomes ( )2

2 27 5 ,a + = which

simplifies to 2 18a = , so we conclude that 18 3 2a = = . 34. Since this is a right triangle, we know from the Pythagorean Theorem that

2 2 2a b c+ = . Using the given information, this becomes 2 2 25 10 ,b+ = which simplifies

to 2 75b = , so we conclude that 75 5 3b = = . 35. If 10x = in., then the hypotenuse of

this triangle has length 10 2 14.14 in.≈

36. If 8x = m, then the hypotenuse of this

triangle has length 8 2 11.31 m≈ .

37. Let x be the length of a leg in the given 45 45 90− − triangle. If the hypotenuse of this triangle has length 2 2 cm, then 2 2 2, so that 2x x= = . Hence, the length of each of the two legs is 2 cm . 38. Let x be the length of a leg in the given 45 45 90− − triangle. If the hypotenuse

of this triangle has length 10 ft., then 10 102 10, so that 522

x x= = = = .

Hence, the length of each of the two legs is 5 ft.

Chapter 1

4

39. The hypotenuse has length

( )2 4 2 in. 8in.= 40. Since 6 2

22 6m 3 2mx x= ⇒ = = ,

each leg has length 3 2 m . 41. Since the lengths of the two legs of the given30 60 90− − triangle are x and 3 x , the shorter leg must have length x. Hence, using the given information, we know that

5mx = . Thus, the two legs have lengths 5 m and 5 3 8.66 m,≈ and the hypotenuse has length 10 m. 42. Since the lengths of the two legs of the given 30 60 90− − triangle are x and 3 x , the shorter leg must have length x. Hence, using the given information, we know that

9ft.x = Thus, the two legs have lengths 9 ft. and 9 3 15.59 ft.,≈ and the hypotenuse has length 18 ft. 43. The length of the longer leg of the given triangle is 3 12x = yards. So,

12 12 3 4 333

x = = = . As such, the length of the shorter leg is 4 3 6.93≈ yards, and

the hypotenuse has length 8 3 13.9≈ yards. 44. The length of the longer leg of the given triangle is 3x n= units. So,

333

n nx = = . As such, the length of the shorter leg is 33

n units, and the hypotenuse

has length 2 33

n units.

45. The length of the hypotenuse is 2 10x = inches. So, 5x = . Thus, the length of the shorter leg is 5 inches, and the length of the longer leg is 5 3 8.66≈ inches. 46. The length of the hypotenuse is 2 8x = cm. So, 4x = . Thus, the length of the shorter leg is 4 cm, and the length of the longer leg is 4 3 6.93≈ cm.

Section 1.1

5

47. For simplicity, we assume that the minute hand is on the 12. Let α =measure of the desired angle, as indicated in the diagram below. Since the measure of the angle formed using two rays emanating from the center of the

clock out toward consecutive hours is always ( )1 360 3012

= , it immediately follows that

( )4 30 120α = ⋅ − = − (Negative since measured clockwise.)

48. For simplicity, we assume that the minute hand is on the 9. Let α =measure of the desired angle, as indicated in the diagram below. Since the measure of the angle formed using two rays emanating from the center of the

clock out toward consecutive hours is always ( )1 360 3012

= , it immediately follows that

( )5 30 150α = ⋅ − = − . (Negative since measured clockwise.)

α

α

Chapter 1

6

49. The key to solving this problem is setting up the correct proportion. Let x = the measure of the desired angle. From the given information, we know that since 1 complete revolution corresponds to360 , we obtain the following proportion:

36030 minutes 12 minutes

x=

Solving for x then yields

12 minutesx = ( ) 36030 minutes

144⎛ ⎞

=⎜ ⎟⎝ ⎠

.

50. The key to solving this problem is setting up the correct proportion. Let x = the measure of the desired angle. From the given information, we know that since 1 complete revolution corresponds to360 , we obtain the following proportion:

36030 minutes 5 minutes

x=

Solving for x then yields

5 minutesx = ( ) 36030 minutes

60⎛ ⎞

=⎜ ⎟⎝ ⎠

.

51. We know that 1 complete revolution corresponds to360 . Let x = time (in minutes) it takes to make 1 complete revolution about the circle. Then, we have the following proportion:

270 36045 minutes x

=

Solving for x then yields ( )( )

270 360 45 minutes360 45 minutes

60 minutes.270

x

x

=

= =

So, it takes one hour to make one complete revolution. 52. We know that 1 complete revolution corresponds to360 . Let x = time (in minutes) it takes to make 1 complete revolution about the circle. Then, we have the following proportion:

72 3609 minutes x

=

Solving for x then yields ( )( )

72 360 9 minutes360 9 minutes

45 minutes.72

x

x

=

= =

So, it takes 45 minutes to make one complete revolution.

