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
  • Publisher/website: OpenStax
  • Book title: Algebra and Trigonometry
  • Publication date: Feb 13, 2015
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
  • Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/chapter-7

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RWM103: Geometry

Unit 7: right triangle trigonometry.

In this unit we, will explore basic trigonometry. We use trigonometry for several types of measuring techniques, such as calculating the height of a building when you know how far away you are standing from a building and the angle of your gaze to the top. Sailors used trigonometry to determine distances and set their course by using the stars as their guide.

Completing this unit should take you approximately 3 hours.

Upon successful completion of this unit, you will be able to:

  • explain and apply the Pythagorean theorem;
  • calculate the three basic trigonometric functions of a given angle; and
  • calculate the sides of a 45-45-90 or 30-60-90 triangle.

7.1: The Pythagorean Theorem

The Pythagorean theorem allows us to calculate the lengths of the sides of a triangle. We use it frequently to solve for the unknown length of a side of a triangle.

unit 7 homework 6 trigonometry review answer key

Watch this introductory video.

Watch this video on how to use the Pythagorean theorem to solve for the length of an unknown side of a triangle.

unit 7 homework 6 trigonometry review answer key

Then, complete these practice problems and check your answers.

7.2: 45-45-90 Triangles

Now that we know the general form of the Pythagorean theorem and how to use it, we can begin exploring special types of triangles and their properties. The first type of triangle we will study is the 45-45-90 right triangle, which is also known as an isosceles triangle.

Read this article and watch the videos to learn about the special properties of isosceles triangles. Pay attention to the examples of finding the length of missing side and solving for unknowns. The article also shows how we can divide squares into two isosceles triangles and then use our knowledge of these triangles to solve for an unknown.

Then, complete review questions 9–12 and check your answers.

7.3: 30-60-90 Right Triangles

The other type of right triangle are 30-60-90 right triangles. These triangles have the special property that their side lengths are always related by a specific ratio.

Read this article and watch the videos. Pay attention to the 30-60-90 theorem, which gives the ratio needed to solve for an unknown side of this type of triangle.

Then, complete review questions 5–8 and check your answers.

7.4: Sine, Cosine, and Tangent

When performing calculations with triangles, as we have seen, we often are interested in ratios between lengths of the sides of the triangle. The study of trigonometry is the study of these ratios.

Read this page and watch the video. This gives a good overview of the three main trigonometric functions: sine, cosine, and tangent. It may be helpful to write the definitions down for yourself to keep track of them.

Then, take this assessment and check your answers.

7.5: Trigonometric Ratios with a Calculator

Finally, we need to learn to use a calculator to solve for sine, cosine, and tangent.

Read this article and watch the videos to learn how to input trigonometric functions into your calculator.

Then, complete review questions 1–3 and check your work to make sure you are able to use your calculator correctly.

IMAGES

  1. Answer Key Trigonometry Worksheet With Answers

    unit 7 homework 6 trigonometry review answer key

  2. Solved Unit 6 Test Study Guide (Right Triangles &

    unit 7 homework 6 trigonometry review answer key

  3. Unit 4 Congruent Triangles Homework 7 Answer Key

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  4. Answer Key Trigonometry Worksheet With Answers

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  5. Answer Key Special Right Triangles Worksheet Answers

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  6. Unit 8 Test Right Triangles And Trigonometry Answer Key

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VIDEO

  1. 10th Class Math Unit No 7 Exercise No 7.4 Question No 11

  2. Exercise 6.3 / The Triangle And Its Properties /Class-7 Maths NCERT Chapter-6 Solution & Explanation

  3. Unit 3, Step 7, Ex. 3 / ГДЗ. 6 класс. Учебник Rainbow English. 1 часть

  4. Geometry (Gina Wilson) 1.6 (Part 1) Angle Relationships. Identify Types of Angles

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  6. 😉 7th Grade, Unit 6, Lesson 5 "Reasoning About Equations and Tape Diagrams (Part 2)" IM Math

COMMENTS

  1. Unit 8: Right Triangles & Trigonometry Homework 6 ...

    The main trigonometric ratios are presented below. Triangle 1. For angle D you will find: For angle E you will find: Triangle 2. The question gives an angle (62°) and the adjacent side (25) from the angle 62° of the right triangle. Therefore, you can find x from the trigonometric ratio of tan (62°): Triangle 3.

