7.2.3 homework answers

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3 π 2 3 π 2

−135 ° −135 °

7 π 10 7 π 10

α = 150° α = 150°

β = 60° β = 60°

7 π 6 7 π 6

215 π 18 = 37.525 units 215 π 18 = 37.525 units

− 3 π 2 − 3 π 2 rad/s

1655 kilometers per hour

7.2 Right Triangle Trigonometry

sin  t = 33 65 , cos  t = 56 65 , tan  t = 33 56 , sec  t = 65 56 , csc  t = 65 33 , cot  t = 56 33 sin  t = 33 65 , cos  t = 56 65 , tan  t = 33 56 , sec  t = 65 56 , csc  t = 65 33 , cot  t = 56 33

sin ( π 4 ) = 2 2 , cos ( π 4 ) = 2 2 , tan ( π 4 ) = 1 , sec ( π 4 ) = 2 , csc ( π 4 ) = 2 , cot ( π 4 ) = 1 sin ( π 4 ) = 2 2 , cos ( π 4 ) = 2 2 , tan ( π 4 ) = 1 , sec ( π 4 ) = 2 , csc ( π 4 ) = 2 , cot ( π 4 ) = 1

adjacent = 10 ; opposite = 10 3 ; adjacent = 10 ; opposite = 10 3 ; missing angle is π 6 π 6

About 52 ft

7.3 Unit Circle

cos ( t ) = − 2 2 , sin ( t ) = 2 2 cos ( t ) = − 2 2 , sin ( t ) = 2 2

cos ( π ) = − 1 , sin ( π ) = 0 cos ( π ) = − 1 , sin ( π ) = 0

sin ( t ) = − 7 25 sin ( t ) = − 7 25

approximately 0.866025403

• ⓐ cos ( 315° ) = 2 2 , sin ( 315° ) = – 2 2 cos ( 315° ) = 2 2 , sin ( 315° ) = – 2 2
• ⓑ cos ( − π 6 ) = 3 2 , sin ( − π 6 ) = − 1 2 cos ( − π 6 ) = 3 2 , sin ( − π 6 ) = − 1 2

( 1 2 , − 3 2 ) ( 1 2 , − 3 2 )

7.4 The Other Trigonometric Functions

sin t = − 2 2 cos t = 2 2 , tan t = − 1 , s e c t = 2 , csc t = − 2 , cot t = − 1 sin t = − 2 2 cos t = 2 2 , tan t = − 1 , s e c t = 2 , csc t = − 2 , cot t = − 1

sin π 3 = 3 2 , cos π 3 = 1 2 , tan π 3 = 3 , s e c π 3 = 2 , c s c π 3 = 2 3 3 , c o t π 3 = 3 3 sin π 3 = 3 2 , cos π 3 = 1 2 , tan π 3 = 3 , s e c π 3 = 2 , c s c π 3 = 2 3 3 , c o t π 3 = 3 3

sin ( − 7 π 4 ) = 2 2 , cos ( − 7 π 4 ) = 2 2 , tan ( − 7 π 4 ) = 1 , sec ( − 7 π 4 ) = 2 , csc ( − 7 π 4 ) = 2 , cot ( − 7 π 4 ) = 1 sin ( − 7 π 4 ) = 2 2 , cos ( − 7 π 4 ) = 2 2 , tan ( − 7 π 4 ) = 1 , sec ( − 7 π 4 ) = 2 , csc ( − 7 π 4 ) = 2 , cot ( − 7 π 4 ) = 1

sin t sin t

cos t = − 8 17 , sin t = 15 17 , tan t = − 15 8 csc t = 17 15 , cot t = − 8 15 cos t = − 8 17 , sin t = 15 17 , tan t = − 15 8 csc t = 17 15 , cot t = − 8 15

sin t = − 1 , cos t = 0 , tan t = Undefined sec t = Undefined, csc t = − 1 , cot t = 0 sin t = − 1 , cos t = 0 , tan t = Undefined sec t = Undefined, csc t = − 1 , cot t = 0

sec t = 2 , csc t = 2 , tan t = 1 , cot t = 1 sec t = 2 , csc t = 2 , tan t = 1 , cot t = 1

≈ − 2.414 ≈ − 2.414

7.1 Section Exercises

Whether the angle is positive or negative determines the direction. A positive angle is drawn in the counterclockwise direction, and a negative angle is drawn in the clockwise direction.

Linear speed is a measurement found by calculating distance of an arc compared to time. Angular speed is a measurement found by calculating the angle of an arc compared to time.

