Solution manual for trigonometry 3rd edition by young

From the given information, we know that since 1 complete revolution corresponds to 360 , we obtain the following proportion: 50.. From the given information, we know that since 1 comple

since it is a clockwise rotation.)

since it is a clockwise rotation.)

since it is a clockwise rotation.)

since it is a clockwise rotation.)

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

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

So, the two angles have measures 36 and 54

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

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

10

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

So, the two angles have measures 60 and 120

18 Since the angles with measures (3x+15) and (10x+10) are assumed to be complementary, we know that (3x+15) (+ 10x+10) =90 Simplifying this yields

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

a +b = Using the given information, this becomes c 62+b2 =10 ,2 which simplifies

to 36+b2 =100 and then to, b2 =64, so we conclude that b= 8

28 Since this is a right triangle, we know from the Pythagorean Theorem that

simplifies to a2 =18, so we conclude that a= 18 3 2=

34 Since this is a right triangle, we know from the Pythagorean Theorem that

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 2 10, so that 10 10 5

22

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

39 The hypotenuse has length

( )

6 2 2

each leg has length 3 2 m

41 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

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 ( )

12 = , it immediately follows that

α = ⋅ − = − (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 ( )

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

to 360 , we obtain the following proportion:

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

to 360 , we obtain the following proportion:

51 We know that 1 complete revolution corresponds to 360

Let x = time (in minutes) it takes to make 1 complete revolution about the circle

Then, we have the following proportion:

45 minutes= x Solving for x then yields

=

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:

9 minutes= x Solving for x then yields

=

So, it takes 45 minutes to make one complete revolution

53 Let d = distance (in feet) the dog runs along the hypotenuse Then, from the

Pythagorean Theorem, we know that

2

30 807,300

85 7,300

d d

54 Let d = distance (in feet) the dog runs along the hypotenuse Then, from the

Pythagorean Theorem, we know that

2

25 10010,625

103 10,625

d d

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 2x=100, so that 100 100 2 50 2

22

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

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 2x=60, so that 60 60 2 30 2

22

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

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

17 (2 )

289 4

289 32893

2899.8

3

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 289

3

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

should be 2 289 19.6299 feet

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

in order to have such stakes support the tree

6060

30 30

59 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 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

100 400300

300 17.3 feet

x x x x

=

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

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

7 49 3 3

7 (2 )

49 34.0

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, x=7 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 102+152 ≠202

64 x2+82 =172 ⇒ x2 =225 ⇒ x=15ft

65 The speed is 170060 =1706 revolutions per second Since each revolution corresponds to

360 , the engine turns ( )170 ( )

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

45 45

71 True Since the angles of a right triangle are α β, , and 90 , and also we know that

90 180

α +β + = , it follows that α +β =90

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, α β+ =90 , 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 ,α β+ =90 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 1

60 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 β α α α+ 1+ 2+ 3 From the above discussion, we know that

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

60the way between the 9 and the 10; 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 9:10, we know that the minute hand is on the 2, but the hour hand

has moved 10 0.5× = clockwise from the 9 towards the 10, thereby leaving an angle of 5

25 between the hour hand and the 10 The picture is as follows:

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

79 Consider the following diagram:

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 32+y2 = , so that 524

y= Next, using the Pythagorean Theorem on ABC yields

( )2

2 2 2 2

80 Consider the following diagram:

Since we seek the length ofDC (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 2

81 Consider the following diagram:

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

x x x

82 Consider the following diagram:

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

12111

x x x

y y y

=

≈

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 = x 2) 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, 2

2

x

AE=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

84 Consider the following diagram:

Using the Pythagorean Theorem, we find that

x= (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

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

B= since corresponding angles are congruent

8 Since G=65 , it follows that F =65 since vertical angles are congruent Since A and

B are supplementary angles, A=115

9 8x=9x−15 (vertical angles), so that x=15 Thus, A=8(15 ) 120= = D

10 9x+ =7 11x−7 (corresponding angles), so that solving for x yields x=7 Thus, (9(7) 7) 70

1.4b = 2.6 Solving for b yields b=2.1

23 First, note that 26.25km=26, 250m Now, observe that by similarity, d a

f = , so that c

1.12.5 26, 250

m= m Solving for a yields

2

2 28,875 2.5

m m

=

24 First, note that 35m=3,500cm Now, observe that by similarity, c b

c m cm

in in. 12

25 in.