  2. Unit 7 - Chapter 6 Trigonometry Review

    6. Point P ( s t ,− w) lies on the terminal arm of angle θ, in standard position. Determine the exact trig ratios for sin 𝜃, cos 𝜃, and tan 𝜃. 7. Determine the exact value of the following angles: a) sin150o b) tan315o 8. Solve for θ. (Find the values of angle θ, 0 ≤𝜃≤ 360 ) a) cos𝜃= −√3 2 b) sin𝜃= −1 2

  3. Answer Key Chapter 7 - Algebra and Trigonometry | OpenStax

    Answer Key Chapter 7 - Algebra and Trigonometry | OpenStax. Our mission is to improve educational access and learning for everyone. OpenStax is part of Rice University, which is a 501 (c) (3) nonprofit. Give today and help us reach more students. Help.

  4. Trigonometry - 6th Edition - Solutions and Answers | Quizlet

    Exercise 96. At Quizlet, we’re giving you the tools you need to take on any subject without having to carry around solutions manuals or printing out PDFs! Now, with expert-verified solutions from Trigonometry 6th Edition, you’ll learn how to solve your toughest homework problems. Our resource for Trigonometry includes answers to chapter ...

  5. Unit 8 - Right Triangles & Trigonometry

    1 2-7.71 peri me-er- - 103. BYS 32 m Directions: Given the side lengths, determine whether the triangle is acute. right, obtuse, or not a triangle. 13. 10, 24, 26 2 Seq O NotaA Acute 12. 20, 23, 41 + > q 201 14. 16. 6. 13, 20 a NotaA a Acute a Right Obtuse O Acute O Right a Obtuse a NotaA Y' Acute a Right Obtuse a a a a Right Obtuse Not a A ...

  6. Unit 7 Right Triangles and Trigonometry Flashcards | Quizlet

    Right Triangle. c^2 = a^2 + b^2. angle of elevation. angle formed by a horizontal line and a line of sight to a point above the line. angle of depression. angle formed by a horizontal line and a line of sight to a point below the line. Study with Quizlet and memorize flashcards containing terms like Pythagorean Theorem, Converse of the ...

  7. Unit 7: Right Triangles & Trigonometry Flashcards | Quizlet

    9-40-41. Pythagorean Triple. 8-15-17. Pythagorean Triple. 45-45-90 Triangle Theorem. in a 45°-45°-90° triangle, the hypotenuse is √2 times as long as each leg and both legs are congruent. 30-60-90 Triangle Theorem. (Smaller leg is x) Longer leg is x times the square root of 3, hypotenuse is 2x. sine.

  8. Solved Nome: Unit 8: Right Triangles & Trigonometry Homework ...

    Question: Nome: Unit 8: Right Triangles & Trigonometry Homework 6: Trigonometry Review Dale: Per: * This is a 2-page document ** Directions: Give each trigratio as a fraction in simplest form. 29 • sin D = • sin E E . cos De . cos E- 20 • tan D = • tan E- Direc ns: Solve for x. Round to the nearest tenth 26 25 11 12 32 45 29 23 15 10.

  9. Unit 7: Right Triangles & Trigonometry - math24seven

    7B. Define the trigonometric ratios for acute angles in a right triangle and calculate sine, cosine, and tangent ratios when given two side lengths. [G.SRT.7] 7C. Use the characteristics of similar figures to justify the trigonometric ratios. [G.SRT.6] 7D. Use trigonometry to solve for missing sides and angles of right triangles. [G.SRT.6] 7E.

  10. RWM103: Geometry, Topic: Unit 7: Right Triangle Trigonometry ...

    Completing this unit should take you approximately 3 hours. Upon successful completion of this unit, you will be able to: explain and apply the Pythagorean theorem; calculate the three basic trigonometric functions of a given angle; and. calculate the sides of a 45-45-90 or 30-60-90 triangle.