Section 1.1

7

53. Let d = distance (in feet) the dog runs along the hypotenuse. Then, from the Pythagorean Theorem, we know that

2 2 2

2

30 807,300

85 7,300

dd

d

+ ==

≈ =

So, 85 feetd ≈ . 54. Let d = distance (in feet) the dog runs along the hypotenuse. Then, from the Pythagorean Theorem, we know that

2 2 2

2

25 10010,625

103 10,625

dd

d

+ ==

≈ =

So, 103 feetd ≈ . 55. Consider the following triangle T.

Since T is a 45 45 90− − triangle, the two legs (i.e., the sides opposite the angles with measure 45 ) have the same length. Call this length x. Since the hypotenuse of such a

triangle has measure 2x , we have that 2 100x = , so that 100 100 2 50 222

x = = = .

So, since lights are to be hung over both legs and the hypotenuse, the couple should buy 50 2 50 2 100 100 100 2 241 feet+ + = + ≈ of Christmas lights.

45 45

Chapter 1

8

56. Consider the following triangle T.

Since T is a 45 45 90− − triangle, the two legs (i.e., the sides opposite the angles with measure 45 ) have the same length. Call this length x. Since the hypotenuse of such a

triangle has measure 2x , we have that 2 60x = , so that 60 60 2 30 222

x = = = .

So, since lights are to be hung over both legs and the hypotenuse, the couple should buy 30 2 30 2 60 60 60 2 145 feet+ + = + ≈ of Christmas lights.

45 45

Section 1.1

9

57. Consider the following diagram:

The dashed line segment AD represents the TREE and the vertices of the triangle ABC represent STAKES. Also, note that the two right triangles ADB and ADC are congruent (using the Side-Angle-Side Postulate from Euclidean geometry). Let x = distance between the base of the tree and one staked rope (measured in feet). For definiteness, consider the right triangle ADC. Since it is a 30 60 90− − triangle, the side opposite the 30 -angle (namely DC) is the shorter leg, which has length x feet. Then, we know that the hypotenuse must have length 2x. Thus, by the Pythagorean Theorem, it follows that:

2 2 2

2 2

2

2

17 (2 )289 4289 32893

2899.83

x xx x

x

x

x

+ =+ =

=

=

≈ =

So, the ropes should be staked approximately 9.8 feet from the base of the tree. 58. Using the computations from Problem 57, we observe that since the length of the

hypotenuse is 2x, and 2893

x = , it follows that the length of each of the two ropes

should be 2892 19.6299 feet3

≈ . Thus, one should have 2 19.6299 39.3 feet× ≈ of rope

in order to have such stakes support the tree.

6060

30 30

Chapter 1

10


The dashed line segment AD represents the TREE and the vertices of the triangle ABC represent STAKES. Also, note that the two right triangles ADB and ADC are congruent (using the Side-Angle-Side Postulate from Euclidean geometry). Let x = distance between the base of the tree and one staked rope (measured in feet). For definiteness, consider the right triangle ADC. Since it is a 30 60 90− − triangle, the side opposite the 30 -angle (namely AD) is the shorter leg, which has length 10 feet. Then, we know that the hypotenuse must have length 2(10) = 20 feet. Thus, by the Pythagorean Theorem, it follows that:

2 2 2

2

2

10 20100 400

300300 17.3 feet

xx

xx

+ =+ =

=

= ≈

So, the ropes should be staked approximately 17.3 feet from the base of the tree. 60. Using the computations from Problem 59, we observe that since the length of the hypotenuse is 20 feet, it follows that the length of each of rope tied from tree to the stake in this manner should be 20 feet in length. Hence, for four stakes, one should have 4 20 80 feet× ≈ of rope.

60 60

30 30

Section 1.1

11

61. The following diagram is a view from one of the four sides of the tented area – note that the actual length of the side of the tent which we are viewing (be it 40 ft. or 20 ft.) does not affect the actual calculation since we simply need to determine the value of x, which is the amount beyond the length or width of the tent base that the ropes will need to extend in order to adhere the tent to the ground.

