Chapter 9 right triangles and trigonometry answer key

1 CHAPTER 9 Right Triangle Trigonometry Chapter Outline 9.1 THE PYTHAGOREAN THEOREM 9.2 CONVERSE OF THE PYTHAGOREAN THEO...

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C HAPTER

9 Right Triangle Trigonometry

Chapter Outline 9.1

T HE P YTHAGOREAN T HEOREM

9.2

C ONVERSE OF THE P YTHAGOREAN T HEOREM

9.3

U SING S IMILAR R IGHT T RIANGLES

9.4

S PECIAL R IGHT T RIANGLES

9.5

TANGENT, S INE AND C OSINE

9.6

I NVERSE T RIGONOMETRIC R ATIOS

9.7

E XTENSION : L AWS OF S INES AND C OSINES

9.8

C HAPTER 9 R EVIEW

9.9

E XTENSION : R ADIAN M EASURE

9.10

R EFERENCES

This chapter explores right triangles in far more depth than previous chapters. Recall that a right triangle is a triangle with exactly one right angle. In this chapter, we will ﬁrst prove the Pythagorean Theorem and its converse, followed by analyzing the sides of certain types of triangles. Then, we will introduce trigonometry, which starts with the tangent, sine and cosine ratios. Finally, we will extend sine and cosine to any triangle, through the Law of Sines and the Law of Cosines.

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Chapter 9. Right Triangle Trigonometry

9.1 The Pythagorean Theorem

Learning Objectives

• • • •

Prove and use the Pythagorean Theorem. Identify common Pythagorean triples. Use the Pythagorean Theorem to ﬁnd the area of isosceles triangles. Use the Pythagorean Theorem to derive the distance formula on a coordinate grid.

Review Queue

1. Draw a right scalene triangle. 2. Draw an isosceles right triangle. 3. Simplify the radical. √ (a) √50 (b) √27 (c) 272 4. Perform the indicated operations on the following numbers. Simplify all radicals. √ √ (a) 2 √10 +√ 160 (b) 5√ 6 · 4 √18 (c) 8 · 12 2

Know What? All televisions’ dimensions refer to the diagonal of the rectangular viewing area. Therefore, for a 52” TV, 52” is the length of the diagonal. High Deﬁnition Televisions (HDTVs) have sides in the ratio of 16:9. What is the length and width of a 52” HDTV? What is the length and width of an HDTV with a y long diagonal?

The Pythagorean Theorem

The Pythagorean Theorem has already been used in this text, but has not been proved. Recall that the sides of a right triangle are called legs (the sides of the right angle) and the side opposite the right angle is the hypotenuse. For the Pythagorean Theorem, the legs are “a” and “b” and the hypotenuse is “c”. 467

9.1. The Pythagorean Theorem

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Pythagorean Theorem: Given a right triangle with legs of lengths a and b and a hypotenuse of length c, then a2 + b2 = c2 . There are several proofs of the Pythagorean Theorem. One proof will be provided within the text and two others in the review exercises. Investigation 9-1: Proof of the Pythagorean Theorem Tools Needed: pencil, 2 pieces of graph paper, ruler, scissors, colored pencils (optional)

1. On the graph paper, draw a 3 in. square, a 4 in. square, a 5 in square and a right triangle with legs of 3 and 4 inches. 2. Cut out the triangle and square and arrange them like the picture on the right.

3. This theorem relies on area. Recall from a previous math class that the area of a square is length times width. But, because the sides are the same you can rewrite this formula as Asquare = length × width = side × side = side2 . So, the Pythagorean Theorem can be interpreted as (square with side a)2 + (square with side b)2 = (square with side c)2 . In this Investigation, the sides are 3, 4 and 5 inches. What is the area of each square? 4. Now, we know that 9 + 16 = 25, or 32 + 42 = 52 . Cut the smaller squares to ﬁt into the larger square, thus proving the areas are equal.

Another Proof of the Pythagorean Theorem

This proof is “more formal,” meaning that we will use letters, a, b, and c to represent the sides of the right triangle. In this particular proof, we will take four right triangles, with legs a and b and hypotenuse c and make the areas equal. 468

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Chapter 9. Right Triangle Trigonometry

For two animated proofs, go to http://www.mathsisfun.com/pythagoras.html and scroll down to “And You Can Prove the Theorem Yourself.” Using the Pythagorean Theorem

The Pythagorean Theorem can be used to ﬁnd a missing side of any right triangle, to prove that three given lengths can form a right triangle, to ﬁnd Pythagorean Triples, to derive the Distance Formula, and to ﬁnd the area of an isosceles triangle. Here are several examples. Simplify all radicals. Example 1: Do 6, 7, and 8 make the sides of a right triangle? Solution: Plug in the three numbers into the Pythagorean Theorem. The largest length will always be the hypotenuse. 62 + 72 = 36 + 49 = 85 = 82 . Therefore, these lengths do not make up the sides of a right triangle. Example 2: Find the length of the hypotenuse of the triangle below.

Solution: Let’s use the Pythagorean Theorem. Set a and b equal to 8 and 15 and solve for c, the hypotenuse.

82 + 152 = c2 64 + 225 = c2 289 = c2

Take the square root o f both sides.

17 = c 469

9.1. The Pythagorean Theorem

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Using the square root opersation usually results in two answers, in this case either +17 or -17. However, because c represents a length, use only the positive answer. Length is never negative. Example 3: Find the missing side of the right triangle below.

Solution: Here, the hypotenuse and a leg are given. Solve for b.

72 + b2 = 142 49 + b2 = 196 b2 = 147 √ √ √ b = 147 = 7 · 7 · 3 = 7 3 √ Example 4: What is the diagonal of a rectangle with sides 10 and 16 5?

Solution: For any square and rectangle, you can use the Pythagorean Theorem to ﬁnd the length of a diagonal. Plug in the sides to ﬁnd d. √ 2 102 + 16 5 = d 2 100 + 1280 = d 2 1380 = d 2 √ √ d = 1380 = 2 345 Pythagorean Triples

In Example 2, the sides of the triangle were 8, 15, and 17. This combination of numbers is referred to as a Pythagorean triple. Pythagorean Triple: A set of three whole numbers that makes the Pythagorean Theorem true. 470

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Chapter 9. Right Triangle Trigonometry

The most frequently used Pythagorean triple is 3, 4, 5, as in Investigation 10-1. Any multiple of a Pythagorean triple is also considered a triple because it would still be three whole numbers. Therefore, 6, 8, 10 and 9, 12, 15 are also sides of a right triangle. Other Pythagorean triples are: 3, 4, 5

5, 12, 13

7, 24, 25

8, 15, 17

There are inﬁnitely many Pythagorean triples. To see if a set of numbers makes a triple, plug them into the Pythagorean Theorem. Example 5: Is 20, 21, 29 a Pythagorean triple? Solution: If 202 + 212 is equal to 292 , then the set is a triple. 202 + 212 = 400 + 441 = 841 292 = 841 Therefore, 20, 21, and 29 is a Pythagorean triple. Area of an Isosceles Triangle

There are many different applications of the Pythagorean Theorem. One way to use The Pythagorean Theorem is to identify the heights in isosceles triangles so you can calculate the area. The area of a triangle is 12 bh, where b is the base and h is the height (or altitude).

If the base and the sides of an isosceles triangle are given, use the Pythagorean Theorem to calculate the height. Example 6: What is the area of the isosceles triangle?

Solution: First, draw the altitude from the vertex between the congruent sides, which will bisect the base (Isosceles Triangle Theorem). Then, ﬁnd the length of the altitude using the Pythagorean Theorem.

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72 + h2 = 92 49 + h2 = 81 h2 = 32 √ √ h = 32 = 4 2 Now, use h and b in the formula for the area of a triangle.

√ √ 1 1 bh = (14) 4 2 = 28 2 units2 2 2

The Distance Formula

Another application of the Pythagorean Theorem is the Distance Formula. The Distance Formula has already been used in this text; it is proved here.

First, draw the vertical and horizontal lengths to make a right triangle. Then, use the differences to ﬁnd these distances. The Pythagorean Theorem to ﬁnd d in this right triangle.

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Chapter 9. Right Triangle Trigonometry

Distance Formula: The distance A(x1 , y1 ) and B(x2 , y2 ) is d =

(x1 − x2 )2 + (y1 − y2 )2 .

Example 7: Find the distance between (1, 5) and (5, 2). Solution: Make A(1, 5) and B(5, 2). Plug into the distance formula.

d= =

(1 − 5)2 + (5 − 2)2

(−4)2 + (3)2 √ √ = 16 + 9 = 25 = 5 You might recall that the distance formula was presented as d = (x2 − x1 )2 + (y2 − y1 )2 , with the ﬁrst and second points switched. It does not matter which point is ﬁrst. In Example 7, A and B could be switched and the formula would still yield the same answer. (5 − 1)2 + (2 − 5)2 = (4)2 + (−3)2 √ √ = 16 + 9 = 25 = 5

Also, just like the lengths of the sides of a triangle, distances are always positive. Know What? Revisited To ﬁnd the length and width of a 52” HDTV, plug in the ratios and 52 into the Pythagorean Theorem. We know that the sides are going to be a multiple of 16 and 9, which we will call n.

(16n)2 + (9n)2 = 522 256n2 + 81n2 = 2704 337n2 = 2704 n2 = 8.024 n = 2.833

Therefore, the dimensions of the TV are 16(2.833) √ by 9(2.833), or 45.3 by 25.5. If the diagonal is y long, it would √ be n 337 long. The extended ratio is 9 : 16 : 337. Review Questions

Find the length of the missing side. Simplify all radicals. 473

9.1. The Pythagorean Theorem

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7. 8. 9. 10.

6. If the legs of a right triangle are 10 and 24, then the hypotenuse is _____________. If the sides of a rectangle are 12 and 15, then the diagonal is _____________. If the legs of a right triangle are x and y, then the hypotenuse is ____________. If the sides of a square are 9, then the diagonal is _____________.

Determine if the following sets of numbers are Pythagorean Triples. 11. 12. 13. 14. 15. 16.