4 π 3 4 π 3

2 π 3 2 π 3

7 π 2 ≈ 11.00 in 2 7 π 2 ≈ 11.00 in 2

81 π 20 ≈ 12.72 cm 2 81 π 20 ≈ 12.72 cm 2

π 2 π 2 radians

−3 π −3 π radians

π π radians

5 π 6 5 π 6 radians

5.02 π 3 ≈ 5.26 5.02 π 3 ≈ 5.26 miles

25 π 9 ≈ 8.73 25 π 9 ≈ 8.73 centimeters

21 π 10 ≈ 6.60 21 π 10 ≈ 6.60 meters

104.7198 cm 2

0.7697 in 2

8 π 9 8 π 9

1320 1320 rad/min 210.085 210.085 RPM

7 7 in./s, 4.77 RPM , 28.65 28.65 deg/s

1 , 809 , 557.37 mm/min = 1 , 809 , 557.37 mm/min = 30.16 m/s 30.16 m/s

5.76 5.76 miles

794 miles per hour

2,234 miles per hour

11.5 inches

7.2 Section Exercises

The tangent of an angle is the ratio of the opposite side to the adjacent side.

For example, the sine of an angle is equal to the cosine of its complement; the cosine of an angle is equal to the sine of its complement.

b = 20 3 3 , c = 40 3 3 b = 20 3 3 , c = 40 3 3

a = 10,000 , c = 10,00.5 a = 10,000 , c = 10,00.5

b = 5 3 3 , c = 10 3 3 b = 5 3 3 , c = 10 3 3

5 29 29 5 29 29

5 41 41 5 41 41

c = 14 , b = 7 3 c = 14 , b = 7 3

a = 15 , b = 15 a = 15 , b = 15

b = 9.9970 , c = 12.2041 b = 9.9970 , c = 12.2041

a = 2.0838 , b = 11.8177 a = 2.0838 , b = 11.8177

a = 55.9808 , c = 57.9555 a = 55.9808 , c = 57.9555

a = 46.6790 , b = 17.9184 a = 46.6790 , b = 17.9184

a = 16.4662 , c = 16.8341 a = 16.4662 , c = 16.8341

498.3471 ft

22.6506 ft  

368.7633 ft

7.3 Section Exercises

The unit circle is a circle of radius 1 centered at the origin.

Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, t , t , formed by the terminal side of the angle t t and the horizontal axis.

The sine values are equal.

60° , 60° , Quadrant IV, sin ( 300° ) = − 3 2 sin ( 300° ) = − 3 2 , cos ( 300° ) = 1 2 cos ( 300° ) = 1 2

45° , 45° , Quadrant II, sin ( 135° ) = 2 2 sin ( 135° ) = 2 2 , cos ( 135° ) = − 2 2 cos ( 135° ) = − 2 2

60° , 60° , Quadrant II, sin ( 120° ) = 3 2 sin ( 120° ) = 3 2 , cos ( 120° ) = − 1 2 cos ( 120° ) = − 1 2

30° , 30° , Quadrant II, sin ( 150° ) = 1 2 sin ( 150° ) = 1 2 , cos ( 150° ) = − 3 2 cos ( 150° ) = − 3 2

π 6 , π 6 , Quadrant III, sin ( 7 π 6 ) = − 1 2 sin ( 7 π 6 ) = − 1 2 , cos ( 7 π 6 ) = − 3 2 cos ( 7 π 6 ) = − 3 2

π 4 , π 4 , Quadrant II, sin ( 3 π 4 ) = 2 2 sin ( 3 π 4 ) = 2 2 , cos ( 4 π 3 ) = − 2 2 cos ( 4 π 3 ) = − 2 2

π 3 , π 3 , Quadrant II, sin ( 2 π 3 ) = 3 2 sin ( 2 π 3 ) = 3 2 , cos ( 2 π 3 ) = − 1 2 cos ( 2 π 3 ) = − 1 2

π 4 , π 4 , Quadrant IV, sin ( 7 π 4 ) = − 2 2 , cos ( 7 π 4 ) = 2 2 sin ( 7 π 4 ) = − 2 2 , cos ( 7 π 4 ) = 2 2

− 15 4 − 15 4

( −10 , 10 3 ) ( −10 , 10 3 )

( –2.778 ,   15.757 ) ( –2.778 ,   15.757 )

