Solution manual for trigonometry by ratti

3, 3 lies in Quadrant I, so a line through that point forms a 45° angle with the positive x-axis.. 1, 0 lies on the positive x-axis, so a line through that point forms a 0° angle with th

Therefore, the 765° and 45° angles are

coterminal All angles that are coterminal

with 45° are 45° + ⋅n 360 ,° where n is an

integer

b Adding 2(360°) = 720° to –570° gives an

angle between 0° and 360°

–570° + 720° = 150°

Therefore, the –570° and 150° angles are

coterminal All angles that are coterminal with 150° are 150° + ⋅n 360 ,° where n is an

integer

7

In the figure, OP is the diagonal of the 3 3×

square PQRO So OP bisects the right angle

QOR Therefore 180 1 90 225

2

θ= ° + ⋅ ° = °

Since there are infinitely many angles with OP

as the terminal side, and those angles differ by 360°, 225θ= ° + ⋅n 360 ,° n any integer

8

Angles 2 and 3 are supplements, and angles 3 and 7 are congruent, so angles 2 and 7 are supplements Therefore,

1.1 A Exercises: Basic Skills and Concepts

1 The degree measure of one complete revolution

is 360°

2 The sum of two complementary angles is 90°

3 An angle is in standard position if the initial

side is the positive x-axis and the vertex is at

the origin

4 For any integer n, an angle of θ° + ⋅n 360° has

the same terminal side as the angle of θ

degrees

5. True

6. False An angle is standard position is

quadrantal if its terminal side lies on a

9. complement: none because the measure of the

angle is greater than 90°

supplement: 180° – 120° = 60°

10. complement: none because the measure of the

angle is greater than 90°

supplement: 180° – 160° = 20°

11. complement: none because the measure of the

angle is greater than 90°; supplement: none

because the measure of the angle is greater than

180°

12. complement: none because the measure of the

angle is negative; supplement: none because the

measure of the angle is negative

52 Since 700° is a positive angle greater than han

360°, subtract 360° to get an angle between 0°

and 360°: 700° – 360° = 340° Therefore, the

700° and 340° angles are coterminal All angles

that are coterminal with 340° are

Therefore, the 1785° and 345° angles are

coterminal All angles that are coterminal with

345° are 345° + ⋅n 360 ,° where n is an integer

54 Since 2064° is a positive angle greater than

1800° = (360°)(5), subtract 1800° to get an

angle between 0° and 360°

2064° – 1800° = 264°

Therefore, the 2064° and 264° angles are

coterminal All angles that are coterminal with

264° are 264° + ⋅n 360 ,° where n is an integer

55 Adding 360° to –50° gives an angle between 0°

and 360°: –50° + 360° = 310°

Therefore, the –50° and 310° angles are

coterminal All angles that are coterminal with

310° are 310° + ⋅n 360 ,° where n is an integer

56 Adding 360° to –225° gives an angle between

0° and 360°: –225° + 360° = 135°

Therefore, the –225° and 135° angles are

coterminal All angles that are coterminal with

135° are 135° + ⋅n 360 ,° where n is an integer

57 Adding 2(360°) = 720° to –400° gives an angle

between 0° and 360°: –400° + 720° = 320°

Therefore, the –400° and 320° angles are

coterminal All angles that are coterminal with

320° are 320° + ⋅n 360 ,° where n is an integer

58 Adding 2(360°) = 720° to –700° gives an angle

between 0° and 360°: –700° + 720° = 20°

Therefore, the –700° and 20° angles are

coterminal All angles that are coterminal with

20° are 20° + ⋅n 360 ,° where n is an integer

For exercises 59–66, review the explanations in

Example 7 and Practice Problem 7

59 (3, 3) lies in Quadrant I, so a line through that

point forms a 45° angle with the positive x-axis

Thus, 45θ = ° + ⋅n 360 ,° n any integer

60. (4, –4) lies in Quadrant IV, so a line through

that point forms a 270° + 45° = 315° angle with

the positive x-axis Thus, θ =315° + ⋅n 360 ,° n

any integer

61 (–5, 5) lies in Quadrant II, so a line through that point forms a 90° + 45° = 135° angle with the positive x-axis Thus, θ =135° + ⋅n 360 ,° n any integer

