TRIG 3e ISM Chapter 1 FINAL - test bank U › sample › Solution-Manual-for... · 2017-12-04 ·...

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 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= = = = .


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


= , 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

Chapter 1

36

38. We have the following triangle:

Observe that 3

5cos A = , and the Pythagorean theorem implies that the height is 4, as shown.

Now we have a similar triangle:

As such,

34 8 in. 6 in.x x= ⇒ =

39. Since 11.75sec 1.75 cosθ θ= ⇒ = , we

have the following triangle:

Note that 741.75 = . Applying the

Pythagorean theorem yields ( )22 337

4 41 z z+ = ⇒ = . Hence,

334

74

33sin7

θ = = .

40. Note that tan 1θ = when 45θ = . So, if tan 1.2,θ = then θ must be greater than 45 . Hence, the hill is too steep.

41. 6 1sin 3012 2

mmmm

θ θ= = ⇒ = 42. 2 2sin 452 2

mmmm

θ θ= = ⇒ =

43. a) 10,000 2tan25,000 5

θ = = b) 1 5cottan 2

θθ

= =

44. a) By the Pythagorean theorem, this equals 2 2s p+ . b) 2 2

cos ss p

θ =+

45. Opposite side has length 3, not 4. 46. opptan

adjx = , not adj

opp.

θ

Section 1.3

37

47. 1seccos

xx

= , not 1sin x

. 48. 1cscsin

yy

= , not 1cos y

.

49. True. It follows from the co-function identity that ( ) ( ) ( )sin 45 cos 90 45 cos 45= − = .

50. True. It follows from the co-function identity that ( ) ( ) ( )sin 60 cos 90 60 cos 30= − = .

51. True. ( )sec60 csc 90 60= − .

52. True. ( ) ( )cot 45 tan 90 45= − 53.

opp 1sin 30hyp 2 2adj 3 3cos30hyp 2 2

xx

xx

= = =

= = =

54. Using Exercise 53, we have 3sin 60 cos30

21cos 60 sin 302

= =

= =

55. Using Exercises 53 and 54, we have:

123

2

sin 30 1 3tan 30cos30 33

= = = = and 3

212

sin 60tan 60 3cos 60

= = = .

(Alternatively, in order to avoid having to use Exercise 54 in the computation of tan 60 , you can instead use the co-function identity to obtain

13

1 1tan 60 cot 30 3tan 30

= = = = .)

56. First, note that ( ) ( ) 1 2sin 45 cos 4522 2

xx

= = = = . Using these values, we see

that ( )( )

222

2

sin 45tan 45 1

cos 45= = = .

57. First, note that 1

2 2

1 1 1sec 45 2cos 45 x

x

= = = = . Using the co-function identity, we

see that csc 45 sec 45 2= = . 58. Using Exercise 55, we see that tan 60 3= and cot 30 tan 60 3= = . 59. 1 sin 1 and 1 cos 1θ θ− ≤ ≤ − ≤ ≤ 60. Any real number 61. sec 1 and csc 1θ θ≥ ≥

62. sinθ increases from 0 to 1 as θ decreases from 0 to 90 .

Chapter 1

38

63. The cofunction of sinθ is cosθ . As θ increases from 0 to 90 , cosθ decreases from 1 to 0. 64. As θ increases from 0 to 90 , cscθ decreases to 1.

65. a. cos 70 0.342≈ and so, 1 1 2.92398cos 70 0.342

≈ ≈

b. The keystroke sequence 70, cos, 1/x yields 2.92380.

66. a. sin 40 0.643≈ and so, 1 1 1.55521sin 40 0.643

≈ ≈

b. The keystroke sequence 40, sin, 1/x yields 1.55572.

67. a. tan 54.9 1.423≈ and so, 1 1 0.70274tan 54.9 1.423

≈ ≈

b. The keystroke sequence 54.9, tan, 1/x yields 0.70281.

68. a. cos18.6 0.948≈ and so, 1 1 1.05485cos18.6 0.948

≈ ≈

b. The keystroke sequence 18.6, cos, 1/x yields 1.05511. Section 1.4 Solutions ---------------------------------------------------------------------------------- 1. a 2. b 3. b 4. a 5. c 6. c

7. 123

2

sin 30 1 3tan 30cos30 33

= = = = 8. 2

22

2


= = =

9. 3

212


= = = 10. 12

1 1csc30 2sin 30

= = =

11. 3

2

1 1 2 2 3sec30cos30 33

= = = = 12. Using Exercise 7, we see that

33

1 1cot 30 3tan 30

= = = .

13. 3

2

1 1 2 2 3csc60sin 60 33

= = = = 14. 12

1 1sec60 2cos60

= = =

15. Using Exercise 9, we see that 1 1 3cot 60

tan 60 33= = = .

16. 2

2

1 1csc 45 2sin 45

= = =

17. 12

1 1sec 45 2cos 45

= = = 18. Using Exercise 8, we see that

1 1cot 45 1tan 45 1

= = = .

19. 0.6018 20. 0.3057 21. 0.1392 22. 0.9278 23. 1.3764 24. 0.9391 25. 1.0098 26. 3.8637

Section 1.4

39

27. 1.0002 28. 1.2868 29. 0.7002 30. 1.8040 31.

5 17 2963 28 35

68 45 64

′ ′′′ ′′+

′ ′′

Remember that whenever the number of minutes or number of seconds is at least 60, you can further simplify the expression. In this case, since 64 1 4′′ ′ ′′= , the completely simplified answer is 68 46 4′ ′′ .

32. 5 17 29

16 11 30

21 28 59

′ ′′′ ′′+

′ ′′

This answer is completely simplified since neither the number of minutes nor the number of seconds is greater than or equal to 60.

33. 63 28 3516 11 30

79 39 65

′ ′′′ ′′+

′ ′′

Remember that whenever the number of minutes or number of seconds is at least 60, you can further simplify the expression. In this case, since 65 1 5′′ ′ ′′= , the completely simplified answer is 79 40 5′ ′′ .

34. 63 28 355 17 29

58 11 6

′ ′′′ ′′−

′ ′′


35. 63 28 3516 11 30

47 17 5

′ ′′′ ′′−

′ ′′


36. In order to compute 16 11 305 17 29

′ ′′′ ′′−

we borrow 1 from16 and add it (in the form of 60′ ) to11′ in order to rewrite the problem in the equivalent form:

15 71 305 17 29

10 54 1

′ ′′′ ′′−

′ ′′


Chapter 1

40


′ ′′′ ′′−

we borrow 1 from90 and add it (in the form of 60′ ) to 0′ in order to rewrite the problem in the equivalent form:

89 60 05 17 29

′ ′′′ ′′−

Next, we borrow 1′ from 60′ and add it (in the form of 60′′ ) to 0′′ to rewrite this problem in the equivalent form:

89 59 605 17 29

84 42 31

′ ′′′ ′′−

′ ′′



′ ′′′ ′′−

we borrow 1 from90 and add it (in the form of 60′ ) to 0′ in order to rewrite the problem in the equivalent form:

89 60 063 28 35

′ ′′′ ′′−

Next, we borrow 1′ from 60′ and add it (in the form of 60′′ ) to 0′′ to rewrite this problem in the equivalent form:

89 59 6063 28 35

26 31 25

′ ′′′ ′′−

′ ′′


39. First, note that 2020 0.3360

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

.

So, 33 20 33.33′ ≈ .


⎛ ⎞′ = =⎜ ⎟⎝ ⎠

.