Now, solving this problem is very similar to solving Problem 57. The two right triangles labeled in the diagram are congruent. So, we can focus on the leftmost one, for definiteness. The side opposite the angle with measure30 is the shorter leg, the length of which is x. So, the hypotenuse has length 2x. From the Pythagorean Theorem, it then follows that

2 2 2

2

7 4933

7 (2 )49 34.0

x xx

x

+ ==

≈ ≈ =

Hence, along any of the four edges of the tent, the staked rope on either side extends approximately 4 feet beyond the actual dimensions of the tent. As such, the actual footprint of the tent is approximately ( ) ( )40 2(4) ft. 20 2(4) ft.+ × + , which is

48ft. 28ft.×

60 60

Chapter 1

12

62. The following diagram is a view from one of the four sides of the tented area – note that the actual length of the side of the tent which we are viewing (be it 80 ft. or 40 ft.) does not affect the actual calculation since we simply need to determine the value of x, which is the amount beyond the length or width of the tent base that the ropes will need to extend in order to adhere the tent to the ground.

Now, solving this problem is very similar to solving Problem 53. The two right triangles labeled in the diagram are congruent. So, we can focus on the leftmost one, for definiteness. Since this is a 45 45 90− − triangle, the lengths of the two legs must be equal. So, 7x = . Hence, along any of the four edges of the tent, the staked rope on either side must extend 7 feet beyond the actual dimensions of the tent. As such, the actual footprint of the tent is approximately ( ) ( )40 2(7) ft. 80 2(7) ft.+ × + , which is

54ft. 94ft.× .

63. The corner is not 90 because 2 2 210 15 20+ ≠ . 64. 2 2 2 28 17 225 15ft.x x x+ = ⇒ = ⇒ = 65. The speed is 1700 170

60 6= revolutions per second. Since each revolution corresponds to

360 , the engine turns ( )( )1706 360 10,200= each second.

66. The speed is 300,00015 20,000= per second. Since each revolution corresponds to

360 , this amounts to 5009 revolutions per second. Since 1 minute = 60 seconds, the speed

is 10,0005009 360 3,333.3× = ≈ RPMs.

67. In a 30 60 90− − triangle, the length opposite the 60 -angle has length ( )3 shorter leg× , not ( )2 shorter leg× . So, the side opposite the 60 -angle has length

10 3 17.3 inches≈ . 68. The length of the hypotenuse must be positive. Hence, the length must be 5 2 cm . 69. False. Each of the three angles of an equilateral triangle has measure 60 . But, in order to apply the Pythagorean theorem, one of the three angles must have measure90 . 70. False. Since the Pythagorean theorem doesn’t apply to equilateral triangles, and equilateral triangles are also isosceles (since at least two sides are congruent), we conclude that the given statement is false.

4545

Section 1.1

13

71. True. Since the angles of a right triangle are , , and 90α β , and also we know that 90 180α β+ + = , it follows that 90α β+ = .

72. False. The length of the side opposite the 60 -angle is 3 times the length of the side opposite the 30 -angle. 73. True. The sum of the angles , ,90α β must be 180 . Hence, 90α β+ = , so that α and β are complementary. 74. False. The legs have the same length x, but the hypotenuse has length 2x . 75. True. Angles swept out counterclockwise have a positive measure, while those swept out clockwise have negative measure. 76. True. Since the sum of the angles , ,90α β must be 180 , 90α β+ = . So, neither angle can be obtuse. 77. First, note that at 12:00 exactly, both the minute and the hour hands are identically on the 12. Then, for each minute that passes, the minute hand moves 1

60 the way around the clock face (i.e., 6 ). Similarly, for each minute that passes, the hour hand moves

160 the way between the 12 and the 1; since there are 1

12 (360 ) 30= between consecutive integers on the clock face, such movement corresponds to 1

60 (30 ) 0.5= . Now, when the time is 12:20, we know that the minute hand is on the 4, but the hour hand has moved 20 0.5 10× = clockwise from the 12 towards the 1. The picture is as follows:

The angle we seek is 1 2 3β α α α+ + + . From the above discussion, we know that

1 2 3 30α α α= = = and 20β = . Thus, the angle at time 12:20 is 110 .

β1α

2α

3α

Chapter 1

14

78. First, note that at 9:00 exactly, the minute is identically on the 12 and the hour hand is identically on the 9. Then, for each minute that passes, the minute hand moves 1

60 the way around the clock face (i.e., 6 ). Similarly, for each minute that passes, the hour hand moves 1

60 the way between the 9 and the 10; since there are 112 (360 ) 30= between

consecutive integers on the clock face, such movement corresponds to 160 (30 ) 0.5= .