in

e e

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

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

=

=

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

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

So, the son’s shadow is23ft 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

300cm (5 ) 60,000

60,000

12, 000 120 5

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

131 0.9

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 11

4 16 2

2 11 16 1 4

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

( )

7 1

=

=

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 x=2.8ft

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

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 2x=2, so that x= 2 in

60

39 The information provided is:

Solving for d yields: 2 s d s =1.9d s+6.65 ⇒ 0.1d s =6.65 ⇒ d s =66.5cm

41 The information provided is:

2 4 , so that 260

42 The information provided is:

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 y

x, 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 x=1 and y= , then the triangles are NOT congruent 2

49 False Such angles are congruent, and hence unless they are both 90 , they need not

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

4

x= , so that x= 2

54 Consider the following diagram:

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 BE

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

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

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 H and 0 H are equal because they are vertical i

angles Hence, using the similar triangles ratio yields 0 0

b) Consider the following diagram:

Again, from similarity if follows that 0

i

H

57 Claim (Lens Law):

0 0 0

1 0 0

i

i i

i i

i i

i i

f

=

as desired ☻

58 We begin with several observations:

1.) m(∠ABE)=90 since BF is parallel to CG, and ∠ABE≈ ∠GCD since they are corresponding angles

2.) m(∠AEB)=m(∠CGA)=m(∠FEG)=60

3.) m(∠CGD)=m(∠CDG)=45 since ΔCDG is an isosceles right triangle

4.) BC and FG have the same length

Hence, ΔABE∼ΔACG∼ΔGFE

59 We begin with several observations:

1.) m(∠ABE)=90 since BF is parallel to CG, and ∠ABE≈ ∠GCD since they are corresponding angles

2.) m(∠AEB)=m(∠CGA)=m(∠FEG)=60

3.) m(∠CGD)=m(∠CDG)=45 since ΔCDG is an isosceles right triangle

4.) BC and FG have the same length

Hence, ΔABE∼ΔACG∼ΔGFE Now, using this information, consider the following triangles:

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

5 3 3

7 Note that by the Pythagorean Theorem,

the hypotenuse has length 5 So,

cos

55

8 Note that by the Pythagorean Theorem,

the hypotenuse has length 5 So,

sin

55

9 Note that by the Pythagorean Theorem,

the hypotenuse has length 5 So,

1cos

5

θ

θ

10 Note that by the Pythagorean Theorem,

the hypotenuse has length 5 So,

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

length 3 So, cos 3 3 34

3434

15 Using the Pythagorean Theorem, we

know that the leg adjacent to angle θ has length 3 So, tan 5

17 Using the Pythagorean Theorem, we

know that the leg adjacent to angle θ has length 3 So, sec 1 34

θ

θ

18 Using the Pythagorean Theorem, we

know that the leg adjacent to angle θ has length 3 So, cot 1 3

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

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

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 tan opp 1

adj

y x

θ = = = , so y = x

Since we are given that 2(x+y)=200

(since a round trip is 200 yards), we see that

100

x+ =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

33 Consider the triangle: Applying the Pythagorean theorem yields

2 2 2

5 +12 =z ⇒ = z 13Thus,

5 13sinθ =

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

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

13cosθ = 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

35 If tanθ = 3, 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 = W ⇒ W = ≈3.46 ft

37 Consider the following triangle: Observe that

3 5

sin B=

θ

38 We have the following triangle:

Observe that 3

5

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

Now we have a similar triangle:

( )2

1 z+ = ⇒ =z Hence,

33 4 7

33sin

mm mm

47 1sec

sin 45 =cos 90 −45 =cos 45

50 True It follows from the co-function identity that

54 Using Exercise 53, we have

3sin 60 cos30

21cos 60 sin 30

( )

2 2 2 2

58 Using Exercise 55, we see that tan 60 = 3 and cot 30 =tan 60 = 3

59 − ≤1 sinθ ≤1 and 1 cos− ≤ θ ≤1

60 Any real number

61 secθ ≥1 and cscθ ≥1

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

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 2.92398

cos 70 ≈ 0.342≈

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

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

sin 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 0.70274

tan 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 1.05485

60, you can further simplify the expression

In this case, since 64′′=1 4′ ′′, the completely simplified answer is

60, you can further simplify the expression

In this case, since 65′′=1 5′ ′′, the completely simplified answer is

to 60