So, 89 45 89.75′ = .


⎛ ⎞′ = =⎜ ⎟⎝ ⎠

.

So, 59 27 59.45′ = .


⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

. So,

72 13 72.22′ ≈ . 43. First, note that

1515 0.0043600

⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

4545 0.7560

⎛ ⎞′ = =⎜ ⎟⎝ ⎠

.

So, 27 45 15 27.754′ ′′ = .


3600⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

55 0.083360

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

.

So, 36 5 30 36.092′ ′′ = .

Section 1.4

41


3600⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

2828 0.466760

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

.

So, 42 28 12 42.470′ ′′ = .


3600⎛ ⎞′′ = =⎜ ⎟⎝ ⎠

1010 0.166760

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

.

So, 63 10 9 63.169′ ′′ = . 47. Observe that

( )15.75 15 0.75

6015 0.751

15 45

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

So, 15.75 15 45′= .

48. Observe that

( )15.50 15 0.50

6015 0.501

15 30

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

So, 15.50 15 30′= . 49. Observe that

( )22.35 22 0.35

6022 0.351

22 21

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

So, 22.35 22 21′= .

50. Observe that

( )

( )

80.47 80 0.476080 0.471

80 28.280 28 0.2

6080 28 0.21

80 28 12

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

′ ′= + +′′⎛ ⎞′ ′= + + ⎜ ⎟′⎝ ⎠

′ ′′= + +

So, 80.47 80 28 12′ ′′= . 51. Observe that

( )

( )

30.175 30 0.1756030 0.1751

30 10.530 10 0.5

6030 10 0.51

30 10 30

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

′ ′= + +′′⎛ ⎞′ ′= + + ⎜ ⎟′⎝ ⎠

′ ′′= + +

So, 30.175 30 10 30′ ′′= .

52. Observe that

( )

( )

25.258 25 0.2586025 0.2581

25 15.4825 15 0.48

6025 15 0.481

25 15 28.8

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

′ ′= + +′′⎛ ⎞′ ′= + + ⎜ ⎟′⎝ ⎠

′ ′′= + +

So, 25.258 25 15 29′ ′′≈ .

Chapter 1

42

53. Observe that

( )

( )

77.535 77 0.5356077 0.5351

77 32.177 32 0.1

6077 32 0.11

77 32 6

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

′ ′= + +′′⎛ ⎞′ ′= + + ⎜ ⎟′⎝ ⎠

′ ′′= + +

So, 77.535 77 32 6′ ′′= .

54. Observe that

( )

( )

5.995 5 0.995605 0.9951

5 59.75 59 0.7

605 59 0.71

5 59 42

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

′ ′= + +′′⎛ ⎞′ ′= + + ⎜ ⎟′⎝ ⎠

′ ′′= + +

So, 5.995 5 59 42′ ′′= .

55. Begin by converting10 25′ to DD:

Since 2525 0.416760

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

, we see

that 10 25 10.4167′ ≈ . Then, observe that ( )sin 10.4167 0.1808≈ .

56. Begin by converting 75 13′ to DD:

Since 1313 0.21760

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

, we see that

75 13 75.217′ ≈ . Then, observe that ( )cos 75.217 0.2552≈ .


Since 1515 0.2560

⎛ ⎞′ = =⎜ ⎟⎝ ⎠

, we see that

22 15 22.25′ = . Then, observe that ( )tan 22.25 0.4091≈ .


Since 2222 0.366760

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

, we see that

68 22 68.3667′ ≈ . Then, observe that

( ) ( )1sec 68.3667 2.7125

cos 68.3667= ≈ .

59. Begin by converting 28 25 35′ ′′ to DD: Since

2525 0.416760

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

3535 0.00973600

⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

,

we see that 28 25 35 28 0.4167 0.0097 28.426′ ′′ ≈ + + ≈ . Then, observe that

( ) ( )1csc 28.426 2.1007

sin 28.426= ≈ .

60. Begin by converting50 20 19′ ′′ to DD: Since

2020 0.333360

⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

1919 0.00533600

⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

,

we see that 50 20 19 50 0.3333 0.0053 50.339′ ′′ ≈ + + ≈ . Then, observe that

( ) ( )1sec 50.339 1.5668

cos 50.339= ≈ .

Section 1.4

43

61. Using 2 sinn dλ θ= , observe that ( )1(1.54) 2 sin 45

21.54 22

1.54 1.09 angstroms2

d

d

d

=

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

= ≈

62. Using 2 sinn dλ θ= , observe that ( )

( )

( )

4(1.67) 2 sin 71.3

6.68 2sin 71.3

6.68 3.53 angstroms2sin 71.3

d

d

d

=

=

= ≈

63. Using ( ) ( )sin sini i r rn nθ θ= , observe that

( ) ( )( )

( )

1.00sin 30 sin 12

1.00sin 302.405 .

sin 12

r

r

n

n

=

= ≈


( ) ( )( )

( )

1.00sin 30 sin 18.5

1.00sin 301.576 .

sin 18.5

r

r

n

n

=

= ≈


( ) ( )( )

( )

1.00sin 30 sin 22

1.00sin 301.335 .

sin 22

r

r

n

n

=

= ≈


( ) ( )( )

( )

1.00sin 30 sin 20

1.00sin 301.462 .

sin 20

r

r

n

n

=

= ≈

67. 90 85 28 4 32′ ′− = 68. 36 19 36 19 72 38′ ′ ′+ = 69. ( )3 23 3

2 215sin 60 4 3 15 4 3 ft.d = + = + =

70. 15csc65

4 17.6ft.d = + ≈

71. ( )22100sin 45 100 50 2 ft.= =

72. ( )32100sin 60 100 50 3 ft.= =

73. ( ) ( )1 160 360040 18 27 40 18 27 40.3075′ ′′′ ′′ ′ ′′= + + ≈

74. First, convert 39 28 37′ ′′ to DD:

( ) ( )1 160 360039 28 37 39 28 37 39.476944′ ′′′ ′′ ′ ′′= + + ≈ .

So, ( )15 tan 39.476944 12.4 ft.h = ≈

75. ( ) 1cos 60 ,2

= not 32

76. 2525 0.4167 0.4260

⎛ ⎞′ = ≈ ≈⎜ ⎟⎝ ⎠

, not

0.25 . So, 36 25 36.42′ ≈ .

Chapter 1

44

77. False.

( )( )

130 1

30

1sec , not cos30cos

=

78. False. Note that sin 50 0.7660≈ , which is not equal to 0.77.

79. True. 80. False. Since ( )tan cot 90 ,θ θ= − we have

( ) ( ) ( )tan 20 50 cot 90 20 50 cot 69 10′ ′ ′= − = 81. Consider the following triangle:

( )2 2

sin 0 0ba b

≈ ≈+

since a b and

so, the denominator is much larger than the numerator.


( )2 2

cos 0 1aa b

≈ ≈+

since a b and so,

2 2a a b≈ + .

83. Using the co-function identity with Exercise 81, we see that

( ) ( )cos 90 sin 0 0= = .

84. Using the co-function identity with Exercise 82, we see that

( ) ( )sin 90 cos 0 1= = . 85. Since

12

322 2

1 3 3sec 45 2 , tan 30 , and cos303 2

= = = = = ,

we have

( )33

32

2 3 3 2 32sec 45 tan 30 3 2 3 2 2 6 2cos30 3 9 33

+++ + += = ⋅ = = .