12, 35, 37 9, 17, 18 10, 15, 21 11, 60, 61 15, 20, 25 18, 73, 75

Find the area of each triangle below. Simplify all radicals.

17. 474

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Chapter 9. Right Triangle Trigonometry

18.

19. Find the distance between each pair of points. 20. 21. 22. 23. 24.

(-1, 6) and (7, 2) (10, -3) and (-12, -6) (1, 3) and (-8, 16) What are the length and width of a 42" HDTV? Round your answer to the nearest thousandth. Standard deﬁnition TVs have a length and width ratio of 4:3. What are the length and width of a 42" Standard deﬁnition TV? Round your answer to the nearest tenth. 25. Challenge An equilateral triangle is an isosceles triangle. If all the sides of an equilateral triangle are s, ﬁnd the area, using the technique learned in this section. Leave your answer in simplest radical form.

26. Find the area of an equilateral triangle with sides of length 8. Pythagorean Theorem Proofs The ﬁrst proof below is similar to the one done earlier in this lesson. Use the picture to answer the following questions.

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27. Find the area of the square with sides (a + b). 28. Find the sum of the areas of the square with sides c and the right triangles with legs a and b. 29. The areas found in the previous two problems should be the same. Set the expressions equal to each other and simplify to get the Pythagorean Theorem. Major General James A. Garﬁeld (former President of the U.S) is credited with deriving this next proof of the Pythagorean Theorem using a trapezoid.

30. Find the area of the trapezoid using the trapezoid area formula: A = 12 (b1 + b2 )h 31. Find the sum of the areas of the three right triangles in the diagram. 32. The areas found in the previous two problems should be the same. Set the expressions equal to each other and simplify to get the Pythagorean Theorem. Review Queue Answers

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(a) (b) (c) (a) (b) (c)

2. √ √ √ 50 = 25 · 2 = 5 2 √ √ √ 9·3 = 3 3√ √27 = √ 272 = √ 16 · 17 =√4 17 √ √ √ 2 √10 +√ 160 =√2 10 + √4 10 = √4 10 √ 5√ 6 · 4 √18 = 5 √6 · 12 2 = 60 12 = 120 3 8 · 12 2 = 12 16 = 12 · 4 = 48

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Chapter 9. Right Triangle Trigonometry

9.2 Converse of the Pythagorean Theorem Learning Objectives

• Understand the converse of the Pythagorean Theorem. • Identify acute and obtuse triangles from side measures.

Review Queue

1. Determine if the following sets of numbers are Pythagorean triples. (a) 14, 48, 50 (b) 9, 40, 41 (c) 12, 43, 44 2. Do the following lengths make a right triangle? How do you know? √ √ 3, 14 (a) 5,√ 8 (b) 6,√ 2 3,√ √ (c) 3 2, 4 2, 5 2 Know What? A friend of yours is designing a building and wants it to be rectangular. One wall is 65 ft. long and the other is 72 ft. long. How can he ensure the walls are going to be perpendicular?

Converse of the Pythagorean Theorem

In the last lesson, you learned about the Pythagorean Theorem and how it can be used. The converse of the Pythagorean Theorem is also true. We touched on this in the last section with Example 1. Pythagorean Theorem Converse: If the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle. With this converse, you can use the Pythagorean Theorem to prove that a triangle is a right triangle, even if you do not know any of the triangle’s angle measurements. Example 1: Determine if the triangles below are right triangles. a) 477

9.2. Converse of the Pythagorean Theorem

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Solution: Check to see if the three lengths satisfy the Pythagorean Theorem. Let the longest sides represent c in the equation. a) a2 + b2 = c2 √ 2 ? 82 + 162 = 8 5 64 + 256 = 64 · 5 320 = 320 ?

The triangle is a right triangle. b) a2 + b2 = c2 ? 222 + 242 = 262 484 + 576 = 676 1060 = 676 The triangle is not a right triangle.

Identifying Acute and Obtuse Triangles

The converse of the Pythagorean Theorem can be extended to determine if a triangle is obtuse or acute. If the sum of the squares of the two smaller sides equals the square of the larger side, then the triangle is right. We can also interpret the outcome if the sum of the squares of the smaller sides does not equal the square of the third. Theorem 9-3: If the sum of the squares of the two shorter sides in a right triangle is greater than the square of the longest side, then the triangle is acute. Theorem 9-4: If the sum of the squares of the two shorter sides in a right triangle is less than the square of the longest side, then the triangle is obtuse. In other words: The sides of a triangle are a, b, and c and c > b and c > a. If a2 + b2 > c2 , then the triangle is acute. If a2 + b2 = c2 , then the triangle is right. If a2 + b2 < c2 , then the triangle is obtuse. Proof of Theorem 9-3 Given: In ABC, a2 + b2 > c2 , where c is the longest side. In LMN, N is a right angle. 478

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Chapter 9. Right Triangle Trigonometry

Prove: ABC is an acute triangle. (all angles are less than 90◦ )

TABLE 9.1: Statement 1. In ABC, a2 + b2 > c2 , and c is the longest side. In LMN, N is a right angle. 2. a2 + b2 = h2 3. c2 < h2 4. c < h 5. C is the largest angle in ABC. 6. m N = 90◦ 7. m C < m N 8. m C < 90◦ 9. C is an acute angle. 10. ABC is an acute triangle.

Reason Given Pythagorean Theorem Transitive PoE Take the square root of both sides The largest angle is opposite the longest side. Deﬁnition of a right angle SSS Inequality Theorem Transitive PoE Deﬁnition of an acute angle If the largest angle is less than 90◦ , then all the angles are less than 90◦ .

The proof of Theorem 9-4 is very similar and is in the review questions. Example 2: Determine if the following triangles are acute, right or obtuse. a)

Solution: Set the shorter sides in each triangle equal to a and b and the longest side equal to c. 479

9.2. Converse of the Pythagorean Theorem

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√ a) 62 + (3 5)2 ? 82 36 + 45 ? 64 81 > 64 The triangle is acute. b) 152 + 142 ? 212 225 + 196 ? 441 421 < 441 The triangle is obtuse. Example 3: Graph A(−4, 1), B(3, 8), and C(9, 6). Determine if ABC is acute, obtuse, or right. Solution: This looks like an obtuse triangle, but the correct conclusion must be justiﬁed. Use the distance formula to ﬁnd the length of each side.

√ √ √ (−4 − 3)2 + (1 − 8)2 = 49 + 49 = 98 = 7 2 √ √ √ BC = (3 − 9)2 + (8 − 6)2 = 36 + 4 = 40 = 2 10 √ √ AC = (−4 − 9)2 + (1 − 6)2 = 169 + 25 = 194 AB =

Now, plug these lengths into the Pythagorean Theorem.

√ 2 √ 2 √ 2 98 + 40 ? 194 98 + 40 ? 194 138 < 194 ABC is an obtuse triangle. Know What? Revisited To make the walls perpendicular, ﬁnd the length of the diagonal. 480

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Chapter 9. Right Triangle Trigonometry

652 + 722 = c2 4225 + 5184 = c2 9409 = c2 97 = c In order to make the building rectangular, both diagonals must be 97 feet. Review Questions

1. The two shorter sides of a triangle are 9 and 12. (a) What would be the length of the third side to make the triangle a right triangle? (b) What is a possible length of the third side to make the triangle acute? (c) What is a possible length of the third side to make the triangle obtuse? 2. The two longer sides of a triangle are 24 and 25. (a) What would be the length of the third side to make the triangle a right triangle? (b) What is a possible length of the third side to make the triangle acute? (c) What is a possible length of the third side to make the triangle obtuse? 3. The lengths of the sides of a triangle are 8x, 15x, and 17x. Determine if the triangle is acute, right, or obtuse. Determine if the following lengths make a right triangle. 4. 15, 20, 25 5. 20, √25, 30√ 6. 8 3, 6, 2 39 Determine if the following triangles are acute, right or obtuse. 7. 8. 9. 10. 11. 12. 13. 14. 15.

7, 8, 9 14, 48, 50 5, 12, 15 13, 84, 85 20, 20, 24 35, 40, 51 39, 80, 89 20, 21, 38 48, 55, 76

Graph each set of points and determine if ABC is acute, right, or obtuse. 16. A(3, −5), B(−5, −8),C(−2, 7) 17. A(5, 3), B(2, −7),C(−1, 5) 18. Writing Explain the two different ways you can show that a triangle in the coordinate plane is a right triangle. The ﬁgure is a rectangular prism. All sides (or faces) are either squares (the front and back) or rectangles (the four around the middle). All sides are perpendicular to each other. 481

9.2. Converse of the Pythagorean Theorem

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19. Find c. 20. Find d.

21. Writing Explain why m A = 90◦ . 22. Fill in the blanks for the proof of Theorem 9-4.

Given: In ABC, a2 + b2 < c2 , where c is the longest side. In LMN, N is a right angle. Prove: ABC is an obtuse triangle. (one angle is greater than 90◦ )

TABLE 9.2: Statement 1. In ABC, a2 + b2 < c2 , and c is the longest side. In LMN, N is a right angle. 2. a2 + b2 = h2 3. c2 > h2 4. 5. C is the largest angle in ABC. 6. m N = 90◦ 7. m C > m N 8. 9. C is an obtuse angle. 10. ABC is an obtuse triangle. 482

Reason

Transitive PoE

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Chapter 9. Right Triangle Trigonometry

Given AB, with A(3, 3) and B(2, −3) determine whether the given point C in problems 23-25 makes an acute, right or obtuse triangle. 23. C(3, −3) 24. C(4, −1) 25. C(5, −2) Given AB, with A(−2, 5) and B(1, −3) ﬁnd at least two possible points, C, such that ABC is 26. 27. 28. 29.

right, with right C. acute, with acute C. obtuse, with obtuse C. Construction (a) (b) (c) (d)

Draw AB, such that AB = 3 in. −→ Draw AD such that BAD < 90◦ . −→ Construct a line through B which is perpendicular to AD, label the intersection C. ABC is a right triangle with right C.