Now, when the time is 9:10, we know that the minute hand is on the 2, but the hour hand has moved 10 0.5 5× = clockwise from the 9 towards the 10, thereby leaving an angle of 25 between the hour hand and the 10. The picture is as follows:

The angle we seek is 1 2 3 4β α α α α+ + + + . From the above discussion, we know that

1 2 3 4 30α α α α= = = = and 25β = . Thus, the angle at time 9:10 is 145 .

β1α 2α 3α

4α

Section 1.1

15


Let x = length of DC and y = length of BD . Ultimately, we need to determine x. We proceed as follows. First, we find y. Using the Pythagorean Theorem on ABD yields 2 2 23 5y+ = , so that

4y = . Next, using the Pythagorean Theorem on ABC yields

( )22 2

2

2

( ) 3 58

(4 ) 9 58(4 ) 49

4 7 so that 11

y x

xx

x x

+ + =

+ + =+ =+ = ± = − or 3

So, the length of DC is 3.

58

Chapter 1

16


Since we seek the length of DC (which we shall denote as DC), we first need only to apply the Pythagorean Theorem to ABD to find the length of BD . Indeed, observe that

2 2 24 ( ) 5BD+ = , so that 3BD = . Next, we apply the Pythagorean Theorem to ABC to find DC:

( ) ( )( ) ( )

( ) ( )

222

2

2

4 3 41

16 9 6 41

6 16 0( 8)( 2) 0

8

DC

DC DC

DC DCDC DC

DC

+ + =

+ + + =

+ − =+ − =

= − , 2

So, the length of DC is 2 .

41

Section 1.1

17


Let x = length of AD and y = length of BD . Ultimately, we need to determine x. We proceed as follows. First, we find y. Using the Pythagorean Theorem on ABC yields

2 2 2

2

18

24 ( 11) 30( 11) 324

11 324

11 187 or 29

yy

y

yy

=

+ + =+ =

+ = ±

= − ±= −

so that 7y = . Next, using the Pythagorean Theorem on ABD yields 2 2 2

2 2 2

2

2424 7

62525

y xxxx

+ =+ =

==

So, the length of AD is 25.

Chapter 1

18


Let x = length of BD and y = length of AC . Ultimately, we need to determine y We proceed as follows. First, we find x. Using the Pythagorean Theorem on ABD yields

2 2 2

2

60 6112111

xxx

+ ===

so that 11x = . Next, using the Pythagorean Theorem on ABC yields 2 2 2

2 2 2

2

60 ( 36)60 47

5,80976.2

x yyyy

+ + =+ =

=≈

So, the length of AC is approximately 76.2. 83. Consider the following diagram:

Observe that BD = AC (in fact, by the Pythagorean Theorem, both = 2x ). Furthermore, since the diagonals of a square bisect each other in a 90 − angle, the measure of each of

the angles AEB, BEC, DEC, and AED is 90 . Further, 22

xAE EC BE ED= = = = .

Hence, by the Side-Side-Side postulate from Euclidean geometry, all four triangles in the above picture are congruent. Regarding their angle measures, since the diagonals bisect the angles at their endpoints, each of these triangles is a 45 45 90− − triangle.

Section 1.1

19


Using the Pythagorean Theorem, we find that

2 2 2

2 2 2

2

(3 2) (2 2)9 12 4 4 8 4

6 20 02 (3 10) 0

0

x x xx x x x x

x xx x

x

+ − = ++ − + = + +

− =− =

= 103 or

So, 103x = (since if x = 0, then there would be no triangle to speak of).

85. Since the lengths of the two legs of the given 30 60 90− − triangle are x and 3 x , the shorter leg must have length x. Hence, using the given information, we know that

16.68ft.x = Thus, the two legs have lengths 16.68 ft. and 16.68 3 28.89 ft.,≈ and the hypotenuse has length 33.36 ft. 86. Since the lengths of the two legs of the given 30 60 90− − triangle are x and 3 x , the shorter leg must have length x. Hence, using the given information, we know that

3 134.75x cm= Thus, the two legs have lengths 134.753

134.75 and 77.80 ,cm cm≈ and the hypotenuse has length155.60 cm. Section 1.2 Solutions ---------------------------------------------------------------------------------- 1. 80B = (vertical angles) 2. 80 180E+ = (interior angles), so that

100E = 3. 80F = (alternate interior angles) 4. 80G = (corresponding angles) 5. 75B = (corresponding angles) 6. Since 75F = and 180D F+ = (interior angles), we know that 105D = . Hence,

105A = (vertical angles).