2 2a b+

b

aθ

2 2a b+

bθ

a

Section 1.4

45

86. Since 3

23 1

2 2

1 2 3 2csc60 , cot 30 3 , and sin 453 2

= = = = = ,

we have

( )( )( )

( )( )( )2

62 22 2 122 3 9

3

3sin 45 cot 30 3 68csc60

= = =

87.

( )322

2

1 6 2 1cos 75 csc 45 cos30 sin15 cos30sin 45 4

6 2 6 6 24 2 4

− ⎛ ⎞⎛ ⎞− = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− − −= − =

88. ( ) ( )

( )( ) 6 32 12 2 2 2

cot 75 cos 45 sin 30 tan15 cos 45 sin 30

2 3 2 1

+ = +

= − + = − + −

89. We begin by converting all to DD: ( ) ( )( ) ( )( ) ( )

1 160 3600

1 160 3600

1 160 3600

25 10 15 25 10 15 25.17083

46 14 26 46 14 26 46.24056

23 17 23 23 17 23 23.28972

′ ′′

′ ′′

′ ′′

′ ′′ ′ ′′= + + ≈

′ ′′ ′ ′′= + + ≈

′ ′′ ′ ′′= + + ≈

Then, we have ( ) ( )

( )tan 25 10 15 sec 46 14 26

0.7575csc 23 17 23

′ ′′ ′ ′′+≈

′ ′′.

90. We begin by converting all to DD: ( ) ( )( ) ( )( ) ( )

1 160 3600

1 160 3600

1 160 3600

33 38 1 33 38 1 33.6336

36 58 6 36 58 6 36.9683

50 33 2 50 33 2 50.5506

′ ′′

′ ′′

′ ′′

′ ′′ ′ ′′= + + ≈

′ ′′ ′ ′′= + + ≈

′ ′′ ′ ′′= + + ≈

Then, we have ( )( ) ( )

cos 33 38 1cot 36 58 6 0.6856

csc 50 33 2

′ ′′′ ′′− ≈ −

′ ′′.

Chapter 1

46

91. a. Step 1: ( )cos 70 0.342≈

Step 2: ( )1 2.92398

cos 70≈

b. ( )( ) 1cos 70 2.92380

−≈

The calculation in part b is more accurate since one does not round twice as in part a.

92. a. Step 1: ( )sin 40 0.643≈

Step 2: ( )1 1.55521

sin 40≈

b. ( )( ) 1sin 40 1.55572

−≈

The calculation in part b is more accurate since one does not round twice as in part a.

93. This calculation sequence results in 3.240.

94. This calculation sequence results in 27 40 59′ ′′ .

Section 1.5 Solutions --------------------------------------------------------------------------------- 1. three 2. four 3. two 4. one 5. 47 6. 10 7. 55 8. 28 9. 83 10. 74 11. Consider this triangle:

Since ( ) adjcos 35

hyp 17 in.a

= = , we have

( ) ( )17 in. cos 35 13.93 in. 14 in.a = ≈ ≈

12. Consider this triangle:

Since ( ) oppsin 35

hyp 17 in.b

= = , we have

( ) ( )17 in. sin 35 9.751 in. 10 in.b = ≈ ≈

35 35

Section 1.5

47


Since ( ) oppsin 55

hyp 22 ft.a

= = , we have

( ) ( )22 ft. sin 55 18.02 ft. 18 ft.a = ≈ ≈


Since ( ) adjcos 55

hyp 22 ft.b

= = , we have

( ) ( )22 ft. cos 55 12.619 ft. 13 ft.b = ≈ ≈


Since ( ) opptan 20.5

adj 14.7 mi.a

= = , we

have ( ) ( )14.7 mi. tan 20.55.496 mi.5.50 mi.

a =

≈

≈



adj 0.752 mi.b

= = , we

have ( ) ( )0.752 mi. tan 69.31.990 mi.1.99 mi.

b =≈

≈

(since the number of significant digits is 3).

55 55

20.5

69.3

Chapter 1

48


Since ( ) adj 11 kmcos 25

hyp c= = , we have

( )11 km 12.137 km 12 km

cos 25c = ≈ ≈



Since ( ) opp 26 kmsin 75

hyp c= = , we have

( )26 km 26.917 km 27 km

sin 75c = ≈ ≈

(since the number of significant digits is 2). 19. Consider this triangle:

Since ( ) opp 15.37 cmsin 48.25

hyp c= = , we

have

( )15.37 cm 20.602 cm 20.60 cm

sin 48.25c = ≈ ≈



Since ( ) adj 16.79 cmcos 29.80

hyp c= = , we

have

( )16.79 cm 19.3485 cm 19.35 cm

cos 29.80c = ≈ ≈


25 75

48.25 29.80

Section 1.5

49


Since ( ) opp 29 mmsin

hyp 38 mmα = = , we have

1 29sin 49.74 5038

α − ⎛ ⎞= ≈ ≈⎜ ⎟⎝ ⎠



Since ( ) adj 89 mmcos

hyp 99 mmβ = = , we have

1 89cos 25.974 2699

β − ⎛ ⎞= ≈ ≈⎜ ⎟⎝ ⎠

(since the number of significant digits is 2). 23. Consider this triangle:

Since ( ) adj 2.3 mcos

hyp 4.9 mα = = , we have

1 2.3cos 62.005 624.9

α − ⎛ ⎞= ≈ ≈⎜ ⎟⎝ ⎠



Since ( ) opp 7.8 msin

hyp 13 mβ = = , we have

1 7.8sin 36.870 3713

β − ⎛ ⎞= ≈ ≈⎜ ⎟⎝ ⎠


α

β

α

β

Chapter 1

50


First, observe that

121 17 21 17 21.283360

α⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠.

Since ( ) opptan 21.2833adj 210.8 yd.

a= = ,

we have ( ) ( )210.8 yd. tan 21.283382.117 yd.82.12 yd.

a =

≈

≈



First, observe that

127 21 27 21 27.3560

β⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠.


b= = , we

have ( ) ( )117.0 yd. tan 27.35

60.52 yd.

b =

≈


α

β

Section 1.5

51


First, observe that

115 20 15 20 15.333360

β⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠.

Since ( ) adj 10.2 kmcos 15.3333hyp c

= = , we

have

( )10.2 km 10.58 km 10.6 km

cos 15.3333c = ≈ ≈



First, observe that

165 30 65 30 65.560

β⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠.

Since ( ) opp 18.6 kmsin 65.5hyp c

= = , we

have

( )18.6 km 20.440 km 20.4 km

sin 65.5c = ≈ ≈


β β

Chapter 1

52


First, observe that

40 28 101 140 28 10 40.469460 3600

α ′ ′′=⎛ ⎞ ⎛ ⎞′ ′′= + + ≈⎜ ⎟ ⎜ ⎟′ ′′⎝ ⎠ ⎝ ⎠

Since ( ) opp 12,522 kmsin 40.4694hyp c

= = ,

we have

( )12,522 km 19, 293.05 km 19, 293 km

sin 40.4694c = ≈ ≈



First, observe that

28 32 501 128 32 50 28.547260 3600

α ′ ′′=⎛ ⎞ ⎛ ⎞′ ′′= + + ≈⎜ ⎟ ⎜ ⎟′ ′′⎝ ⎠ ⎝ ⎠

Since ( ) adj 17,986 kmcos 28.5472hyp c

= = ,

we have

( )17,986 km 20, 475.32 km 20, 475 km

cos 28.5472c = ≈ ≈



First, observe that

( )180 90 32 58β = − + = .