30. Is the triangle you made unique? In other words, could you have multiple different outcomes with the same AB? Why or why not? You may wish to experiment to ﬁnd out. 31. Why do the instructions speciﬁcally require that BAD < 90◦ ? 32. Describe how this construction could be changed so that B is the right angle in the triangle. Review Queue Answers

(a) (b) (c) (a) (b) (c)

Yes Yes No Yes No Yes

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9.3 Using Similar Right Triangles

Learning Objectives

• Identify similar triangles inscribed in a larger triangle. • Evaluate the geometric mean. • Find the length of an altitude or leg using the geometric mean.

Review Queue

1. 2. 3. 4.

If two triangles are right triangles, does that mean they are similar? Explain. If two triangles are isosceles right triangles, does that mean they are similar? Explain. x Solve the ratio: 3x = 27 . If the legs of an isosceles right triangle are 4, ﬁnd the length of the hypotenuse. Draw a picture and simplify the radical.

Know What? In California, the average home price increased 21.3% in 2004 and another 16.0% in 2005. What is the average rate of increase for these two years?

Inscribed Similar Triangles

You may recall that if two objects are similar, corresponding angles are congruent and their sides are proportional in length. Look at a right triangle, with an altitude drawn from the right angle. There are three right triangles in this picture, ADB, CDA, and CAB. Both of the two smaller triangles are similar to the larger triangle because they each share an angle with ADB. That means all three triangles are similar to each other.

Theorem 9-5: If an altitude is drawn from the right angle of any right triangle, then the two triangles formed are similar to the original triangle and all three triangles are similar to each other. The proof of Theorem 9-5 is in the review questions. Example 1: Write the similarity statement for the triangles below. 484

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Chapter 9. Right Triangle Trigonometry

Solution: If m E = 30◦ , then m I = 60◦ and m T RE = 60◦ . m IRT = 30◦ because it is complementary to T RE. Line up the congruent angles in the similarity statement. IRE ∼ IT R ∼ RT E The side proportions can also be used to ﬁnd the length of the altitude. Example 2: Find x.

Solution: First, separate the triangles to ﬁnd the corresponding sides.

Now set up a proportion.

shorter leg in EDG hypotenuse in EDG = shorter leg in DFG hypotenuse in DFG 6 10 = x 8 48 = 10x 4.8 = x Example 3: Find x. 485

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Solution: Set up a proportion. shorter leg in SV T hypotenuse in SV T = shorter leg in RST hypotenuse in RST x 4 = x 20 x2 = 80 √ √ x = 80 = 4 5 Example 4: Find y in RST above. Solution: Use the Pythagorean Theorem. √ 2 y2 + 4 5 = 202 y2 + 80 = 400 y2 = 320 √ √ y = 320 = 8 5 The Geometric Mean

Many students are familiar with arithmetic mean, which divides the sum of n numbers by n. This is commonly used to determine an average test score for a group of students. The geometric mean is a different sort of average, which takes the nth root of the product of n numbers. In this text, two numbers will be compared, so the square root of the product of two numbers will be used. Geometric mean is commonly used with rates of increase or decrease. Geometric Mean: The geometric mean of two positive numbers a and b is the number x, such that √ and x = ab.

a x

x b

or x2 = ab

Example 5: Find the geometric mean of 24 and 36. √ √ √ Solution: x = 24 · 36 = 12 · 2 · 12 · 3 = 12 6 Example 6: Find the geometric mean of 18 and 54. √ √ √ Solution: x = 18 · 54 = 18 · 18 · 3 = 18 3 Notice that in both of these examples, the two numbers were not actually multiplied together, but kept separate. This made it easier to simplify the radical. A practical application of the geometric mean is to ﬁnd the altitude of a right triangle. Example 7: Find x. 486

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Chapter 9. Right Triangle Trigonometry

Solution: Using similar triangles, the following proportion is true: shortest leg of smallest longer leg of smallest = shortest leg of middle longer leg of middle x 9 = x 27 x2 = 243 √ √ x = 243 = 9 3 x In Example 7, 9x = 27 is in the deﬁnition of the geometric mean. So, the altitude is the geometric mean of the two segments that it divides the hypotenuse into.

Theorem 9-6: In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of these two segments. Theorem 9-7: In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

In other words Theorem 9-6: Theorem 9-7:

BC AC BC AB

= =

AC DC AB DB

and

DC AD

AD DB

Both of these theorems are proved using similar triangles. Example 8: Find x and y.

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Solution: Use theorem 9-7 to solve for x and y. 20 x = x 35 2 x = 20 · 35 √ x = 20 · 35 √ x = 10 7

15 y = y 35 2 y = 15 · 35 √ y = 15 · 35 √ y = 5 21

The Pythagorean Theorem could also be used to solve for y, once x has been solved for.

√ 2 10 7 + y2 = 352 700 + y2 = 1225 √ √ y = 525 = 5 21

Either method is acceptable. Know What? Revisited The average rate of increase can be found by using the geometric mean. We use this rather than arithmetic mean because the 2004 average home price was 1.213 times the 2003 average home price, and the 2005 average home price was 1.16 times the 2004 average home price. These are products, not sums, so geometric mean is appropriate.

√ 0.213 · 0.16 = 0.1846

Over the two year period, housing prices increased 18.46%. Review Questions

Use the diagram to answer questions 1-4.

1. 2. 3. 4.

Write the similarity statement for the three triangles in the diagram. If JM = 12 and ML = 9, ﬁnd KM. Find JK. Find KL.

Find the geometric mean between the following two numbers. Simplify all radicals. 5. 16 and 32 488

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Chapter 9. Right Triangle Trigonometry

45 and 35 10 and 14 28 and 42 40 and 100 51 and 8

Find the length of the missing variable(s). Simplify all radicals.

11.

12.

13.

14.

15.

16.

17. 489

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18.

19. 20. Write a proof for Theorem 9-5.

Given: ABD with AC⊥DB and DAB is a right angle. Prove: ABD ∼ CBA ∼ CAD 21. Fill in the blanks for the proof of Theorem 9-7.

Given: ABD with AC⊥DB and DAB is a right angle. AB Prove: BC AB = DB

TABLE 9.3: Statement 1. ABD with AC⊥DB and DAB is a right angle. 2. ABD ∼ CBA ∼ CAD AB 3. BC AB = DB

Reason

22. Last year Poorva’s rent increased by 5% and this year her landlord wanted to raise her rent by 7.5%. What is the average rate at which her landlord has raised her rent over the course of these two years? 23. Mrs. Smith teaches AP Calculus. Between the ﬁrst and second years she taught the course her students’ average score improved by 12%. Between the second and third years, the scores increased by 9%. What is the average rate of improvement in her students’ scores? 24. According to the US Census Bureau, http://www.census.gov/ipc/www/idb/country.php the rate of growth of the US population was 0.8% and in 2009 it was 1.0%. What was the average rate of population growth during that time period? Algebra Connection A geometric sequence is a sequence of numbers in which each successive term is determined by multiplying the previous term by the common ratio. An example is the sequence 1, 3, 9, 27, ... Here each term is 490

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Chapter 9. Right Triangle Trigonometry

multiplied by 3 to get the next term in the sequence. Another way to look at this sequence is to compare the ratios of the consecutive terms. 25. Find the ratio of the 2nd to 1st terms and the ratio of the 3rd to 2nd terms. What do you notice? Is this true for the next set (4th to 3rd terms)? 26. Given the sequence 4, 8, 16,..., if we equate the ratios of the consecutive terms we get: 84 = 16 8 . This means that 8 is the ___________________ of 4 and 16. We can generalize this to say that every term in a geometric sequence is the ___________________ of the previous and subsequent terms. Use what you discovered in problem 26 to ﬁnd the middle term in the following geometric sequences. 27. 28. 29. 30.

5, ____, 20 4, ____, 100 2, ____, 12 We can use what we have learned in this section in another proof of the Pythagorean Theorem. Use the diagram to ﬁll in the blanks in the proof below.

TABLE 9.4: Statement ? and 1. ae = d+e

2. 3. 4. 5. 6.

d b

b ?

a2 = e(d + e) and b2 = d(d + e) a2 + b2 =? ? c = d +e ?

Reason The length of each leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg ? Combine equations from #2. Distributive Property ? Substitution PoE

Review Queue Answers

1. No, another angle besides the right angles must also be congruent. 2. Yes, the three angles in an isosceles right triangle are 45◦ , 45◦ , and 90◦ . Isosceles right triangles will always be similar. x → x2 = 81 → x = ±9 3. 3x = 27 4 2 = h2 √ 4. 42 + √ h = 32 = 4 2 491

9.3. Using Similar Right Triangles

492

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Chapter 9. Right Triangle Trigonometry

9.4 Special Right Triangles

Learning Objectives

• Identify and use the ratios involved with isosceles right triangles. • Identify and use the ratios involved with 30-60-90 triangles.

Review Queue

Find the value of the missing variable(s). Simplify all radicals.

3. √ 4. Do the lengths 6, 6,√and 6 2 make a right triangle? 5. Do the lengths 3, 3 3, and 6 make a right triangle?

Know What? The Great Giza Pyramid is a pyramid with a square base and four isosceles triangles that meet at a point. It is thought that the original height was 146.5 meters and the base edges were 230 meters. First, ﬁnd the length of the edge of the isosceles triangles. Then, determine if the isosceles triangles are also equilateral triangles. Round your answers to the nearest tenth. You can assume that the height of the pyramid is from the center of the square base and is a vertical line. 493

9.4. Special Right Triangles

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Isosceles Right Triangles

There are two types of special right triangles. The ﬁrst special right triangle is an isosceles right triangle. Here, the legs are congruent and, by the Base Angles Theorem, the base angles will also be congruent. Therefore, the angle measures will be 90◦ , 45◦ ,and 45◦ . You will also hear an isosceles right triangle called a 45-45-90 triangle. Because the three angles are always the same, all isosceles right triangles are similar.