Chapter 1

20

7. Since 65 , it follows that 65G F= = since vertical angles are congruent. Hence, 65B = since corresponding angles are congruent.

8. Since 65 , it follows that 65G F= = since vertical angles are congruent. Since A and B are supplementary angles, 115A = . 9. 8 9 15x x= − (vertical angles), so that 15x = . Thus, 8(15 ) 120A D= = = . 10. 9 7 11 7x x+ = − (corresponding angles), so that solving for x yields 7x = . Thus,

(9(7) 7) 70B F= + = = . 11. Since 180A C+ = and C = G, it follows that 180A G+ = . As such, observe that

(12 14) (9 2) 18021 12 180

21 1688

x xx

xx

+ + − =+ =

==

Thus, ( )12(8) 14 110A = + = and ( )9(8) 2 70G = − = . 12. We know that (22 3) (30 5) 180x x+ + − = (interior angles). Solving for x yields

18252

(22 3) (30 5) 18052 2 180

52 1823.5

x xx

xx

+ + − =− =

== =

Thus, ( )22(3.5) 3 80C = + = and ( )30(3.5) 5 100E = − = . 13. b 14. c 15. a 16. f 17. d 18. e

19. By similarity, a cd f= , so that 4 6

2 f= . Solving for f yields 3 .f =

20. By similarity, a bd e= , so that 12 9

3d= . Solving for d yields 4 .d =

21. By similarity, b ae d= , so that

3

7.52.5 5

a

=

= . Solving for a yields 15 .a =

22. By similarity, b ce f= , so that 3.9

1.4 2.6b

= . Solving for b yields 2.1.b =

23. First, note that 26.25 26,250km m= . Now, observe that by similarity, d af c= , so that

1.12.5 26,250

m am m= . Solving for a yields

2

2

28,8752.5

(2.5 ) 28,87511,550 11.55m

m

m a ma m km=

= = =

Section 1.2

21

24. First, note that 35 3,500m cm= . Now, observe that by similarity, c bf e= , so that

3,50014 10

cm bcm cm

= . Solving for b yields

2

2

35,00014

(14 ) 35,0002,500 25cm

cm

cm b cmb cm m=

= = =

25. By similarity, b ec f= , so that

45

5 32 2

in.in. in.

e= . Solving for e yields

( ) ( )( )( )( )

( )34

5 25

2

5 342 5 2

in. in. 1225in.

in. in. in.

in.

e

e

=

= =

26. By similarity, d ae b= , so that

716

514 4

mmmm mm

a= . Solving for a yields

( ) ( )( )( )( )

( )7 5

16 41

4

7 514 16 4

mm mm 3516mm

mm mm mm

mm

a

e

=

= =

27. Note that 1414 14.25= . Now, consider the following two diagrams.

Let y = height of the tree (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain

2

2

14.25 ft.4 ft. 1.5 ft.

(1.5 ft.) 57 ft.57 ft. 38 ft.1.5 ft.

y

y

y

=

=

= =

So, the tree is 38 feet tall.

Chapter 1

22

28. Note that 34 0.75= . Now, consider the following two diagrams.

Let y = height of the flag pole (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain

2

2

15 ft.2 ft. 0.75 ft.

(0.75 ft.) 30 ft.30 ft. 40 ft.0.75 ft.

y

y

y

=

=

= =

So, the flag pole is 40 feet tall. 29. Consider the following two diagrams.

Let y = height of the lighthouse (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain

2

2

5 ft. 1.2 ft.48 ft.

(1.2 ft.) 240 ft.240 ft. 200 ft.1.2 ft.

yy

y

=

=

= =

So, the lighthouse is 200 feet tall.

48 ft.

Section 1.2

23

30. Consider the following two diagrams.

Let y = length of the son’s shadow (in feet). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain

22 2

36 ft. 1 ft. 4 ft. so that (6 ft.) 4 ft. and hence, ft.4 ft. 6 ft.

y yy

= = = =

So, the son’s shadow is 23 ft. long. Since 1 ft. 12 in.= , this is equivalent to 8 in.

31. First, make certain to convert all quantities involved to a common unit. In order to avoid using decimals, use the smallest unit – in this particular problem, use cm. Now, consider the following two diagrams:

Let y = height of the lighthouse (in cm). Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain

2

2

200 5300cm

(5 ) 60,000 60,000 12,000 120

5

cm cmy

cm y cmcmy cm m

cm

=

=

= = =

So, the lighthouse is 120 m tall.