Since ( ) oppsin 32hyp 12 ft.

a= = , we have

( )( )12 ft. sin 32 6.4 ft.a = ≈

Similarly, since ( ) adjcos 32hyp 12 ft.

b= = , we

have ( )( )12 ft. cos32 10 ft.b = ≈


First, observe that

( )180 90 65 25β = − + = .


a= = , we have

( )( )37 ft. sin 65 34 ft.a = ≈


b= = , we

have ( )( )37 ft. cos 65 16 ft.b = ≈

α α

32

β

65

β

Section 1.5

53


First, observe that

( )180 44 90 46β = − + = .

Since ( ) adj 2.6 cmcos 44hyp c

= = , we have

( )2.6 cm 3.6 cm

cos 44c = ≈

Similarly, since ( ) opptan 44adj 2.6 cm

a= = ,

we have ( ) ( )2.6 cm tan 44 2.5 cma = ≈ .


First, observe that

( )180 12 90 78β = − + = .

Since ( ) adj 10 mcos 12hyp c

= = , we have

( )10 m 10.2 m

cos 12c = ≈

Similarly, since ( ) opptan 12adj 10 m

a= = , we

have ( ) ( )10 m tan 12 2.13 ma = ≈ .

44

β β

12

Chapter 1

54


First, observe that

( )180 60 90 30β = − + = .

Since ( ) adjcos 60hyp 5 in.

b= = , we have

( ) ( )5 in. cos 60 3 in.b = ≈

Since ( ) oppsin 60hyp 5 in.

a= = , we have

( )( )5 in. sin 60 4 in.a = ≈


First, observe that

( )180 9.67 90 80.33β = − + = .

Since ( ) adjcos 9.67hyp 5.38 in.

b= = , we have

( ) ( )5.38 in. cos 9.67 5.30 in.b = ≈

Since ( ) oppsin 9.67hyp 5.38 in.

a= = , we have

( ) ( )5.38 in. sin 9.67 0.904 in.a = ≈ 37. Consider this triangle:

First, observe that

( )180 90 72 18α = − + = .

Since ( ) oppsin 72hyp 9.7 mm

b= = , we have

( )( )9.7 mm sin 72 9.2 mmb = ≈ .

Similarly, since ( ) adjcos 72hyp 9.7 mm

a= = ,

we have ( )( )9.7 mm cos 72 3.0 mma = ≈ .


First, observe that

( )180 90 45 45α = − + = .


b= = , we have

( )( )7.8 mm sin 45 5.5 mmb = ≈ .

Similarly, since ( ) adjcos 45hyp 7.8 mm

a= = ,

we have ( )( )7.8 mm cos 45 5.5 mma = ≈ .

α

72

α

45

β

60

β

9.67

Section 1.5

55


First, observe that

( )180 90 54.2 35.8β = − + = .

Since ( ) opp 111 mi.sin 54.2hyp c

= = , we have

( )111 mi. 137 mi.

sin 54.2c = ≈

Similarly, since

( ) opp 111 mi.tan 54.2adj b

= = , we have

( )111 mi. 80.1 mi.

tan 54.2b = ≈


First, observe that

( )180 90 47.2 42.8α = − + = .

Since ( ) adj 9.75 mi.cos 47.2hyp c

= = , we

have

( )9.75 mi. 14.4 mi.

cos 47.2c = ≈

Similarly, since

( ) opptan 47.2adj 9.75 mi.

b= = , we have

( ) ( )9.75 mi. tan 47.2 10.5 mi.b = ≈

54.2

β

α

47.2

Chapter 1

56


First, observe that

( )180 45 90 45α = − + = . Since the triangle is isosceles, we know that 10.2 kma = and 10.2 2 14kmc = ≈ .


First, observe that

( )180 85.5 90 4.5α = − + = .

Since ( ) opp 14.3 ft.sin 85.5hyp c

= = , we have

( )14.3 ft. 14.344 ft.

sin 85.5c = ≈

Similarly, since ( ) opp 14.3 ft.tan 85.5adj a

= = ,

we have ( )

14.3 ft. 1.13 ft.tan 85.5

a = ≈ .

45

α α

85.5

Section 1.5

57


First, observe that

( )180 90 28 23 61 37

161 37 61.6260

β ′ ′= − + =

⎛ ⎞′= + ≈⎜ ⎟′⎝ ⎠

128 23 28 23 28.3860

α⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠

Since ( ) adj 1,734 ft.cos 28.38hyp c

= = , we

have

( )1,734 ft. 1,971 ft.

cos 28.38c = ≈

Similarly, since

( ) opptan 28.38adj 1,734 ft.

a= = , we have

( ) ( )1,734 ft. tan 28.38 936.9 ft.a = ≈


First, observe that

( )180 90 72 59 17 1

117 1 17.0260

β ′ ′= − + =

⎛ ⎞′= + ≈⎜ ⎟′⎝ ⎠

172 59 72 59 72.9860

α⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠

Since ( ) opp 2,175 ft.tan 72.98adj b

= = , we

have

( )2,175 ft. 665.7 ft.

tan 72.98b = ≈

Similarly, since

( ) opp 2,175 ft.sin 72.98hyp c

= = , we have

( )2,175 ft. 2,275 ft.

sin 72.98c = ≈

α

β

α

β

Chapter 1

58


First, since 42.5 ft.tan

28.7 ft.α = , we obtain

1 42.5tan 56.028.7

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

So, ( )180 90 56.0 34.0β ≈ − + ≈ . Next, using the Pythagorean Theorem, we obtain

( ) ( )2 22 228.7 ft. 42.5 ft. 2,629.94 ft.c = + = and so, 51.3 ft.c ≈


First, since 19.8 ft.sin


1 19.8sin 24.048.7

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Similarly, since 19.8 ft.cos48.7 ft.

β = , we obtain

1 19.8cos 66.048.7

β − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Next, using the Pythagorean Theorem, we obtain

( ) ( )2 22

2 2

19.8 ft. 48.7 ft.1,979.65 ft.44.5 ft.

bbb

+ =

=≈

β

α α

β

Section 1.5

59


First, since 35,236 kmsin

42,766 kmα = , we obtain

1 35,236sin 55.48042,766

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Similarly, since 35,236 kmcos42,766 km

β = , we

obtain 1 35,236cos 34.520

42,766β − ⎛ ⎞= ≈⎜ ⎟

⎝ ⎠.


( ) ( )2 22

2 2

35,236 km 42,766 km587,355,060 km24,235 km

bbb

+ =

=≈


First, since 0.1245 mmcos

0.8763 mmα = , we obtain

1 0.1245cos 81.830.8763

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Similarly, since 0.1245 mmsin0.8763 mm

β = , we

obtain 1 0.1245sin 8.17

0.8763β − ⎛ ⎞= ≈⎜ ⎟

⎝ ⎠.


( ) ( )2 22

2 2

0.1245 mm 0.8763 mm0.752401 mm0.8674 mm

aaa

+ =

=≈

β β

α α

Chapter 1

60

49. Consider this diagram:

Observe that 5 ft.tan1

x= , so that

5 ft. 286 ft.tan1

x = ≈ .


Observe that 5 ft.tan 3

x= , so that

5 ft. 95 ft.tan 3

x = ≈ .


Observe that sin 36

150 ft.a

= , so that

( )( )sin 36 150 ft. 88 ft.a = ≈ So, the altitude between the two planes should be approximately 88 ft.