Investigation 9-2: Properties of an Isosceles Right Triangle Tools Needed: Pencil, paper, compass, ruler, protractor

1. Construct an isosceles right triangle with 2 in legs. Use the SAS construction that you learned previously.

2. Find the measure of the hypotenuse. What is it? Simplify the radical. 3. Now, let’s say the legs are of length x and the hypotenuse is h. Use the Pythagorean Theorem to ﬁnd the hypotenuse. What is it? How is this similar to your answer in #2? 494

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Chapter 9. Right Triangle Trigonometry

x 2 + x 2 = h2 2x2 = h2 √ x 2=h 45-45-90 Corollary: If a triangle is an isosceles right triangle, then its sides are in the extended ratio x : x : x

√

Step 3 in the above investigation proves the 45-45-90 Triangle Theorem. So, anytime√you have a right triangle with congruent legs or congruent angles, then the sides will always be in the ratio x : x : x 2. The hypotenuse is always √ x 2 because that is the longest length. This is a speciﬁc case of the Pythagorean Theorem, so it will still always work as well. Example 1: Find the length of the missing sides. a)

Solution: Use the x : x : x

√ 2 ratio.

√ a) TV = 6 because it is equal to ST . So, SV = 6 2 . √ √ √ b) AB = 9 2 because it is equal to AC. So, BC = 9 2 · 2 = 9 · 2 = 18. Example 2: Find the length of x. a) 495

9.4. Special Right Triangles

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√ Solution: Again, use the x : x : x 2 ratio, but in these two we are given the hypotenuse. We need to solve for x in the ratio. √ √ a) 12 2 = x 2 12 = x √ b) x 2 = √ 16 √ √ 16 x = √ · √2 = 16 2 2 = 8 2 2 2 In part b, the denominator was rationalized. Whenever there is a radical in the denominator of a fraction, multiply the top and bottom by that radical. This will cancel out the radical from the denominator and reduce the fraction. Some teachers require students to rationalize denominators, while others do not; follow your teacher’s instructions. 30-60-90 Triangles

The second special right triangle is called a 30-60-90 triangle, after the three angles. To construct a 30-60-90 triangle, start with an equilateral triangle. Investigation 9-3: Properties of a 30-60-90 Triangle Tools Needed: Pencil, paper, ruler, compass 1. Construct an equilateral triangle with 2 in sides.

2. Draw or construct the altitude from the top vertex to the base for two congruent triangles. 496

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Chapter 9. Right Triangle Trigonometry

3. Find the measure of the two angles at the top vertex and the length of the shorter leg.

The top angles are each 30◦ and the shorter leg is 1 in because the altitude of an equilateral triangle is also the angle and perpendicular bisector. 4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical. 5. Now, let’s say the shorter leg is length x and the hypotenuse is 2x. Use the Pythagorean Theorem to ﬁnd the longer leg. What is it? How is this similar to your answer in #4?

x2 + b2 = (2x)2 x2 + b2 = 4x2 b2 = 3x2 √ b=x 3 30-60-90 Corollary: If a triangle is a 30-60-90 triangle, then its sides are in the extended ratio x : x

√

3 : 2x.

Step 5 in√the above investigation proves the 30-60-90 Corollary. The shortest leg is always x, the longest leg is always x 3, and the hypotenuse is always 2x. Note the Pythagorean Theorem can always be used as well. Example 3: Find the length of the missing sides. a)

b) 497

9.4. Special Right Triangles

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Solution: In part a, the shortest leg is given and in part b, the hypotenuse is given. √ a) If x = 5, then the longer leg, b = 5 3, and the hypotenuse, c = 2(5) = 10. √ b) Now, 2x = 20, so the shorter leg, f = 10, and the longer leg, g = 10 3. Example 4: Find the value of x and y. a)

Solution: In part a, the longer leg is given and in part b, the hypotenuse is given. √ a) x 3 = √ 12 √ √ 3 12 12 x= √ · √ = 3 3 =4 3 3 3 Then,the hypotenuse would be √ √ y=2 4 3 =8 3 √ b) 2x =√15 6 x = 15 2 6 The,the √ longer leg would√be √ √ 15 18 15 6 45 2 y= = · 3= 2

Example 5: Find the measure of x.

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Chapter 9. Right Triangle Trigonometry

Solution: Think of this trapezoid as a rectangle between a 45-45-90 triangle and a 30-60-90 triangle.

From this picture, x = a + b + c. First, ﬁnd a, which is a leg of an isosceles right triangle. √ √ √ 24 2 24 2 = 12 2 a= √ · √ = 2 2 2 a = d, so we can use this to ﬁnd c, which is the shorter leg of a 30-60-90 triangle. √ √ √ √ 12 2 3 12 6 c= √ · √ = =4 6 3 3 3 √ √ b = 20, so x = 12 2 + 20 + 4 6. Nothing simpliﬁes, so this is how we leave our answer. Know What? Revisited The line that the vertical height is perpendicular to is the diagonal of the square base. This length (blue) is the same as the hypotenuse of an isosceles right triangle because half of a square is an isosceles right √ √ triangle. So, the diagonal is 230 2. Therefore, the base of the right triangle with 146.5 as the leg is half of 230 2 √ or 115 2. Use the Pythagorean Theorem to ﬁnd the edge.

edge =

√ 2 115 2 + 146.52 ≈ 218.889 m

In order for the sides to be equilateral triangles, this length should be 230 meters. It is not, so the triangles are isosceles. Review Questions

1. In an isosceles right triangle, if a leg is x, then the hypotenuse is __________. 499

9.4. Special Right Triangles

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2. In a 30-60-90 triangle, if the shorter leg is x, then the longer leg is __________ and the hypotenuse is ___________. 3. A square has sides of length 15. What is the length of the diagonal? 4. A square’s diagonal is 22. What is the length of each side? √ 5. A rectangle has sides of length 4 and 4 3. What is the length of the diagonal? 6. A baseball diamond is a square with 90 foot sides. What is the distance from home base to second base? (HINT: It’s the length of the diagonal).

For questions 7-18, ﬁnd the lengths of the missing sides.

10.

11.

12. 500

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Chapter 9. Right Triangle Trigonometry

13.

14.

15.

16.

17.

√ √18. √ 19. Do the lengths 8 √2, 8 √6, and 16 √ 2 make a special right triangle? If so, which one? 20. Do the lengths 4 3, 4 6 and 8 3 make a special right triangle? If so, which one? 21. Find the measure of x.

22. Find the measure of y. 501

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23. What is the ratio of the sides of a rectangle if the diagonal divides the rectangle into two 30-60-90 triangles? 24. What is the length of the sides of a square with diagonal 8 in? For questions 25-28, it might be helpful to recall #25 from section 9.1. 25. What is the height of an equilateral triangle with sides of length 3 in? 26. What is the area of an equilateral triangle with sides of length 5 ft? 27. A regular hexagon has sides of length 3 in. What is the area of the hexagon? (Hint: the hexagon is made up a 6 equilateral triangles.) √ 28. The area of an equilateral triangle is 36 3. What is the length of a side? 29. If a road has a grade of 30◦ , this means that its angle of elevation is 30◦ . If you travel 1.5 miles on this road, how much elevation have you gained in feet (5280 ft = 1 mile)? 30. Four isosceles triangles are formed when both diagonals are drawn in a square. If the length of each side in the square is s, what are the lengths of the legs of the isosceles triangles? Review Queue Answers

1. 42 + 42 = x2 32 = x2√ x=4 2 2. 32 + z2 = 62 z2 = 27√ z=3 3 3. x2 + x2 = 102 2x2 = 100 x2 = 50√ x = 5 2

√ 2 3 3 + 92 = y2 108 = y2√ y=6 3

√ 2 = 6 2 → 36 + 36 = 72 4. Yes, √ 2 5. Yes, 32 + 3 3 = 62 → 9 + 27 = 36 62 + 62

502

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Chapter 9. Right Triangle Trigonometry

9.5 Tangent, Sine and Cosine Learning Objectives

• • • •

Use the tangent, sine and cosine ratios in a right triangle. Understand these trigonometric ratios in special right triangles. Use a calculator to ﬁnd sines, cosines, and tangents. Use trigonometric ratios in real-life situations.

Review Queue

1. 2. 3. 4.

The legs of an isosceles right triangle have length 14. What is the hypotenuse? Do the lengths 8, 16, 20 make a right triangle? If not, is the triangle obtuse or acute? In a 30-60-90 triangle, what do the 30, 60, and 90 refer to? Find the measure of the missing lengths.

(a)

(b) Know What? A restaurant needs to build a wheelchair ramp for its customers. The angle of elevation for a ramp is recommended to be 5◦ . If the vertical distance from the sidewalk to the front door is two feet, what is the horizontal distance that the ramp will take up (x)? How long will the ramp be (y)? Round your answers to the nearest thousandth.

What is Trigonometry?

The word trigonometry comes from two words meaning triangle and measure. In this lesson we will deﬁne three trigonometric (or trig) functions. Once we have deﬁned these functions, we will be able to solve problems like the 503

9.5. Tangent, Sine and Cosine

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Know What? above. Trigonometry: The study of the relationships between the sides and angles of triangles. In trigonometry, sides are named in reference to a particular angle. The hypotenuse of a triangle is always the same, but the terms adjacent and opposite depend on which angle you are referencing. A side adjacent to an angle is the leg of the triangle that helps form the angle. A side opposite to an angle is the leg of the triangle that does not help form the angle. Never reference the right angle when referring to trig ratios. a is ad jacent to B. b is ad jacent to A.

a is opposite A. b is opposite B.

c is the hypotenuse.

Sine, Cosine, and Tangent Ratios

The three basic trig ratios are called sine, cosine and tangent. At this point, we will only take the sine, cosine and tangent of acute angles. However, these ratios can be used with obtuse angles as well. Sine Ratio: For an acute angle x in a right triangle, the sin x is equal to the ratio of the side opposite the angle over the hypotenuse of the triangle. Using the triangle above, sin A =

a c

and sin B = bc .