62 (–2, –2) lies in Quadrant III, so a line through that point forms a 180° + 45° = 225° angle with the positive x-axis Thus, θ =225° + ⋅n 360 ,° n

any integer

63 (1, 0) lies on the positive x-axis, so a line through that point forms a 0° angle with the positive x-axis Thus, θ = ° + ⋅0 n 360 ,° n any integer

64 (0, 2) lies on the positive y-axis, so a line through that point forms a 90° angle with the positive x-axis Thus, θ = ° + ⋅90 n 360 ,° n any integer

65 (–3, 0) lies on the negative x-axis, so a line through that point forms a 180° angle with the positive x-axis Thus, θ =180° + ⋅n 360 ,° n any integer

66 (0, –4) lies on the negative y-axis, so a line through that point forms a 270° angle with the positive x-axis Thus, θ =270° + ⋅n 360 ,° n any integer

1.1 B Exercises: Applying the Concepts

min360

6⋅360° =300°

in the given time period

70 There are 4 hours 30 minutes from 3:15 pm to

7:45 pm The hour hand travels 360° in 12

hours, so it travels 4.5 360 135

12 ⋅ ° = ° in the given time period

Since α and γ are vertical angles, they are

equal and γ =135 ° Since β and θ are

vertical angles, they are equal and θ = ° 45

75 The angle vertical with α is the interior angle

on the same side of the transversal as β so the ,angles are supplementary

76 The angle vertical with α is the interior angle

on the same side of the transversal as β so the ,angles are supplementary

78

6030

m DCE

Angles CEF and DCE are interior angles on the

same side of the transversal, so they are

The sum of the measures of angles AOC, COE,

and BOE is 180°, so we have

m BOE∠ = ⋅ ° = ° Angles AOC and

DOB are vertical angles, so their measures are

equal m DOB∠ = = °y 20 Angles DOB and

DOA are supplements, so

MP bisects angle AMN, MQ bisects angle BMN,

NQ bisects angle DNM, and NP bisects angle CNM Therefore, ∠AMP= ∠NMP,

DNQ MNQ NMQ NMQ

MNQ MNQ NMQ MNQ

(continued from page 7)

Now we must show that angles PMQ and PNQ

are also right angles We know that angles

AMN and DNM are equal since they are

alternate interior angles Each angle is bisected,

so we can conclude that angle PMN = angle

MNQ Similarly, we can deduce that angle

QMN = angle MNP Adding equations (1) and

(2) then substituting, we have

180

9090

NMQ MNQ MNP

PNM NMQ PNM MNQ MNP

From exercise 81, we know that MPNQ is a

rectangle Since MN⊥AB, angles AMN and

BMN each measure 90° MP bisects angle

AMN , so angle PMN = 45° MQ bisects angle

BMN , so angle QMN = 45° Thus

QMN ≅ PMN

△ △ by AAS, and therefore,

MP = MQ From geometry, we know that a

rectangle with two consecutive equal sides is a

n n−

1.2 Triangles 1.2 Practice Problems

3

3420 3420 2

2 3420

22

a Because BC is parallel to DE, AB and AC are

transversals Thus, ∠ADE= ∠ABC and

AED ACB

Both triangles contain A, so triangles ADE

and ABC are similar because they have equal

103

1.2 A Exercises: Basic Skills and Concepts

1 The sum of the measures of the three angle of a triangle is 180°

2 In an isosceles triangle, the two angles opposite

equal sides are equal or congruent

3 In a 30°-60°-90° triangle, the hypotenuse is two

time the length of the shortest side and the side opposite the 60° angle is 3 time the length of the shortest side

4 In similar triangles, the lengths of the

corresponding sides are proportional

5 True

6 False The corresponding sides of two similar

triangles are proportional while the corresponding sides of two congruent triangles are equal Note that congruent triangles are also similar