Observe that 100 ft.sin 36

H= , so that

100 ft. 170 ft.sin 36

H = ≈

So, the hose should be approximately 170 ft. long.

1 3

36 36

Section 1.5

61


Observe that sin 3

5,000 ft.a

= , so that

( )( )sin 3 5,000 ft. 262 ft.a = ≈ So, rounding to two significant digits, we see that the altitude should be approximately 260 ft.


Let θ = glide slope angle. Observe that

450 ft.sin5,200 ft.

θ = , so that

1 450sin 55,200

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Since an angle of 3 will allow the pilot to see both red and white lights while higher angles enable him to see only white lights, we conclude that in this case the pilot will see only white lights.



3,000 ft.sin15,500 ft.

θ = , so that

1 3,000sin 11.015,500

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

This angle is way too low (since the typical range is 18 20− ).



2,500 ft.sin7,800 ft.

θ = , so that

1 2,500sin 18.07,800

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

So, she is within safety specifications to land.

36θ

θ θ

Chapter 1

62


Since tan15

150 ft.r

= , we have

( )( )150 ft. tan15 40 ft.r = ≈ So, the diameter of the circle is 80 ft.


Since tan15

500 ft.r

= , we have

( )( )4 500 ft. tan15 134 ft.= ≈ So, the diameter of the circle is 268 ft., which we round to 270 ft.


First, note that 1 0.0002778θ ′′= ≈ .

Since ( )tan 0.000277835,000 km

x= , we

have that ( ) ( )35,000 km tan 0.0002778

0.1697 km 170 m

x =

≈ ≈


First, note that 1 0.000138892

θ′′⎛ ⎞= ≈⎜ ⎟

⎝ ⎠.

Since ( )tan 0.0001388935,000 km

x= , we

have that ( ) ( )35,000 km tan 0.00013889

0.08484 km 85m

x =

≈ ≈

15 15 1515

θ θ

Section 1.5

63


First, note that 0.010 km = 10 m.

Since 0.010 kmtan35,000 km

θ = , we have that

1 0.010tan 0.00001635,000

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.


First, note that 0.030 km = 30 m.

Since 0.030 kmtan35,000 km

θ = , we have that

1 0.030tan 0.00004935,000

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.


Observe that 15 ft.tan 30

x= , so that

15 ft. 26 ft.tan 30

x = ≈


Observe that 50 ft.cos55

x= , so that

50 ft. 87 ft.cos55

x = ≈

θ θ

30

55

Chapter 1

64


Observe that 4,000 ft.sin 65

x= , so that

4,000 ft. 4, 414 ft.sin 65

x = ≈ .


Observe that ( ) 10 mi.tan 33

6 mi.θ+ = , so that

1

1

1033 tan6

10tan 33 266

θ

θ

−

−

⎛ ⎞+ = ⎜ ⎟⎝ ⎠⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

So, the bearing would be 26N E . 67. Consider this diagram:



1

1

835 tan38tan 35 343

θ

θ

−

−

⎛ ⎞+ = ⎜ ⎟⎝ ⎠⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

So, the bearing would be 34N W .




1

1

1748 tan6

17tan 48 236

θ

θ

−

−

⎛ ⎞+ = ⎜ ⎟⎝ ⎠⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

So, the bearing would be 23N E

65 33

θ

θ35 3 mi

8 mi17mi

6mi

θ

48

Section 1.5

65

69. We need to first consider a triangle that lies in the tennis court whose hypotenuse is the shadow of the path of the serve. This triangle has the following dimensions:

Here, S denotes the shadow of the serve in the court. Using the Pythagorean Theorem, we see that 2 2 2164 720 S+ = , so that 738.44 in. 61.54 ft.S ≈ ≈ Next, we consider the triangle that has as its hypotenuse the actual path of the serve, and whose base is the hypotenuse of the above triangle, and whose height is the height of the ball above ground the instant it is struck – as such, this triangle sticks up out of the court, perpendicular to it. This is illustrated below:

We seek the measure of angle θ . Observe that 9 ft.tan

61.54 ft.θ = , so that

1 9 ft.tan 8.3261.54 ft.

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

θ

Chapter 1

66

70. We need to first consider a triangle that lies in the tennis court whose hypotenuse is the shadow of the path of the serve. This triangle has the following dimensions:

Here, S denotes the shadow of the serve in the court. Using the Pythagorean Theorem, we see that 2 2 2164 720 S+ = , so that 738.44 in. 61.54 ft.S ≈ ≈ Next, we consider the triangle that has as its hypotenuse the actual path of the serve, and whose base is the hypotenuse of the above triangle, and whose height is the height of the ball above ground the instant it is struck – as such, this triangle sticks up out of the court, perpendicular to it. This is illustrated below:

We seek the height h at which the ball must be struck so that it hits the ground at an angle

of 8 . Observe that tan861.54 ft.

h= , so that ( )61.54 ft. tan8 8.65 ft.h = ≈

1 9 ft.tan 8.3261.54 ft.

θ − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

8

Section 1.5

67


First, observe that 2.219tan

2.354α = , so that

1 2.219tan 43.3092.354

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

. Since

180α θ+ = , we conclude that 136.69θ ≈ .


First, observe that 2.097tan

2.354α = , so that

1 2.097tan 41.6952.354

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

. Since

180α θ+ = , we conclude that 138.30θ ≈ .


Observe that

100sin 75 100sin 75 97 milesh h= ⇒ = ≈ .


Observe that

25150tan 9.462θ θ= ⇒ ≈ .

So, the heading is 80.5 east of north.

αθ

α

θ

75 θ

Chapter 1

68

75. The original canal has the following dimensions:

Observe that

2tan 305 10

w w= = .

Solving for w yields 1 10 310 tan 30 10 .

33w ft⎛ ⎞= = =⎜ ⎟

⎝ ⎠

The diagram for the present day is:

Using similar triangles and solving for d yields

54 10 3 3

20(3) 6 3.5 .10 3 3

d

d ft

=

= = ≈


Observe that

( ) ( ) ( )5

1 12 5 52 2 8 8

5tan tan 2 tan 1.117198 radians4 8

θ θ θ− −= = ⇒ = ⇒ = ≈ .

Converting to degrees (by multiplying by 180 π ) results in approximately 64 . So, yes, erosion has widened the shape of the canal since the angle has increased by 4 degrees.

30 30

10 3 3 .ft

2θ

2θ

Section 1.5

69

77. Using the right-triangle suggested by the diagram (with the femur being the hypotenuse 18 inches in length and h denoting the height of elevation), we see that

( )sin15 18in. sin15 4.7 in.18in.

h h= ⇒ = ≈

78. Using the right-triangle suggested by the diagram (with the femur being the hypotenuse 18 inches in length, θ being the angle of elevation, and the height of elevation 6.2 in.), we see that

16.2in. 6.2sin sin 2018in. 18

θ θ − ⎛ ⎞= ⇒ = ≈⎜ ⎟⎝ ⎠

79. Should be 1tan (80)β −= , not tan(80) . 80. The calculation of b is correct. However, this value should not be then used to determine the value of a since doing so would introduce TWO computational errors (due

to round off). One should, instead, compute a using cos5615a

= (as in Example 1).

81. True 82. True 83. False. Knowing the measures of all angles in a right triangle would not enable you to find any of the side lengths. In fact, you could produce infinitely many similar triangles with these same angles.