Cosine Ratio: For an acute angle x in a right triangle, the cos x is equal to the ratio of the side adjacent to the angle over the hypotenuse of the triangle. Using the triangle above, cos A =

b c

and cos B = ac .

Tangent Ratio: For an acute angle x, in a right triangle, the tan x is equal to the ratio of the side opposite to the angle over the side adjacent to x. Using the triangle above, tan A =

a b

and tan B = ba .

There are a few important things to note about the way we write these ratios. First, keep in mind that the abbreviations sin x, cos x, and tan x are all functions. Each ratio can be considered a function of the angle (see Chapter 10). Second, be careful when using the abbreviations that you still pronounce the full name of each function. When we write sin x it is still pronounced sine, with a long “i”. When we write cos x, we still say co-sine. And when we write tan x, we still say tangent. An easy way to remember ratios is to use the mnemonic SOH-CAH-TOA.

Example 1: Find the sine, cosine and tangent ratios of A. 504

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Chapter 9. Right Triangle Trigonometry

Solution: First, use the Pythagorean Theorem to ﬁnd the length of the hypotenuse.

52 + 122 = h2 13 = h So, sin A =

12 13 , cos A

=, and tan A =

12 5.

A few important points: • • • •

Always reduce ratios when you can. Use the Pythagorean Theorem to ﬁnd the missing side (if there is one). The tangent ratio can be bigger than 1 (the other two cannot). If two right triangles are similar, then their sine, cosine, and tangent ratios will be the same (because they will reduce to the same ratio). • If there is a radical in the denominator, rationalize the denominator if your teacher has instructed you to do so. Example 2: Find the sine, cosine, and tangent of B.

Solution: Find the length of the missing side.

AC2 + 52 = 152 AC2 = 200 √ AC = 10 2 Therefore, sin B =

√ √ 2 = 2 2 , cos B = 3

10 15

5 15

= 13 , and tan B =

√ 5

2 = 2 √2.

Example 3: Find the sine, cosine and tangent of 30◦ .

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√ Solution: This is a special right triangle, a 30-60-90 triangle. So, if the short leg is 6, then the long leg is 6 3 and the hypotenuse is 12. √ √ √ √ 6 6 = √1 · √3 = 3 3 . sin 30◦ = 12 = 12 , cos 30◦ = 6 12 3 = 2 3 , and tan 30◦ = √ 6 3 3 3 In Example 3, we knew the angle measure of the angle we were taking the sine, cosine and tangent of. This means that the sine, cosine and tangent for an angle are ﬁxed. Sine, Cosine, and Tangent with a Calculator

We now know that the trigonometric ratios are not dependent on the sides, but the ratios. Therefore, there is one ﬁxed value for every angle, from 0◦ to 90◦ . A scientiﬁc or graphing calculator can calculate the values of the sine, cosine and tangent of all of these angles. Depending on the calculator, all have [SIN], [COS], and [TAN] buttons. Use these to ﬁnd the sine, cosine, and tangent of any angle. Example 4: Find the indicated trigonometric value, using your calculator. a) sin 78◦ b) cos 60◦ c) tan 15◦ Solution: Depending on your calculator, enter the degree ﬁrst, and then press the correct trig button or the other way around. For TI-83s and TI-84s, press the trig button ﬁrst, followed by the angle. Also, make sure the mode of the calculator is in DEGREES. a) sin 78◦ = 0.9781 b) cos 60◦ = 0.5 c) tan 15◦ = 0.2679 Finding the Sides of a Triangle using Trig Ratios

One application of the trigonometric ratios is to use them to ﬁnd the missing sides of a right triangle. All that is needed is one angle (other than the right angle) and one side. Example 5: Find the value of each variable. Round the answer to the nearest thousandth.

Solution: The hypotenuse is given, so use sine to ﬁnd b, because it is opposite 22◦ , and cosine to ﬁnd a, because it is adjacent to 22◦ .

b 30 30 · sin 22◦ = b sin 22◦ =

b ≈ 11.823 506

a 30 30 · cos 22◦ = a cos 22◦ =

a ≈ 27.816

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Chapter 9. Right Triangle Trigonometry

Example 6: Find the value of each variable. Round the answer to the nearest thousandth.

Solution: Here, the leg adjacent to 42◦ is given. To ﬁnd c, use cosine and to ﬁnd d, use tangent.

9 c c · cos 42◦ = 9 cos 42◦ =

d 9 9 · tan 42◦ = d tan 42◦ =

9 ≈ 12.111 cos 42◦

d ≈ 8.104

Notice in both of these examples, only use the information that you are given. For example, it is best NOT to use the found value of b to ﬁnd a (in Example 5) because b is an approximation (also sometimes referred to as a calculated value). Use given values to give the most accurate answers.

Angles of Depression and Elevation

Another practical application of the trigonometric functions is to ﬁnd lengths that you cannot measure directly. Very frequently, angles of depression and elevation are used in these types of problems. Angle of Depression: The angle measured from the horizon or horizontal line, down.

Angle of Elevation: The angle measure from the horizon or horizontal line, up. Example 7: An inquisitive math student is standing 25 feet from the base of the Washington Monument. The angle of elevation from her horizontal line of sight is 87.4◦ . If her “eye height” is 5ft, how tall is the monument? 507

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Solution: Find the height of the monument by using the tangent ratio and then adding the eye height of the student. h 25 h = 25 · tan 87.4◦ = 550.543

tan 87.4◦ =

Adding 5 ft, the total height of the Washington Monument is 555.543 ft. According to the National Park Service, the actual height of the monument is 555.320 ft. Know What? Revisited To ﬁnd the horizontal length and the actual length of the ramp, use the tangent and sine.

tan 5◦ = x=

2 x 2 = 22.86 tan 5◦

Review Questions

Use the diagram to ﬁll in the blanks below.

1. 2. 3. 4. 5. 6. 508

tan D = ?? sin F = ?? tan F = ?? cos F = ?? sin D = ?? cos D = ??

sin 5◦ = y=

2 y 2 = 22.947 sin 5◦

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Chapter 9. Right Triangle Trigonometry

From questions 1-6, we can conclude the following. Fill in the blanks. = sin F and sin = cos F 7. cos 8. The sine of an angle is ____________ to the cosine of its ______________. 9. tan D and tan F are ___________ of each other. Use your calculator to ﬁnd the value of each trig function below. Round to four decimal places. 10. 11. 12. 13.

sin 24◦ cos 45◦ tan 88◦ sin 43◦

Find the sine, cosine and tangent of A. Reduce all fractions and radicals.

14.

15.

16. Find the length of the missing sides. Round your answers to the nearest thousandth.

17.

18.

19. 509

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20.

21.

22. 23. Kristin is swimming in the ocean and notices a coral reef below her. The angle of depression is 35◦ and the depth of the ocean at that point is 250 feet. How far away is she from the reef?

24. The Leaning Tower of Pisa currently “leans” at a 4◦ angle and has a vertical height of 55.86 meters. How tall was the tower when it was originally built?

25. The angle of depression from the top of an apartment building to the base of a fountain in a nearby park is 72◦ . If the building is 78 ft tall, how far away is the fountain? 26. William spots a tree directly across the river from where he is standing. He then walks 20 ft upstream and determines that the angle between his previous position and the tree on the other side of the river is 65◦ . How wide is the river? 27. Diego is ﬂying his kite one afternoon and notices that he has let out the entire 120 ft of string. The angle his string makes with the ground is 52◦ . How high is his kite at this time? 510

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Chapter 9. Right Triangle Trigonometry

28. A tree struck by lightning in a storm breaks and falls over to form a triangle with the ground. The tip of the tree makes a 36◦ angle with the ground 25 ft from the base of the tree. What was the original height of the tree to the nearest foot? 29. Upon descent an airplane is 20,000 ft above the ground. The air trafﬁc control tower is 200 ft tall. It is determined that the angle of elevation from the top of the tower to the plane is 15◦ . To the nearest mile, ﬁnd the ground distance from the airplane to the tower. 30. Critical Thinking Why are the sine and cosine ratios always be less than 1? Review Queue Answers

1. 2. 3. 4. 5.

√ The hypotenuse is 14 2. No, 82 + 162 < 202 , the triangle is obtuse. 30◦ , 60◦ , and√90◦ refer to the angle measures in the special right triangle. x = 2,√ y=2 3 √ x = 6 3, y = 18, z = 18 3, w = 36

511

9.6. Inverse Trigonometric Ratios

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9.6 Inverse Trigonometric Ratios

Learning Objectives

• Use the inverse trigonometric ratios to ﬁnd an angle in a right triangle. • Solve a right triangle. • Apply inverse trigonometric ratios to real-life situations and special right triangles.

Review Queue

Find the lengths of the missing sides. Round your answer to the nearest thousandth.

2. 3. Draw an isosceles right triangle with legs of length 3. What is the hypotenuse? 4. Use the triangle from #3, to ﬁnd the sine, cosine, and tangent of 45◦ . 5. Explain why tan 45◦ = 1. Know What? The longest escalator in North America is at the Wheaton Metro Station in Maryland. It is 230 feet long and is 115 ft high. What is the angle of elevation, x◦ , of this escalator?