Chapter 1

24

32. Let H = the height of the pyramid (in meters). (Important! Don’t confuse this with the slant height.) Using the fact that the height is the length of the segment that extends from the apex of the pyramid down to the center of the square base, we have the following diagram. Also, we have

Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain

( )1310.9

(115 16)1 0.9m

0.9 131145.56 146m

m

H mm

m H mH m m

+=

== = ≈

So, the pyramid is approximately 146 m tall. 33. Let x = length of the planned island (in inches). (Note that 2 4 28in.′ ′′ = ) Using similarity, we obtain

( ) ( )( )

1 114 16

211 116 4

21116

14

28 in. in. in.

28 in. in.28 in.

77 in. 6 ft. 5 in. in.

x

x

x

=

=

= = =

So, the planned island is 6ft. 5in. long.

Section 1.2

25

34. Let x = width of the pantry (in inches). Using similarity, we obtain

( ) ( )( )

714 16

27 116 4

2716

14

28 in. in. in.

28 in. in.28 in.

49 in. 4 ft. 1 in. in.

x

x

x

=

=

= = =

So, the pantry is 4ft. 1in. long. 35. Consider the following two triangles:

The triangles are similar by AAA, so that 5 3.5

4 x= . Solving for x yields 2.8ft.x = 36. Consider the following diagram: Since this is a 30 60 90− − triangle and x

is the longer side (being opposite the larger of the two acute angles), we know that

15 3 26ft.x = ≈

37. The two triangles are similar by AAA. So, we have

( ) ( )( )1.6 in. 2.2 in. 1.6 in. 1.2 in. 0.87 in.2.2 in. 1.2 in.

x x x= ⇒ = ⇒ ≈

38. If a right triangle is isosceles, then it must be a 45 45 90− − triangle. Let x = length of a leg. Then, the hypotenuse is 2 2x = , so that 2 in.x =

60

Chapter 1

26

39. The information provided is: 2 8 , 2.5 , 80d t sf cm d cm d cm= = = .

We must determine the value of 2 sf . Using similar triangles, we have the following proportion:

80 2.5 804

s s t

s d s

d d df f f

+ += ⇒ = .

Solving for sf yields: 32082.5 320 3.8787882.5s sf f= ⇒ = ≈

Thus, 2 7.8sf cm≈ . 40. The information provided is:

2 4 , so that 23.5

2 3.8 , so that 1.9

d d

t

s s

f cm f cmd cm

f cm f cm

= === =

We must determine the value of sd . Using similar triangles, we have the following proportion:

3.51.9 2

s s t s s

s d

d d d d df f

+ += ⇒ = .

Solving for sd yields: 2 1.9 6.65 0.1 6.65 66.5s s s sd d d d cm= + ⇒ = ⇒ = . 41. The information provided is:

2 4 , so that 260

2 3.75 , so that 1.875

d d

s

s s

f cm f cmd cm

f cm f cm

= === =

We must determine the value of td . Using similar triangles, we have the following proportion:

60601.875 2

s s t t

s d

d d d df f

+ += ⇒ = .

Solving for td yields: 120 112.5 1.875 7.5 1.875 4t t td d d cm= + ⇒ = ⇒ = .

Section 1.2

27

42. The information provided is: 2 4.15 , 4.5 , 60s t sf cm d cm d cm= = = .

We must determine the value of 2 df . Using similar triangles, we have the following proportion:

60 60 4.52.075

s s t

s d d

d d df f f

+ += ⇒ = .

Solving for df yields: 133.837560 133.8375 2.230660d df f= ⇒ = ≈

Thus, 2 4.5df cm≈ .

43. The ratio is set up incorrectly. It should be A BD E= .

44. In order to find F, one needs C instead of B. In such case, A CD F= .

45. True. Two similar triangles must have equal corresponding angles, by definition. 46. True. If two triangles are congruent, then the ratio of corresponding sides equals 1, for all three pairs of sides, and all corresponding angles are equal. Hence, they must be similar. (Conversely, two similar triangles need not be congruent – see Problem 36.) 47. False. The third angle in such case would have measure ( )180 82 67 31− + = . 48. True. Consider two equilateral triangles, one whose sides have length x and a second whose sides have length y. Observe then that the ratio of corresponding sides is always x

y

(or yx , depending on which you refer to as Triangle 1 and Triangle 2). Since all angles of

an equilateral triangle have measure 60 , we conclude that they must be similar. However, if 1x = and 2y = , then the triangles are NOT congruent. 49. False. Such angles are congruent, and hence unless they are both 90 , they need not be supplementary. 50. True. 51. False. You need the corresponding angles to be the same to ensure similarity of two isosceles triangles. 52. False. All we can say here is that corresponding sides of similar triangles are in proportion.