84. False. The angle opposite the hypotenuse in a right triangle is 90 . If we only knew this and no other side length, we could not determine the measures of either of the two remaining angles.

85. False. There are only four significant digits here.

86. True

87. True 88. False. Use the Pythagorean theorem.

Chapter 1

70


Ultimately, we wish to find x. We set up a system of two equations in x and y which we solve simultaneously:

tan 38 ( ) tan 51 ( )1,900x x

y y= =

+1 2

Equations (1) and (2) can be written equivalently as:

( ) ( )( )tan 38 1,900 tan 38 ( )

tan 51 0 ( )

x y

x y

− =

− =

3

4

Subtracting (3) – (4) yields: ( ) ( ) ( )tan 38 tan 51 1,900 tan 38y y− + =

so that ( )

( ) ( )1,900 tan 38

3, 272.50 ft.tan 51 tan 38

y = ≈−

Substitute this into (4) to find that x is approximately 4,000 ft.


Let H = height of the shuttle (in miles) after 10 seconds of flight.

Observe that tan158 mi.

H= , so that

( )( )8 mi. tan15 2.14 mi.H = ≈ Rounding to one significant digit, we see that the shuttle is approximately 2 miles above ground after 10 seconds of flight.


Observe that

12

12

cos 45 12cos 45

cos 65 12cos 65

y

x

y

x

= ⇒ =

= ⇒ =

So, the width of the sidewalk is 12 cos 45 cos 65 3.4 ft.y x ⎡ ⎤− = − ≈⎣ ⎦

3851

15 406545

65

Section 1.5

71


Note all angles shown – those not labeled are right angles. The angles were found using complementary and supplementary angles. Observe that

10cos 48 10cos 48 6.691yz yz= ⇒ = ≈ . Also, we have

10cos 42 10cos 42 7.4314xy xy= ⇒ = ≈ . Since we are given that xz = 10, we conclude that the perimeter of the triangle A is approximately 24 units.


Observe that

5.33tan 35 5.33tan 35y y= ⇒ = . Hence,

5.33tan355.33 5.33tan 47 xx y ++= = ,

so that 5.33 tan 47 tan 35 1.98x ⎡ ⎤= − ≈⎣ ⎦ .

94. Consider the following diagram: Observe that ( )1250 250

8 8tan tan 88θ θ −= ⇒ = ≈

95. ( )1sin sin 40 40− = 96. ( )1cos cos17 17− =

12

35

θ

θ

Chapter 1

72

97. ( )1cos cos 0.8 0.8− = 98. ( )1sin sin 0.3 0.3− =

99. ( )1sin sinθ θ− = , for any 0 90θ≤ ≤ 100. ( )1cos cosθ θ− = , for any 0 90θ≤ ≤ Chapter 1 Review Solutions ----------------------------------------------------------------------- 1. a) complement: 90 28 62− =

b) supplement: 180 28 152− =

2. a) complement: 90 17 73− =


b) supplement: 180 35 145− =

4. a) complement: 90 78 12− =

b) supplement: 180 78 102− = 5. Since 180α β γ+ + = , we know that

155

120 35 180 and so, 25γ γ=

+ + = = .


130

105 25 180 and so, 50γ γ=

+ + = = .


9

7 180 and so, 20β

β β β β=

+ + = = .

Thus, 7 140 and 20α β γ β= = = = .


( ) ( )8

6 180 and so, 22.5β

β β β β=

+ + = = .

Thus, 6 135 and 22.5α β γ β= = = = . 9. 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 12 ,b+ = which simplifies

to 2 216 144 and then to, 128b b+ = = , so we conclude that 128 8 2b = = . 10. 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 29 15 ,a + = which simplifies to 2 281 225 and then to, 144a a+ = = , so we conclude that 12a = . 11. 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 4 ,c+ = which simplifies

to 2 65c = , so we conclude that 65c = . 12. 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 2 210 8 ,c+ = which simplifies

to 2 164c = , so we conclude that 164 2 41c = = .

Chapter 1 Review

73

13. If the length of a leg in a 45 45 90− − triangle is 12 yards, then the hypotenuse of

this triangle has length 12 2 yd.

14. Let x be the length of a leg in a given 45 45 90− − triangle. If the hypotenuse of

this triangle has length 8 ft., then 82 8, so that 4 22

x x= = = = . Hence, the

length of each of the two legs is 2 ft. 15. 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 3 ft.x = . Thus, the other leg has length 3 3 ft., and the hypotenuse has length 6 ft. 16. The length of the hypotenuse is 2 12x = km. So, 6x = . Thus, the length of the shorter leg is 6 km, and the length of the longer leg is 6 3 km. 17. Consider the following diagram:

Let α =measure of the desired angle between the hour hand and minute hand. Since the measure of the angle formed using two rays emanating from the center of the


= , it immediately follows

that 5 30 150α = ⋅ = .

Chapter 1

74





that 3 30 90α = ⋅ = . 19. 75G = (vertical angles) 20. 105D = (interior angles) 21. 75C = (alternate interior angles) 22. 105E = (supplementary angles) 23. 75B = (corresponding angles) 24. 105A = (since 75B = , and A and B

are supplementary )

25. By similarity, A BD E= , so that 10 8

5 E= . Solving for E yields 4 .E =

26. By similarity, A BD E= , so that 15 12

4D= . Solving for D yields 5 .D =

27. First, convert all quantities involved to the common unit km. Observe that by

similarity, A CD F= , so that 81 km

0.0045 km 0.0082 kmC

= . Solving for C yields

( ) 81 km0.0082 km 147.6 km0.0045 km

C ⎛ ⎞= =⎜ ⎟⎝ ⎠

28. First, convert all quantities involved to the common unit m. Observe that by

similarity, B CE F= , so that 800 cm

8cm 14 cmB

= . Solving for B yields

( ) 8 cm800 cm 457 cm= 4.57 m14 cm

B ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Chapter 1 Review

75

29. Consider the following two diagrams:

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

( )4 m 4 m so that 9.6 m 32 m9.6 m 1.2 m 1.2 m

T T ⎛ ⎞= = =⎜ ⎟⎝ ⎠

.

So, the tree is 32 m tall. 30. Note that 1 ft. 9 in. 1.75 ft.= . Now, consider the following two diagrams.

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

( )4 ft. 4 ft. so that 1.75 ft. 7 ft.1.75 ft. 1 ft. 1 ft.

M M ⎛ ⎞= = =⎜ ⎟⎝ ⎠

So, the man is 7 feet tall.

Chapter 1

76

31. Let x = width of the built-in refrigerator (in inches). Using similarity, we obtain 4 in.1 in. 4 3 ft.3 so that in. 4 ft. 48 in.

3 ft. 3 1 in.x

x⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

So, the built-in refrigerator is 48 inches wide. 32. Let x = width of the pantry (in inches). Using similarity, we obtain

( )5 in.1 in. 5 3 ft. 15 154 so that in. ft. 12 in. 45 in.

3 ft. 4 1 in. 4 4x

x⎛ ⎞⎛ ⎞ ⎛ ⎞= = = = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

So, the pantry is 45in. long. 33. Consider the following diagram:

From the Pythagorean Theorem, we see that 2 23 2 13h = + = . Thus,

2 2 13cos1313

θ = = .

34. Using the diagram and value of h in Exercise 33, we see that

3 3 13sin1313

θ = = .

35. Using the diagram and information from Exercise 33, we see that

1 13seccos 2

θθ

= = .