512

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Chapter 9. Right Triangle Trigonometry

Inverse Trigonometric Ratios

The word inverse is probably familiar. In mathematics, once you learn how to do an operation, you also learn how to “undo” it. For example, addition and subtraction are considered inverse operations. Multiplication and division are also inverse operations. In algebra, inverse operations are used to solve equations and inequalities. When the word inverse is applied to the trigonometric ratios, one can ﬁnd the acute angle measures within a right triangle. Normally, if given an angle and a side of a right triangle, one can ﬁnd the other two sides using sine, cosine or tangent. With the inverse trig ratios, an angle measure can be found if given two sides. Inverse Tangent: If the opposite side and adjacent side of an angle in a right triangle are known, use inverse tangent to ﬁnd the measure of the angle. Inverse tangent is also called arctangent and is labeled tan−1 or arctan. The “-1” indicates inverse. Inverse Sine: If the opposite side of an angle and the hypotenuse in a right triangle are known, use inverse sine to ﬁnd the measure of the angle. Inverse sine is also called arcsine and is labeled sin−1 or arcsin. Inverse Cosine: If the adjacent side of an angle and the hypotenuse in a right triangle are known, use inverse cosine to ﬁnd the measure of the angle. Inverse cosine is also called arccosine and is labeled cos−1 or arccos. Using the triangle below, the inverse trigonometric ratios look like this:

b tan = m B a b sin−1 = m B c a = m B cos−1 c −1

tan−1

= m A b a = m A sin−1 c −1 b cos = m A c

In order to actually ﬁnd the measure of the angles, use a calculator. On most scientiﬁc and graphing calculators, the buttons look like [SIN−1 ], [COS−1 ], and [TAN−1 ]. Typically, a shift or 2nd button must be used to access these functions. For example, on the TI-83 and 84, [2nd ][SIN] is [SIN−1 ]. Again, make sure the mode is in degrees. When ﬁnding the inverse of a trigonometric function, put the word arc in front of it. So, the inverse of a tangent is called the arctangent (or arctan for short). Think of the arctangent as a tool that can be used like any other inverse operation when solving a problem. If tangent gives the ratio of the lengths of the sides opposite and adjacent to an angle, then tangent inverse gives the measure of an angle with a given ratio. Example 1: Use the sides of the triangle and a calculator to ﬁnd the value of A. Round the answer to the nearest thousandth of a degree. 513

9.6. Inverse Trigonometric Ratios

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Solution: In reference to A, the opposite leg and the adjacent leg are given. This means the tangent ratio will be used. 4 −1 4 = m A. Use a calculator. tan A = 20 = , therefore tan 25 5 5 If using a TI-83 or 84, the keystrokes would be: [2nd ][TAN] 45 [ENTER] and the screen looks like:

So, m A = 38.66◦ Example 2: degree.

A is an acute angle in a right triangle. Use a calculator to ﬁnd m A to the nearest thousandth of a

a) sin A = 0.68 b) cos A = 0.85 c) tan A = 0.34 Solution: a) m A = sin−1 0.68 = 42.844◦ b) m A = cos−1 0.85 = 31.788◦ c) m A = tan−1 0.34 = 18.778◦ Solving Triangles

Now that the use of inverse trigonometric ratios to ﬁnd the measure of the acute angles in a right triangle is more clear, the topic can be extended to solve right triangles. To solve a right triangle, ﬁnd all sides and angles in a right triangle, using any method. When solving a right triangle, use sine, cosine or tangent, inverse sine, inverse cosine, or inverse tangent, or the Pythagorean Theorem. Remember when solving right triangles to only use the given values when possible. Example 3: Solve the right triangle.

Solution: To solve this right triangle, ﬁnd AB, m C and m B. Use AC and CB to give the most accurate answers. AB: Use the Pythagorean Theorem. 514

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Chapter 9. Right Triangle Trigonometry

242 + AB2 = 302 576 + AB2 = 900 AB2 = 324 √ AB = 324 = 18 m B: Use the inverse sine ratio.

24 4 sin B = = 30 5 4 sin−1 = 53.13◦ = m B 5 m C: Use the inverse cosine ratio.

24 4 = cosC = 30 5 4 cos−1 = 36.87◦ = m C 5 Example 4: Solve the right triangle.

Solution: To solve this right triangle, ﬁnd AB, BC and m A. AB: Use sine ratio.

25 AB 25 AB = sin 62◦ AB ≈ 28.314

sin 62◦ =

BC: Use tangent ratio.

25 BC 25 BC = tan 62◦ BC ≈ 13.293

tan 62◦ =

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9.6. Inverse Trigonometric Ratios

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m A: Use Triangle Sum Theorem 62◦ + 90◦ + m A = 180◦ m A = 28◦ Example 5: Solve the right triangle.

Solution: Even though there are no angle measures given, we know that the two acute angles are congruent, making them both 45◦ . Therefore, this is a 45-45-90 triangle. Use the trigonometric ratios or the special right triangle ratios (your teacher may have a preference). Trigonometric Ratios 15 BC 15 BC = = 15 tan 45◦

tan 45◦ =

15 AC 15 AC = ≈ 21.213 sin 45◦

sin 45◦ =

45-45-90 Triangle Ratios √ BC = AB = 15, AC = 15 2 ≈ 21.213 Real-Life Situations

Much like the trigonometric ratios, the inverse trig ratios can be used in several real-life situations. Example 6: A 25 foot tall ﬂagpole casts a 42 feet shadow. What is the angle that the sun hits the ﬂagpole?

Solution: First, draw a picture. The angle that the sun hits the ﬂagpole is the acute angle at the top of the triangle, x◦ . From the picture, the inverse tangent ratio is needed. 516

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Chapter 9. Right Triangle Trigonometry

tan x = tan−1

42 25

42 ≈ 59.237◦ = x 25

Example 7: Elise is standing on the top of a 50 foot building and spots her friend, Molly across the street. If Molly is 35 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height is 4.5 feet.

Solution: Because of parallel lines and the alternate interior angle theorem, the angle of depression is equal to the angle at Molly, or x◦ . Use the inverse tangent ratio.

tan

−1

54.5 30

= 61.169◦ = x

Know What? Revisited To ﬁnd the escalator’s angle of elevation, use the inverse sine ratio.

−1

sin

115 230

= 30◦

The angle of elevation is 30◦ .

Review Questions

Use your calculator to ﬁnd m A to the nearest thousandth of a degree.

2. 517

9.6. Inverse Trigonometric Ratios

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Let A be an acute angle in a right triangle. Find m A to the nearest thousandth of a degree.

7. sin A = 0.5684 8. cos A = 0.1234 9. tan A = 2.78

Solving the following right triangles. Find all missing sides and angles.

10.

11. 518

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12.

13.

14.

15. 16. Writing Explain when to use a trigonometric ratio to ﬁnd a side length of a right triangle and when to use the Pythagorean Theorem.

Real-Life Situations Use what you know about right triangles to solve for the missing angle. If needed, draw a picture. Round all answers to the nearest thousandth of a degree.

17. A 75 foot building casts an 82 foot shadow. What is the angle that the sun hits the building? 18. Over 2 miles (horizontal), a road rises 300 feet (vertical). What is the angle of elevation? 19. A boat is sailing and spots a shipwreck 650 feet below the water. A diver jumps from the boat and swims 935 feet to reach the wreck. What is the angle of depression from the boat to the shipwreck? 20. Elizabeth wants to know the angle at which the sun hits a tree in her backyard at 3 pm. She ﬁnds that the length of the tree’s shadow is 24 ft at 3 pm. At the same time of day, her shadow is 6 ft 5 inches. If Elizabeth is 4 ft 8 inches tall, ﬁnd the height of the tree and hence the angle at which the sunlight hits the tree. 21. Alayna is trying to determine the angle at which to aim her sprinkler nozzle to water the top of a 5 ft bush in her yard. Assuming the water takes a straight path and the sprinkler is on the ground 4 ft from the tree, at what angle of inclination should she set it? 22. Science Connection Would the answer to number 20 be the same every day of the year? What factors would inﬂuence this answer? How about the answer to number 21? What factors might inﬂuence the path of the water? 23. Tommy was solving the triangle below and made a mistake. What did he do wrong? 519

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tan−1

21 28

≈ 36.87◦

24. Tommy then continued the problem and set up the equation: cos 36.87◦ = 21 h . By solving this equation he found that the hypotenuse was 26.25 units. Did he use the correct trigonometric ratio here? Is his answer correct? Why or why not? 25. How could Tommy have found the hypotenuse in the triangle another way and avoided making his mistake? Examining Patterns Below is a table that shows the sine, cosine, and tangent values for eight different angle measures. Answer the following questions.

TABLE 9.5: Sine Cosine Tangent

10◦ 0.1736 0.9848 0.1763

20◦ 0.3420 0.9397 0.3640

30◦ 0.5 0.8660 0.5774

40◦ 0.6428 0.7660 0.8391

50◦ 0.7660 0.6428 1.1918

60◦ 0.8660 0.5 1.7321

70◦ 0.9397 0.3420 2.7475

80◦ 0.9848 0.1736 5.6713

What value is equal to sin 40◦ ? What value is equal to cos 70◦ ? Describe what happens to the sine values as the angle measures increase. Describe what happens to the cosine values as the angle measures increase. What two numbers are the sine and cosine values between? Find tan 85◦ , tan 89◦ , and tan 89.5◦ using your calculator. Now, describe what happens to the tangent values as the angle measures increase. 32. Explain why all of the sine and cosine values are less than one. (hint: think about the sides in the triangle and the relationships between their lengths) 26. 27. 28. 29. 30. 31.

Review Queue Answers

1. sin 36◦ = 7y cos 36◦ = 7x y = 4.114 x = 5.663 ◦= y tan 12.7 2. cos 12.7◦ = 40 x 40 x = 41.003 y = 9.014

4. 520

sin 45◦

3 √

√ = 22

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Chapter 9. Right Triangle Trigonometry

√ 3 = 22 cos 45◦ = √ 3 2 tan 45◦ = 33 = 1 5. The tangent of 45◦ equals one because it is the ratio of the opposite side over the adjacent side. In an isosceles right triangle, or 45-45-90 triangle, the opposite and adjacent sides are the same, making the ratio always 1.

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9.7 Extension: Laws of Sines and Cosines Learning Objectives

• Identify and use the Law of Sines and Cosines. In this chapter, we have only applied the trigonometric ratios to right triangles. However, you can extend what we know about these ratios and derive the Law of Sines and the Law of Cosines. Both of these laws can be used with any type of triangle to ﬁnd any angle or side within it. That means we can ﬁnd the sine, cosine and tangent of angle that are greater than 90◦ , such as the obtuse angle in an obtuse triangle. Law of Sines

Law of Sines: If ABC has sides of length, a, b, and c, then

sin A a

sin B b

sinC c .

Looking at a triangle, the lengths a, b, and c are opposite the angles of the same letter.

The proof will be saved for a later course. Example 1: Solve the triangle using the Law of Sines. Round decimal answers to the nearest thousandth.