Chapter 1

28

53. We label the given diagram as follows:

Observe that ACB DCE= (vertical angles). Since we are given that CBA = CDE , the third pair of angles must have the same measures. Hence, ABC is similar to CDE .

The corresponding sides between these two triangles are identified as follows: AB corresponds to ED, AC corresponds to EC, BC corresponds to DC Using the ratio for similar triangles, we have AB AC

ED EC= , so that 3 6

4x= , so that 2x = .


Observe that ABE and ACD are similar triangles (since we are given that

ABE ACD= and the triangles share A . Corresponding sides of these triangles are identified as follows: AB corresponds to AC, AE corresponds to AD, BE corresponds to CD

Using the ratio for similar triangles, we know that AB AE BEAC AD CD

= = , so that

4 5 14 18 20 4

yx y= = =

+ +(1) .

We now use different equalities from (1) to find x and y.

Find x: 4 1 so that 16 4 and hence, 12 .4 4

x xx= = + =

+

Find y: 1 so that 4 18 and hence, 6 .18 4

y y y yy

= = + =+

6

18

Section 1.2

29

55. Triangles 1 and 2 are similar because they both have a 90 angle, and the angles formed at the horizontal have the same measure since they are vertical angles. Similarly, Triangles 3 and 4 are similar. 56. a) Consider the following diagram:

The two right triangles above are similar since they both have a 90 , and the measures of the angles opposite the sides with lengths 0H and iH are equal because they are vertical

angles. Hence, using the similar triangles ratio yields 0 0

i

H D fH f

−= .

b) Consider the following diagram:

Again, from similarity if follows that 0

i i

H fH D f

=−

.

0H

0D f−f

iH

0HiD f−

fiH

Chapter 1

30

57. Claim (Lens Law): 0

1 1 1

iD D f+ = .

Proof. Observe that from Problem 56, we obtain the following identities:

0 001

i i

H Hf f f DH H

⎡ ⎤ ⎛ ⎞+ = + =⎜ ⎟⎢ ⎥

⎣ ⎦ ⎝ ⎠ (from (a))

0 0

1i ii

H Hf f f DH H⎡ ⎤ ⎛ ⎞

+ = + =⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

(from (b))

Hence, observe that

0

0 0

0

0

0

0

0

0

2 0 0

0 0

1

0

0

1 1

1 1

1 1

2

1

21

i

i i

i

i

i

i

i

i

i i

i i

i

i

D DD D D D

H Hf fH H

H Hf fH H

H HfH H

H H H HfH H H H

H HH H

f

=

++ =

⎡ ⎤ ⎡ ⎤+ + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦=⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪ ⎪ ⎪+ ⋅ +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭ ⎩ ⎭

⎡ ⎤+ +⎢ ⎥

⎣ ⎦=⎡ ⎤⎢ ⎥

⋅ + + +⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤

+ +⎢ ⎥⎣ ⎦

=0

0

2 i

i

H HH H

⎡ ⎤+ +⎢ ⎥

⎣ ⎦1 ,f

=

as desired. ☻

Section 1.2

31

58. We begin with several observations: 1.) ( ) 90m ABE∠ = since BF is parallel to CG, and ABE GCD∠ ≈ ∠ since they are corresponding angles. 2.) ( ) ( ) ( ) 60m AEB m CGA m FEG∠ = ∠ = ∠ =

3.) ( ) ( ) 45m CGD m CDG∠ = ∠ = since CDGΔ is an isosceles right triangle. 4.) BC and FG have the same length. Hence, ABE ACG GFEΔ Δ Δ∼ ∼ . 59. We begin with several observations: 1.) ( ) 90m ABE∠ = since BF is parallel to CG, and ABE GCD∠ ≈ ∠ since they are corresponding angles. 2.) ( ) ( ) ( ) 60m AEB m CGA m FEG∠ = ∠ = ∠ =

3.) ( ) ( ) 45m CGD m CDG∠ = ∠ = since CDGΔ is an isosceles right triangle. 4.) BC and FG have the same length. Hence, ABE ACG GFEΔ Δ Δ∼ ∼ . Now, using this information, consider the following triangles:

Since ACGΔ is a 30 60 90− − triangle, we know that ( ) 3 ( ),m AC m CG= ⋅ and so

5 33( )m CG = . Hence, 10 3

3( ) 2 ( )m AG m CG= ⋅ = . Next, since ACG GFEΔ Δ∼ , we know that

10 3 5 33 35 52 3 and 3

3 3y x

y x= ⇒ = = ⇒ = .