[ –1 ,   1 ] [ –1 ,   1 ]

sin t = 1 2 , cos t = − 3 2 sin t = 1 2 , cos t = − 3 2

sin t = − 2 2 , cos t = − 2 2 sin t = − 2 2 , cos t = − 2 2

sin t = 3 2 , cos t = − 1 2 sin t = 3 2 , cos t = − 1 2

sin t = − 2 2 , cos t = 2 2 sin t = − 2 2 , cos t = 2 2

sin t = 0 , cos t = − 1 sin t = 0 , cos t = − 1

sin t = − 0.596 , cos t = 0.803 sin t = − 0.596 , cos t = 0.803

sin t = 1 2 , cos t = 3 2 sin t = 1 2 , cos t = 3 2

sin t = − 1 2 , cos t = 3 2 sin t = − 1 2 , cos t = 3 2

sin t = 0.761 , cos t = − 0.649 sin t = 0.761 , cos t = − 0.649

sin t = 1 , cos t = 0 sin t = 1 , cos t = 0

− 6 4 − 6 4

( 0 , –1 ) ( 0 , –1 )

37.5 seconds, 97.5 seconds, 157.5 seconds, 217.5 seconds, 277.5 seconds, 337.5 seconds

7.4 Section Exercises

Yes, when the reference angle is π 4 π 4 and the terminal side of the angle is in quadrants I and III. Thus, a x = π 4 , 5 π 4 , x = π 4 , 5 π 4 , the sine and cosine values are equal.

Substitute the sine of the angle in for y y in the Pythagorean Theorem x 2 + y 2 = 1. x 2 + y 2 = 1. Solve for x x and take the negative solution.

The outputs of tangent and cotangent will repeat every π π units.

2 3 3 2 3 3

− 2 3 3 − 2 3 3

− 3 3 − 3 3

sin t = − 2 2 3 sin t = − 2 2 3 , sec t = − 3 sec t = − 3 , csc t = − 3 2 4 csc t = − 3 2 4 , tan t = 2 2 tan t = 2 2 , cot t = 2 4 cot t = 2 4

sec t = 2 , sec t = 2 , csc t = 2 3 3 , csc t = 2 3 3 , tan t = 3 , tan t = 3 , cot t = 3 3 cot t = 3 3

− 2 2 − 2 2

sin t = 2 2 sin t = 2 2 , cos t = 2 2 cos t = 2 2 , tan t = 1 tan t = 1 , cot t = 1 cot t = 1 , sec t = 2 sec t = 2 , csc t = 2 csc t = 2

sin t = − 3 2 sin t = − 3 2 , cos t = − 1 2 cos t = − 1 2 tan t = 3 tan t = 3 , cot t = 3 3 cot t = 3 3 , sec t = − 2 sec t = − 2 , csc t = − 2 3 3 csc t = − 2 3 3

sin ( t ) ≈ 0.79 sin ( t ) ≈ 0.79

csc t ≈ 1.16 csc t ≈ 1.16

sin t cos t = tan t sin t cos t = tan t

13.77 hours, period: 1000 π 1000 π

3.46 inches

Review Exercises

− 7 π 6 − 7 π 6

10.385 meters

2 π 11 2 π 11

1036.73 miles per hour

a = 10 3 , c = 2 106 3 a = 10 3 , c = 2 106 3

a = 5 3 2 , b = 5 2 a = 5 3 2 , b = 5 2

369.2136 ft

all real numbers

cosine, secant

Practice Test

6.283 centimeters

3.351 feet per second, 2 π 75 2 π 75 radians per second

a = 9 2 , b = 9 3 2 a = 9 2 , b = 9 3 2

real numbers

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Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
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• Book title: Algebra and Trigonometry
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CHAPTER 7 KEYS:

Reviews & skills keys:.

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Lesson 7.1.1, lesson 7.1.2, lesson 7.1.3, lesson 7.1.4, lesson 7.1.5, lesson 7.2.1, lesson 7.2.2, lesson 7.2.3.

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7.2.3.5 Packet Tracer – Troubleshooting EIGRP for IPv4 Instructions Answers

Packet tracer – troubleshooting eigrp for ipv4 (instructor version).

Instructor Note : Red font color or Gray highlights indicate text that appears in the instructor copy only.

Addressing Table

In this activity, you will troubleshoot EIGRP neighbor issues. Use show commands to identify errors in the network configuration. Then, you will document the errors you discover and implement an appropriate solution. Finally, you will verify full end-to-end connectivity is restored.

Troubleshooting Process

a.  Use testing commands to discover connectivity problems in the network and document the problem in the Documentation Table.

b.  Use verification commands to discover the source of the problem and devise an appropriate solution to implement. Document the proposed solution in the Documentation Table.

c.  Implement each solution one at a time and verify if the problem is resolved. Indicate the resolution status in the Documentation Table.

d.  If the problem is not resolved, it may be necessary to first remove the implemented solution before returning to Step 2.

e.  Once all identified problems are resolved, test for full end-to-end connectivity.

Documentation Table

Download Packet Tracer (.pka) file:

7.2.3.5 packet tracer - troubleshooting eigrp for ipv4.pka 136.61 kb 3855 downloads.