Solution: First, to ﬁnd m A, use the Triangle Sum Theorem.

m A + 85◦ + 38◦ = 180◦ m A = 57◦ Now, use the Law of Sines to set up ratios for a and b. 522

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Chapter 9. Right Triangle Trigonometry

sin 57◦ sin 85◦ sin 38◦ = = a b 12 sin 57◦ sin 38◦ = a 12 ◦ a · sin 38 = 12 · sin 57◦ 12 · sin 57◦ a= ≈ 16.347 sin 38◦

sin 85◦ sin 38◦ = b 12 ◦ b · sin 38 = 12 · sin 85◦ 12 · sin 85◦ b= ≈ 19.417 sin 38◦

Example 2: Solve the triangle using the Law of Sines. Round decimal answers to the nearest thousandth.

Solution: Set up the ratio for B using Law of Sines. sin 95◦ sin B = 27 16 27 · sin B = 16 · sin 95◦ 16 · sin 95◦ sin B = → sin−1 27

16 · sin 95◦ 27

= 36.181◦

To ﬁnd m C use the Triangle Sum Theorem. m C + 95◦ + 36.181◦ = 180◦ → m C = 48.819◦ To ﬁnd c, use the Law of Sines again.

sin 95◦ 27

sin 48.819◦ c

c · sin 95◦ = 27 · sin 48.819◦ 27 · sin 48.819◦ c= ≈ 20.399 sin 95◦ Law of Cosines

Law of Cosines: If ABC has sides of length a, b, and c, then a2 = b2 + c2 − 2bc cos A

b2 = a2 + c2 − 2ac cos B c2 = a2 + b2 − 2ab cosC Even though there are three formulas, they are all very similar. First, notice that whatever angle is in the cosine, the opposite side is on the other side of the equal sign. Example 3: Solve the triangle using Law of Cosines. Round answers to the nearest thousandth. 523

9.7. Extension: Laws of Sines and Cosines

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Solution: Use the second equation to solve for B. b2 = 262 + 182 − 2(26)(18) cos 26◦ b2 = 1000 − 936 cos 26◦ b2 = 158.7288 b ≈ 12.599 To ﬁnd m A or m C, either the Law of Sines or Law of Cosines can be used. The Law of Sines is used here. sin A sin 26◦ = 12.599 18

sin−1

18·sin 26◦ 12.599

12.599 · sin A = 18 · sin 26◦ 18 · sin 26◦ sin A = 12.599

≈ 38.778◦ To ﬁnd m C, use the Triangle Sum Theorem. 26◦ + 38.778◦ + m C = 180◦ m C = 115.222◦

Unlike the previous sections in this chapter, with the Laws of Sines and Cosines, calculated values are needed in addition to given values. Keep in mind to always wait until the very last step to put anything into a calculator, ensuring the most accurate answer. Example 4: Solve the triangle. Round answers to the nearest thousandth.

Solution: When given only the sides, use the Law of Cosines to ﬁnd one angle and then use the Law of Sines to ﬁnd another. 152 = 222 + 282 − 2(22)(28) cos A 225 = 1268 − 1232 cos A −1043 = −1232 cos A −1043 = cos A → cos−1 −1232 524

1043 1232

≈ 32.157◦

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Chapter 9. Right Triangle Trigonometry

Now that an angle and its opposite side are known, use the Law of Sines. sin 32.157◦ sin B = 15 22

sin−1

22·sin 32.157◦ 15

15 · sin B = 22 · sin 32.157◦ 22 · sin 32.157◦ sin B = 15

≈ 51.318◦ To ﬁnd m C, use the Triangle Sum Theorem. 32.157◦ + 51.318◦ + m C = 180◦ m C = 96.525◦

To Summarize

Use Law of Sines when given: • An angle and its opposite side. • Any two angles and one side. • Two sides and the non-included angle. Use Law of Cosines when given: • Two sides and the included angle. • All three sides. Review Questions

Use the Law of Sines or Cosines to solve ABC. If you are not given a picture, draw one. Round all decimal answers to the nearest thousandth.

3. 525

9.7. Extension: Laws of Sines and Cosines

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10. m A = 74◦ , m B = 11◦ , BC = 16 11. mA = 64◦ , AB = 29, AC = 34 12. m C = 133◦ , m B = 25◦ , AB = 48

Use the Law of Sines to solve ABC below. 13. m A = 20◦ , AB = 12, BC = 5 Recall that when we learned how to prove that triangles were congruent we determined that SSA (two sides and an angle not included) did not determine a unique triangle. When we are using the Law of Sines to solve a triangle and we are given two sides and the angle not included, we may have two possible triangles. Problem 14 illustrates this. 14. Let’s say we have ABC as we did in problem 13. In problem 13 you were given two sides and the not included angle. This time, you have two angles and the side between them (ASA). Solve the triangle given that m A = 20◦ , m C = 125◦ , AC = 8.4 526

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Chapter 9. Right Triangle Trigonometry

15. Does the triangle that you found in problem 14 meet the requirements of the given information in problem 13? How are the two different m C related? Draw the two possible triangles overlapping to visualize this relationship. It is beyond the scope of this text to determine when there will be two possible triangles, but the concept of the possibility is something worth noting at this time.

527

9.8. Chapter 9 Review

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9.8 Chapter 9 Review Keywords Theorems • • • • • • • • • • • • • • • • • • • • • • • • •

Pythagorean Theorem Pythagorean Triple Distance Formula Pythagorean Theorem Converse Theorem 9-3 Theorem 9-4 Theorem 9-5 Geometric Mean Theorem 9-6 Theorem 9-7 45-45-90 Corollary 30-60-90 Corollary Trigonometry Adjacent (Leg) Opposite (Leg) Sine Ratio Cosine Ratio Tangent Ratio Angle of Depression Angle of Elevation Inverse Tangent Inverse Sine Inverse Cosine Law of Sines Law of Cosines

Review Questions

Solve the following right triangles using the Pythagorean Theorem, the trigonometric ratios, and the inverse trigonometric ratios. When possible, simplify the radical. If not, round all decimal answers to the nearest thousandth.

2. 528

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Chapter 9. Right Triangle Trigonometry

9. Determine if the following lengths make an acute, right, or obtuse triangle. If they make a right triangle, determine if the lengths are a Pythagorean triple. 10. 11. 12. 13.

11, 12, 13 16, 30, 34 20,√25, 42 √ 10 6, 30, 10 15 529

9.8. Chapter 9 Review

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14. 22, 25, 31 15. 47, 27, 35 Find the value of x.

16.

17.

18. 19. The angle of elevation from the base of a mountain to its peak is 76◦ . If its height is 2500 feet, what is the distance a person would climb to reach the top? Round your answer to the nearest thousandth. 20. Taylor is taking an aerial tour of San Francisco in a helicopter. He spots ATT Park (baseball stadium) at a horizontal distance of 850 feet and down (vertical) 475 feet. What is the angle of depression from the helicopter to the park? Round your answer to the nearest thousandth. Use the Law of Sines and Cosines to solve the following triangles. Round your answers to the nearest thousandth.

21.

22. Texas Instruments Resources

In the CK-12 Texas Instruments Geometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/ﬂexr/chapter/9693 .

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Chapter 9. Right Triangle Trigonometry

9.9 Extension: Radian Measure Learning Objectives

A student will be able to: • Deﬁne radian measure. • Convert angle measure from degrees to radians and from radians to dgrees. • Calculate the values of the 3 trigonometric functions for special angles in terms of radians or degrees. Introduction

In this lesson students will be introduced to the radian as a common unit of angle measure in trigonometry. It is important that they become proﬁcient converting back and forth between degrees and radians. Eventually, much like learning a foreign language, students will become comfortable with radian measure when they can learn to “think” in radians instead of always converting from degree measure. Finally, students will review the calculations of the basic trigonometry functions of angles based on 30, 45, and 60 degree rotations. Understanding Radian Measure

Many units of measure come from seemingly arbitrary and archaic roots. Some even change over time. The meter, for example, was originally intended to be based on the circumference of the earth and now has an amazingly complicated scientiﬁc deﬁnition! See the resources for further reading. We typically use degrees to measure angles. Exactly what is a degree? A degree is 1/360th of a complete rotation around a circle. Radians are alternate units used to measure angles in trigonometry. Just as it sounds, a radian is based on the radius of a circle. One radian is the angle created by bending the radius length around the arc of a circle. Because a radian is based on an actual part of the circle rather than an arbitrary division, it is a much more natural unit of angle measure for upper level mathematics and will be especially useful when you move on to study calculus.

531

9.9. Extension: Radian Measure

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What if we were to rotate all the way around the circle? Continuing to add radius lengths, we ﬁnd that it takes a little more than 6 of them to complete the rotation.

But the arc length of a complete rotation is really the circumference! The circumference is equal to the 2π times the length of the radius. 2π is approximately 6.28, so the circumference is a little more than 6 radius lengths. Or, in terms of radian measure, a complete rotation (360 degrees) is 2π radians.

360 degrees = 2π radians With this as our starting point, we can ﬁnd the radian measure of other angles . Half of a rotation, or 180 degrees, must therefore be π radians, and 90 degrees must be π2 . Complete the table below:

TABLE 9.6: Angle in Degrees 90 45 30 60 75 Because 45 is half of 90, half of

Angle in Radians π 2

π 2

is π4 . 30 is one-third of a right angle, so multiplying gives: π 1 π × = 2 3 6

and because 60 is twice as large as 30:

2× Here is the completed table:

532

π 2π π = = 6 6 3

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Chapter 9. Right Triangle Trigonometry

TABLE 9.7: Angle in Degrees 90 45 30 60 75

Angle in Radians π 2 π 4 π 6 π 3 5π 12

The last value was found by adding the radian measures of 30 and 45: π π 3×π 2×π + = + 4 6 3×4 2×6 3π 2π 5π + = 12 12 12 There is a formula to help you convert between radians and degrees that you may already have discovered and we will discuss shortly, however, most angles that you will commonly use can be found easily from the values in this table, so learning them based on the circumference should help increase your comfort level with radians greatly. For example, most students ﬁnd it easy to remember 30 and 60. 30 is π6 and 60 is π3 . If you know these angles, you can ﬁnd any of the special angles that have reference angles of 30 and 60 because they will all have the same denominators. The same is true of multiples of π4 (45 degrees) and π2 (90 degrees). ”Count”ing in Radians

Do you remember as a child watching the Count on Sesame Street? He would count objects like apples, “one apple, two apples, three apples. . .” and then laugh ﬁendishly as lightning and thunder erupted around him. Well, to be successful with radian measure, you need to learn to count all over again using radians instead of apples. Let’s start counting right angles, which are really π2 radians. “ one π over 2, two π over 2 (really just π), three π over 2 (a ha, ha, ha, ha!!!), four π over 2 (which is really 2π)”

FIGURE 9.1 http%3A//www.ck12.org/rotations%20expressed%20in%20radian%20measure.