So, ( ) 3 , ( ) 2 3m EF cm m EG cm= = . 60. Since 5 3

3( )m CG cm= and DCGΔ is 45 45 90− − , we know that

( )5 3 5 63 3( ) 2m DG cm cm= = .

Chapter 1

32

Section 1.3 Solutions ----------------------------------------------------------------------------------

1. 8 4sin10 5

θ = = 2. 6 3cos10 5

θ = =

3. 1 1 5csc 4sin 45

θθ

= = = 4. 1 1 5sec 3cos 35

θθ

= = =

5. 8 4tan6 3

θ = = 6. 1 1 3cot 4tan 43

θθ

= = =

7. Note that by the Pythagorean Theorem, the hypotenuse has length 5 . So,

1 5cos55

θ = = .


2 2 5sin55

θ = = .


1 1sec 51cos5

θθ

= = = .


1 1 5csc 2sin 25

θθ

= = = .

11. 2tan 21

θ = = 12. 1 1cottan 2

θθ

= =

13. 5 5 34sin3434

θ = =

14. Using the Pythagorean Theorem, we know that the leg adjacent to angle θ has

length 3. So, 3 3 34cos3434

θ = = .


length 3. So, 5tan3

θ = .

16. 1 34cscsin 5

θθ

= =


length 3. So, 1 34seccos 3

θθ

= = .


length 3. So, 1 3cottan 5

θθ

= = .

19. ( ) ( ) ( )sin 60 cos 90 60 cos 30= − = 20. ( ) ( ) ( )sin 45 cos 90 45 cos 45= − =

Section 1.3

33

21. ( ) ( )cos sin 90x x= − 22. ( ) ( )cot tan 90A A= −

23. ( ) ( ) ( )csc 30 sec 90 30 sec 60= − = 24. ( ) ( )sec csc 90B B= −

25. ( ) ( )( )

( )sin cos 90

cos 90

x y x y

x y

+ = − +

= − −

26. ( ) ( )( )

( )sin 60 cos 90 60

cos 30

x x

x

− = − −

= +

27. ( ) ( )( )

( )cos 20 sin 90 20

sin 70

A A

A

+ = − +

= −

28. ( ) ( )( )

( )cos sin 90

sin 90

A B A B

A B

+ = − +

= − −

29. ( ) ( )( )

( )cot 45 tan 90 45

tan 45

x x

x

− = − −

= +

30. ( ) ( )( )

( )sec 30 csc 90 30

csc 60

θ θ

θ

− = − −

= +

31. From the given information, we obtain the following triangle:

Hence, a round trip on the ATV corresponds to twice the length of the hypotenuse, which is

( )2 5 miles 10 miles× = .

32. Consider the following triangle:

Observe that opptan 1

adjyx

θ = = = , so y = x.

Since we are given that ( )2 200x y+ = (since a round trip is 200 yards), we see that

100x y+ = and hence, x = y = 50. As such, by the Pythagorean Theorem, the hypotenuse has length 50 2 . So, the round trip of the ATV is 100 2 yards 141 yards≈ .

θ

θ

Chapter 1

34

33. Consider the triangle:

Applying the Pythagorean theorem yields 2 2 25 12 13z z+ = ⇒ = .

Thus, 5

13sinθ = .

34. The scenario can be described by the following two triangles:

Applying the Pythagorean theorem yields 13 and 193z w= = . Hence, we have:

Bob: 1213cosθ = Neighbor: 12

193cosθ =

The value of cosθ is larger for Bob’s roof. The steeper the roof for the same horizontal distance, the longer the hypotenuse and hence denominator used in computing cosθ .

θ

θ θ

Bob Neighbor

Section 1.3

35

35. If tan 3θ = , we have the following diagram:

Since the hypotenuse has length 2 by the Pythagorean theorem, this is a 30 60 90− − triangle. Hence, since θ is opposite the longer leg, it must be 60 .

36. Since 60θ = we have the following triangle:

Observe that 3 ft. 3

cos30cos30 3.46 ft.W W= ⇒ = ≈

37. Consider the following triangle: Observe that 35sin B = .

θ

trig 3e ism chapter 1 final · the key to solving this problem is setting up the correct...

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