36. Using Exercise 34, we see that 1 13csc

sin 3θ

θ= = .

37. First, note that using the diagram and value of h in Exercise 33, we see that

3sin13

θ = .

As such, we have 3

sin 313tan 2cos 213

θθθ

= = = .

38. Using Exercise 37, we see that 1 2cot

tan 3θ

θ= = .

39. ( ) ( )cos 90 30 cos 60− = 40. ( ) ( )cos sin 90A A= −

θ

Chapter 1 Review

77

41.

( ) ( ) ( )tan 45 cot 90 45 cot 45= − = 42.

( ) ( ) ( )csc 60 sec 90 60 sec 30= − =

43. ( ) ( )( )

( )sin 30 cos 90 30

cos 60

x x

x

− = − −

= +

44. ( ) ( )( )

( )cos 55 sin 90 55

sin 35

A A

A

+ = − +

= −

45. ( ) ( )( )

( )csc 45 sec 90 45

sec 45

x x

x

− = − −

= +

46. ( ) ( )( )

( )sec 60 csc 90 60

csc 30

θ θ

θ

− = − −

= +

47. b 48. a 49. b 50. a 51. c 52. c

53. 123

2

sin 30 1 3tan 30cos30 33

= = = = 54. 2

22

2


= = =

55. 3

212


= = = 56. 12

1 1csc30 2sin 30

= = =

57. 2

2

1 1csc 45 2sin 45

= = = 58. 3

2

1 1 2 2 3csc60sin 60 33

= = = =

59. 3

2

1 1 2 2 3sec30cos30 33

= = = = 60. 12

1 1sec 45 2cos 45

= = =

61. 12

1 1sec60 2cos 60

= = = 62. 3

3

1 1cot 30 3tan 30

= = =

63. 1 1cot 45 1tan 45 1

= = = 64. 1 1 3cot 60tan 60 33

= = =

65. 0.6691 66. 0.5446 67. 0.9548 68. 0.4706 69. 1.5399 70. 1.0446 71. 1.5477 72. 2.7837


⎛ ⎞′ = ≈⎜ ⎟⎝ ⎠

.

So, 39 17 39.28′ ≈ .


⎛ ⎞′ = =⎜ ⎟⎝ ⎠

.

So, 68 15 68.25′ = .

Chapter 1

78


3600⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

3030 0.5060

⎛ ⎞′ = =⎜ ⎟⎝ ⎠

.

So, 29 30 25 29.507′ ′′ = .


3600⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

4545 0.7560

⎛ ⎞′ = =⎜ ⎟⎝ ⎠

.

So, 25 45 15 25.754′ ′′ = . 77. Observe that

( )42.25 42 0.25

6042 0.251

42 15

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

So, 42.25 42 15′= .

78. Observe that

( )60.45 60 0.45

6060 0.451

60 27

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

So, 60.45 60 27′= . 79. Observe that

( )

( )

30.175 30 0.1756030 0.1751

30 10.530 10 0.5

6030 10 0.51

30 10 30

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

′ ′= + +′′⎛ ⎞′ ′= + + ⎜ ⎟′⎝ ⎠

′ ′′= + +

So, 30.175 30 10 30′ ′′= .

80. Observe that

( )

( )

25.258 25 0.2586025 0.2581

25 15.4825 15 0.48

6025 15 0.481

25 15 28.8

= +′⎛ ⎞= + ⎜ ⎟

⎝ ⎠′= +

′ ′= + +′′⎛ ⎞′ ′= + + ⎜ ⎟′⎝ ⎠

′ ′′= + +

So, 25.258 25 15 29′ ′′≈ . 81. First, observe that

1537 15 37 37.2560

⎛ ⎞′ = + =⎜ ⎟⎝ ⎠

.

So, ( )sin 37.25 0.6053≈ .

82. First, observe that 3542 35 42 42.58360

⎛ ⎞′ = + =⎜ ⎟⎝ ⎠

.

So, ( )cos 42.583 0.7363≈ . 83. First, observe that

4861 48 61 61.8060

⎛ ⎞′ = + =⎜ ⎟⎝ ⎠

.

So, ( )cos 61.80 0.4726≈ .

84. First, observe that 1720 17 20 20.28360

⎛ ⎞′ = + =⎜ ⎟⎝ ⎠

.

So, ( )sin 20.283 0.3467≈ .

Chapter 1 Review

79


( ) ( )( )

( )

1.00sin 60 sin 35.26

1.00sin 601.50 .

sin 35.26

r

r

n

n

=

= ≈


( ) ( )( )

( )

1.00sin 60 sin 36.09

1.00sin 601.47 .

sin 36.09

r

r

n

n

=

= ≈


Since ( ) adjcos 25

hyp 15 in.a

= = , we have

( ) ( )15 in. cos 25 13.59 in. 14 in.a = ≈ ≈


Since ( ) oppsin 50

hyp 27 ft.a

= = , we have

( ) ( )27 ft. sin 50 20.68 ft. 21 ft.a = ≈ ≈

25

50

Chapter 1

80



adj 21.9 mi.a

= = , we

have ( ) ( )21.9 mi. tan 33.5 14.5 mi.a = ≈


Since ( ) opp 19.22 cmsin 47.45

hyp c= = , we

have

( )19.22 cm 26.09 cm

sin 47.45c = ≈


First, observe that

137 45 37 45 37.7560

β⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠.


b= = , we

have ( ) ( )120.0 yd. tan 37.75 92.91 yd.b = ≈


First, observe that

175 10 75 10 75.1666760

β⎛ ⎞

′ ′= = + ≈⎜ ⎟′⎝ ⎠.

Since ( ) opp 96.5 kmsin 75.16667hyp c

= = ,

we have ( )96.5 km 99.8 km

sin 75.16667c = ≈

33.5 47.45

37.75β

Chapter 1 Review

81


First, observe that

( )180 90 30 60β = − + = .


a= = , we have

( )( )21 ft. sin 30 11 ft.a = ≈


b= = ,

we have ( )( )21 ft. cos30 18 ft.b = ≈


First, observe that

( )180 90 65 25α = − + = .


b= = , we have

( )( )8.5 mm sin 65 7.7 mmb = ≈ . Similarly, since

( ) adjcos 65hyp 8.5 mm

a= = , we have

( )( )8.5 mm cos 65 3.6 mma = ≈ .

30

β

α

65

Chapter 1

82


First, observe that

( )180 90 48.5 41.5β = − + = .

Since ( ) opp 215 mi.sin 48.5hyp c

= = , we

have

( )215 mi. 287 mi.

sin 48.5c = ≈

Similarly, since

( ) opp 215 mi.tan 48.5adj b

= = , we have

( )215 mi. 190 mi.

tan 48.5b = ≈


First, observe that

( )180 90 30 15 59 45

159 45 59.7560

β ′ ′= − + =

⎛ ⎞′= + =⎜ ⎟′⎝ ⎠

130 15 30 15 30.2560

α⎛ ⎞

′ ′= = + =⎜ ⎟′⎝ ⎠

Since ( ) adj 2,154 ft.cos 30.25hyp c

= = , we

have

( )2,154 ft. 2, 494 ft.

cos 30.25c = ≈

Similarly, since

( ) opptan 30.25adj 2,154 ft.

a= = , we have

( ) ( )2,154 ft. tan 30.25 1,256 ft.a = ≈

48.5

β β

30.25

Chapter 1 Review

83


First, since 30.5 ft.tan


1 30.5tan 33.745.7

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

So, ( )180 90 33.7 56.3β ≈ − + ≈ . Next, using the Pythagorean Theorem, we obtain

( ) ( )2 22 230.5 ft. 45.7 ft. 3,018.74 ft.c = + = and so, 54.9 ft.c ≈


First, since 11,798 kmsin

32,525 kmα = , we obtain

1 11,798sin 21.26832,525

α − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.