25 Since each leg has length 4, the hypotenuse has length 4 2

26. Since each leg has length 5, the hypotenuse has length 5 2

27. Since each leg has length 12, the hypotenuse has length 22

28. Since each leg has length 35, the hypotenuse has length 3 2

31. Since the hypotenuse has length 4, each leg has

In exercises 33–382, recall that in a 30°-60°-90°

triangle, if the length of the shorter leg (the leg

opposite the 30° angle) is x, then the length of the

longer leg (the angle opposite the 60° angle) is x 3,

and the length of the hypotenuse is 2x

33 The length of the shortest side is 4, so the

length of the longer leg is 4 3, and the length

of the hypotenuse is 8

34. The length of the shortest side is 6, so the

length of the longer leg is 6 3, and the length

of the hypotenuse is 12

35 The length of the side opposite the 60° angle

(the longer leg) is 4, so the length of the shorter

leg is 4 4 33

3= , and the length of the

hypotenuse is 8 33

36. The length of the side opposite the 60° angle

(the longer leg) is 6, so the length of the shorter

leg is 6 6 33

3= =2 3, and the length of the

hypotenuse is 4 3

37. The length of the hypotenuse is 4, so the length

of the shorter leg (the leg opposite the 30°

angle) is 2, and the length of the longer leg (the

leg opposite the 60° angle) is 2 3

38. The length of the hypotenuse is 6, so the length

of the shorter leg (the leg opposite the 30°

angle) is 3, and the length of the longer leg (the

leg opposite the 60° angle) is 3 3

Similarly, 73 ∠ = ° Thus, the triangles are D

similar since the corresponding angles are

Similarly, 35 ∠ = ° Thus, the triangles are M

similar since the corresponding angles are

11.112

x

x

y y

(continued from page 13)

AC = Distance the second car traveled after 2

Let AB represent the ladder leaning against the

wall BC By the properties of 45-45-90 triangle,

Let B denote the location of the balloon and let

AB represent the cable

Let BC represent the building and let AC

represent the shadow of the building Then,

BC= = = In a 45°-45°-90° triangle, the legs are equal, so

Let AC be the height of the building

(continued on next page)

Not drawn to scale

(continued from page 14)

Not drawn to scale

Since the angle of elevation of the sun is the

same in each triangle, the triangles are similar

Let AC be the height of the tower Then we

First, we must find the distance the player is

standing from the net

The length of the base of the larger triangle is

15 + 22.5 = 37.5 ft Now use the Pythagorean

theorem to find d

37.5 +7.5 =d ⇒1462.5=d ⇒ ≈d 38.2

The ball traveled approximately 38.2 feet

1.2 C Exercises: Beyond the Basics

65

Apply the Pythagorean theorem to triangles

ABD and ACD

(1)(2)

Apply the Pythagorean theorem to triangles

ABD and ACD

(1)(2)

69

Given right triangle PMN, with legs a and b

and hypotenuse x Thus, a2+b2=x2 (1)

Right triangle: If c2=a2+b2, then

c =x ⇒ = Therefore, c x △PMN≅△ABC

by SSS ∠ ≅ ∠ because corresponding C N

parts of congruent triangles are congruent, so

angle C is a right angle

Obtuse triangle: If c2>a2+b2, then using

equation (1), c2>x2⇒ > From geometry, c x

we know that in two triangles if two sides from

one triangle are equal to two corresponding

sides from another triangle, then the included

angle opposite the longer side is greater than

the included angle opposite the shorter side

That is, c> ⇒ ∠ > ∠ Therefore, angle x C N C

is obtuse and triangle ABC is an obtuse triangle

Acute triangle: If c2<a2+b2, then using

equation (1), c2<x2⇒ < From geometry, c x

we know that in two triangles if two sides from

one triangle are equal to two corresponding

sides from another triangle, then the included

angle opposite the longer side is greater than

the included angle opposite the shorter side

That is, x> ⇒ ∠ > ∠ Therefore, angle c N C C

is actue and triangle ABC is an acute triangle

70 We will use the results from exercise 69

The longest side of the triangle is 16 cm, so

D is the midpoint of AB and E is the midpoint

of AC Then, AD=12AB and AE=12AC.Since ,∠ = ∠ ABC A A △ ∼△ADE by SAS Because corresponding sides of similar triangles are proportional, DE=12BC.Because corresponding angles of similar triangles are equal, ∠ = ∠ These are D B.corresponding angles formed by a transversal