Similarly, since 11,798 kmcos32,525 km

β = , we

obtain 1 11,798cos 68.732

32,525β − ⎛ ⎞= ≈⎜ ⎟

⎝ ⎠.


( ) ( )2 22

2 2

11,798 km 32,525 km918,682,821 km30,310 km

bbb

+ =

=≈

β

α

β

α

Chapter 1

84


Observe that sin 30

150 ft.a

= , so that

( )( )sin 30 150 ft. 75 ft.a = ≈ . So, the altitude should be about 75 ft.


Observe that 100 ft.sin 30

H= , so that

100 ft. 200 ft.sin 30

H = ≈ .

So, the hose should be about 200 ft. long. 101. Consider this diagram:



1

1

1015 tan3

10tan 15 583

θ

θ

−

−

⎛ ⎞+ = ⎜ ⎟⎝ ⎠⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

So, the bearing would be 58N E




1

1

920 tan59tan 20 415

θ

θ

−

−

⎛ ⎞+ = ⎜ ⎟⎝ ⎠⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

So, the bearing would be 41N W .

30 30

10mi

3mi

θ

15θ

20 5mi

9mi

Chapter 1 Review

85

103. 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 41.32ft.x = Thus, the two legs have lengths 41.32 ft. and 41.32 3 71.57 ft.,≈ and the hypotenuse has length 82.64 ft. 104. 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 87.65x cm= Thus, the two legs have lengths

87.653

87.65 and 50.60 ,cm cm≈ and the hypotenuse has length 101.20 cm.

105. a. sin 78.4 0.980≈ and so, 1 1 1.02041sin 78.4 0.980

≈ ≈

b. The keystroke sequence 78.4, sin, 1/x yields 1.02085.

106. a. tan 34.8 0.695≈ and so, 1 1 1.43885tan 34.8 0.695

≈ ≈

b. The keystroke sequence 34.8, tan, 1/x yields 1.43881. 107. 2.612 108. 0.125 109. -1 110. Error – dividing by zero. Chapter 1 Practice Test Solutions ---------------------------------------------------------------- 1. Let x = measure of smallest angle in the triangle. Then,

(1) implies that 5x = measure of the largest angle in the triangle, (2) implies that 3x = measure of the third angle in the triangle.

So, 3 5 180 so that 9 180 and thus, 20 .x x x x x+ + = = = Thus, the three angles are

20 , 60 , 100 . 2. Consider the following triangle:

The angle opposite the 30 angle in a 30 60 90− − triangle is the smallest of the three angles. So, the other leg must have length 5 3 cm and the hypotenuse must have length 10 cm.

30

60

Chapter 1

86

3. Consider the following two diagrams:

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

( )1 12 2

5 ft. 5 ft. so that 600 ft. 6,000 ft.ft. 600 ft. ft.

H H⎛ ⎞

= = =⎜ ⎟⎝ ⎠

So, the Grand Canyon is 6,000 ft. tall. 4. Consider the following triangle:

From the Pythagorean Theorem, we see that 2 23 1 10c = + = .

a. 1 10sin1010

θ = =

b. 3 3 10cos1010

θ = =

c. 1

sin 110tan 3cos 310

θθθ

= = =

d. 1 10seccos 3

θθ

= =

e. 1 10csc 10sin 1

θθ

= = =

f. 1cot 3tan

θθ

= =

θθ

θ

Chapter 1 Practice Test

87

5. a. ( ) 3 10sin 90 cos10

θ θ− = = (by Exercise 4b)

b. ( )sec 90 csc 10θ θ− = = (by Exercise 4e)

c. ( ) 1cot 90 tan3

θ θ− = = (by Exercise 4c)

6. In the following computations, we use sin 1 1 1tan , cot , sec , csccos tan cos sin

θθ θ θ θθ θ θ θ

= = = =

θ sinθ cosθ tanθ cotθ secθ cscθ

30 12

32

33

3 2 3

3

2

45 22

22

1 1 2 2

60 32

12

3 33

2 2 3

3

7. ( ) ( )1sec 42.8 1.3629

cos 42.8= ≈ 8. The first value ( 2

3cosθ = ) is an exact value of the cosine function, while the second ( cos 0.66667θ = ) would serve as an approximation to the first.


3600⎛ ⎞′′ = ≈⎜ ⎟⎝ ⎠

4545 0.7560

⎛ ⎞′ = =⎜ ⎟⎝ ⎠

.

So, 33 45 20 33.756′ ′′ ≈ .


Since ( ) opp 10 cmtan 20

adj b= = , we have

( )10 cm 27 cm

tan 20b = ≈ .

20

Chapter 1

88

11. Consider this triangle: Since 9.2 kmcos

23 kmβ = , we obtain

1 9.2cos 6623

β − ⎛ ⎞= ≈⎜ ⎟⎝ ⎠

.


Since 12 mmcos50 c= , we see that

12 mmcos50

19 mmc = ≈ .


Since 13.3

14.0sin β = , we obtain

( )1 13.314.0sin 71.8β −= ≈ .


First, we convert α to DD:

( )16033 10 33 10 33.17′′ ′= + ≈

Since ( ) 47msin 33.17 a= , we see that

( ) ( )47m sin 33.17 26ma = ≈ .


Since 3.45

6.78tan ,α = we obtain

( )1 3.456.78tan 27.0α −= ≈ .

50

α

β

α α

β

Chapter 1 Practice Test

89





that 30α = . 17. Using ( ) ( )sin sini i r rn nθ θ= , observe that

( ) ( )( )

( )

1.00sin 25 sin 16

1.00sin 251.53 .

sin 16

r

r

n

n

=

= ≈


Since tan 20150 ft.

r= , we have

( )( )150 ft. tan 20 54.6 ft.r = ≈ So, the diameter of the circle is approximately 109 ft.

20 20

Chapter 1

90

19. ( ) ( ) ( ) ( )60 6011

44.27 44 0.27 44 16.20 44 16 0.20 44 16 12′ ′′′

′′ ′ ′ ′′= + = + = + + =

20. ( ) ( )1 160 360022 10 23 22 10 23 22.1731′ ′′′ ′′ ′ ′′= + + ≈

21. Since 1 46 2sin 75

csc75+

= = , we have 6 24cos15 sin 75 += = .

22. 3

21

2

22cos 45 3 3 2 1 3 6 3 3 6sin 60

cot 30 2 2 2 2 6 63+

+ = + = + ⋅ = + =

23. 1

12 40 55 49 67 89 67 (60 29) 67 60 29 68 29=

′ ′ ′ ′ ′ ′ ′+ = = + + = + + = .

24. ( ) ( )82 27 35 39 81 87 35 39 87 35 87 39 46 48′′ ′ ′ ′ ′− = − = − + − = 25. Consider the following triangle: We have

200 yardstan 72 h= , so that

( )200 yards tan 72 620 yardsh = ≈

72

θ

TRIG 3e ISM Chapter 1 FINAL - test bank U › sample › Solution-Manual-for... · 2017-12-04 ·...

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