(AB) cutting BC and DE, so BC DE

74

D is the midpoint of AB, E is the midpoint of

AC , and F is the midpoint of BC Then,

Since ,∠ = ∠B B △ABC∼△DBF by SAS

Since ,∠ = ∠C C △ABC∼△EFC by SAS

Since each side in triangle DEF is 12 the

corresponding side in triangle ABC,

ABC∼ FDE

Use the figure below for exercises 75 and 76

75 We are given that △ABC∼△XYZ Therefore,

( )( )

( )( ) ( )( ) ( )( ) ( ( )( ) )( )

( ) ( )

BC AD BC AD AD AD

YZ XW XW XW

YZ XW

AD XW

The length of the side is 45 cm

78 Using the result of exercise 75, we have

2 2

from one vertex of an n-gon is n – 2 Since the

sum of the angles of a triangle is 180°, the sum

of the interior angles of an n-gon is 180(n – 2)

In a regular polygon, the interior angles are equal, so the measure of each interior angle is

36

V

∠ = °

81 From exercise 79, we know that the measure of

an interior angle of a regular n-gon is

82 From exercise 81, we know that the measure of

each angle of the star is 180 720 ,

n

°

° − so the sum of the angles is

83 Statement (iv) is always true This is a

restatement of the Triangle Inequality, which

states that the sum of the lengths of two sides of

a triangle is greater than the length of the third

01

11

1

y r r y x r r x

0

y x x y

θθ

10

11

01

00

1

y r r y x r r x y x x y

θθθθθθ

sin1170 sin 90 1csc1170 csc 90 1cos1170 cos 90 0sec1170 sec 90 , undefinedtan1170 tan 90 , undefinedcot1170 cot 90 0

( ) ( ) ( ) ( ) ( ) ( )

sin 630 sin 90 1csc 630 csc 90 1

(continued on next page)

(continued from page 18)

2csc 405 csc 45 2

2cos 405 cos 45

2sec 405 sec 45 2

9 Because tanθ < and cos0 θ > , θ lies in 0

quadrant IV tan 4

θ = = −y = −

r

41sec

The graph of h is a parabola with the

x-coordinate of the vertex

The ball remains in flight for about 6.19

seconds, and d=6.19(140) cos 45° ≈613 ft

Thus, the ball reaches a maximum height of

about 153 ft and has a range of about 613 ft

1.3 A Exercises: Basic Skills and Concepts

1 For a point P(x, y) on the terminal side of an

angle θ in standard position, we let r =

sinθ equals sinθ

5 False The value of a trigonometric function for any angle is the same for any point on the terminal side of θ

6. False In each quadrant, cosine and secant are either both positive or both negative, and sine and cosecant are either both positive or both negative

For exercises 7–22, recall that

45 Quadrant III 46 Quadrant III

47 Quadrant II 48 Quadrant III

49 Quadrant IV 50 Quadrant III

51 Quadrant II 52 Quadrant III

Since θ lies in quadrant III, x = –12

54 If θ lies in quadrant IV, x = 12

Since θ lies in quadrant I, y = 24

56. Since θ lies in quadrant IV, y = –24

1.3 B Exercises: Applying the Concepts

Use the figure below for exercises 65−72

2

1

44 sin 30 7.5625 ft64

44 sin 30

1.375 sec16

44 sin 30 cos 30

52.39 ft16

44 sin 45

1.94 sec16

44 sin 45 cos 45

60.5 ft16

44 sin 60 cos 60

52.39 ft16

44 sin 90 cos 90

0 ft16

2

16 sec 45tan 45

80

1

20080

For exercises 80−83, refer to the following figure

80 Since sin( )− = −θ sin ,θ the y-value of the point on the unit circle on the terminal side of θ

− equals the opposite of y-value of the point

on the unit circle on the terminal side of θ Since cos( )− =θ cos ,θ the x-value of the point

on the unit circle on the terminal side of − θequals the x-value of the point on the unit circle

on the terminal side of θ Thus, the coordinates

of the point on the unit circle are (x, −y), and the point on the unit circle is S The triangle we are seeking is SOM