You just covered all the angles that are multiples of 90 degrees in one rotation. Here is the drawing for 45−degree angles: Notice that the additional angles in the drawing all have reference angles of 45 degrees and their radian measures are all multiples of π4 . Complete the following radian measures by counting in multiples of π3 and π6 : Notice that all of the angles with 60−degree reference angles are multiples of π3 , and all of those with 30−degree reference angles are multiples of π6 . If you can learn to count in these terms, rather than constantly having to convert back to degrees, it will help you to be effective dealing with most radian measures that you will encounter. 533

9.9. Extension: Radian Measure

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FIGURE 9.2 http%3A//www.ck12.org/degree%20rotations

FIGURE 9.3 http%3A//www.ck12.org/degree%20reference%20angles

FIGURE 9.4 http%3A//www.ck12.org/degree%20reference%20angles

FIGURE 9.5

http%3A//www.ck12.org/degree%20reference%20angle%20radian%20measure%20through%

For other examples there is a formula. Remember that:

π radians = 180 degrees If you divide both sides of this equality by 180 you will uncover the formula for easy conversion: 534

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Chapter 9. Right Triangle Trigonometry

FIGURE 9.6

http%3A//www.ck12.org/degree%20reference%20angle%20radian%20measure%20through%

π radians = 1 degree 180 so

radians ×

π = degrees 180

If we have a degree measure and wish to convert it to radians, then manipulating the equation above gives:

degrees ×

180 = radians π

Example 1 Convert

11π 3

to degree measure

Recognize that this angle is a multiple of

π 3

(or 60 degrees), so there are 11,

π ◦ 3 s in this angle, 3 ×11 = 60×11 = 660 .

Here is what it would look like using the formula:

180 = degrees π 60 π 180 11Z = 11 × 60 = 660◦ × 3 π Z

radians ×

Example 2 Convert −120◦ to radian measure.Leave the answer in terms of π. Using the formula: π = radians 180 π −120π −120 × = 180 180

degrees ×

and reducing to lowest terms gives: 535

9.9. Extension: Radian Measure

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− However, 120 is 2 × 60. Since 60◦ is

π 3

2π 3

radians, then 120 is 2,

3 s,

2π 3 .

Make it negative , − 2π 3 .

Example 3 Express

11π 12

radians in degree measure. radians ×

180 = degrees π

π Note: Sometimes students have trouble remembering if it is 180 π or 180 . It might be helpful to remember that radian measure is almost always expressed in terms of π. If converting from radians to degrees, the π will likely cancel out when multiplying, so it must be in the denominator.

11π 180 × = 12 π 15 π 180 11Z = 11 × 15 = 165◦ × 12 π Z Radians, Degrees, and a Calculator

Most scientiﬁc and graphing calculators have a [MODE] setting that will allow you to either convert between the two, or to ﬁnd approximations for trig functions using either measure. It is important that if you are using your calculator to estimate a trig function that you know which mode you are using. Look at the following screen:

If you entered this expecting to ﬁnd the sine of 30 degrees you would realize based on the last chapter that something is wrong because it should be 12 . In fact, as you may have suspected, the calculator is interpreting this as 30 radians. In this case, changing the mode to degrees and recalculating we give the expected result.

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Chapter 9. Right Triangle Trigonometry

Scientiﬁc calculators will usually have a 3−letter display that shows either DEG or RAD to tell you which mode you are in. Always check before calculating a trig ratio!! Example 4 Find the tangent of

3π 4

nd First of all, note that no claculator is needed! You should know this one! 3π 4 is a 2 quadrant angle with a reference π π angle of 4 (45 degrees). The tangent of 4 is 1, and because tangent is negative in quadrant II, the answer is −1. To verify this on your calculator, make sure the mode is set to Radians, and evaluate the tan 3π 4 .

Example 5 Find the value of cos

11π 6

to four decimal places.

π Again, ﬁnd the exact value based on your previous work. 11π and the sign 6 has a reference angle of 6 (30 degrees) √ √ of π is 3 . Because 11π is in the 4th quadrant, the cosine is positive and so the exact answer is 3 . Using the 6

calculator gives:

Which, when rounded, is 0.8660. Verify that it is indeed a very good approximation of the exact answer using the calculator as well.

Example 6 Convert 1 radian to degree measure. Many students get so used to using π in radian measure that they incorrectly think that 1 radian means 1π radians. While it is more convenient and common to express radian measure in terms of π, don’t lose sight of the fact that π radians is actually a number! It speciﬁes an angle created by a rotation of approximately 3.14 radius lengths. So 1 radian is a rotation created by an arc that is only a single radius in length. Look back at Figure 1.1. What would you estimate the degree measure of this angle to be? It is certainly acute and appears similar to a 60◦ angle. To ﬁnd a closer approximation, use the formula and a calculator. 537

9.9. Extension: Radian Measure

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radians ×

180 = degrees π

So 1 radian would be 180 π degrees. Using any scientiﬁc or graphing calculator will give a reasonable approximation for this degree measure, which is approximately 57.296◦ .

Example 7 Find the radian measure of an acute angle θ with a sin θ = 0.7071. First of all, it is important to understand that a calculator will most likely not give you radian measure in terms of π, but a decimal approximation instead. In this case, use the inverse sine function.

√ This answer may not look at all familiar, but 0.7071 may sound familiar. It is an approximation of 2 2 . So this is really a 45◦ angle. Sure enough, evaluating π4 will show that the calculator is giving its best approximation of the radian measure.

note that these are not exactly the same. Remember that 0.7071 is only an approximation of π4 , so there is already some rounding error. Lesson Summary

Angles can be measure in degrees or radians. A radian is the angle deﬁned by an arc length equal to the radius length bent around the circle. One complete rotation around a circle, or 360◦ is equal to 2π radians. To convert from degrees to radians, use the following formula:

degrees × 538

π = radians 180

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Chapter 9. Right Triangle Trigonometry

To convert from radians to degrees the formula becomes:

radians ×

180 = degrees π

Much like learning a foreign language where you have to memorize vocabulary to be successful, it will be very helpful to understand and be able to communicate in radian measure if you become familiar with the radian measures of ◦ = 2π and special angles 30◦ = π , 45◦ = π , 60◦ = π , 360 the quadrant angles 90◦ = π2 , 180◦ = π, 270◦ = 3π 2 6 4 3 Further Reading

• http://www.mel.nist.gov/div821/museum/timeline.htm http://www.mel.nist.gov/div821/museum/timeline.htm • http://en.wikipedia.org/wiki/Degree_(angle) http://en.wikipedia.org/wiki/Degree_(angle) • http://www.joyofpi.com/ http://www.joyofpi.com/ Review Questions

1. Figure 9-7 is a sign for a store that sells cheese.

FIGURE 9.7

(a) Estimate the degree measure of the angle of the circle that is missing. (b) Convert that measure to radians. (c) What is the radian measure of the part of the cheese that remains? 2. Convert the following degree measures to radians. Give exact answers in terms of π, not decimal approximations. (a) 240◦ 539

9.9. Extension: Radian Measure (b) (c) (d) (e) (f) (g) (h) (i) (j)

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270◦ 315◦ −210◦ 120◦ 15◦ −450◦ 72◦ 720◦ 330◦

3. Convert the following radian measures to degrees: (a) π2 (b) 11π 5 (c) 2π 3 (d) 5π (e) 7π 2 (f) 3π 10 (g) 5π 12 (h) − 13π 6 (i) π8 (j) 4π 15 4. The drawing shows all the quadrant angles as well as those with reference angles of 30◦ , 45◦ , and 60◦ . On the inner circle, label all angles with their radian measure in terms of πand on the outer circle, label all the angles with their degree measure.

5. Using a calculator, ﬁnd the approximate degree measure (to the nearest tenth) of each angle expressed in radians. (a) 6π 7 (b) 1 radian (c) 3 radian (d) 20π 11 6. Gina wanted to calculate the cosine of 210and got the following answer on her calculator: 540

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Chapter 9. Right Triangle Trigonometry

(a) Write the correct answer. (b) Explain what she did wrong.

Fortunately, Kylie saw her answer and told her that it was obviously incorrect. 7. Complete the following chart. Write your answers in simplest radical form.

Sin(x)

Cos(x)

Tan(x)

5π/4 11π/6 2π/3 π/2 7π/2

Review Answers

(a) (b) (c) (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

Answer may vary, but 120◦ seems reasonable. Based on the answer in part a., the ration masure would be Again, based on part a., 4π 3

2π 3

4π 3 3π 2 7π 4

− 7π 6 2π 3 π 12

− 5π 2 π 5

4π 11π 6 90◦

396◦ 120◦ 540◦ 630◦ 54◦ 75◦ −210◦ 1440◦ 48◦ 541

9.9. Extension: Radian Measure

(a) (b) (c) (d) (a) (b)

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154.3◦

57.3 171.9 327.3 The correct answer is − 12 Her calculator was is the wrong mode and she calculated the sine of 210 radians.

7. x

π/2

Sin(x) √ 2 − 2 1 − √2 3 2 1

Cos(x) √ 2 − 2 √ 3 2 1 − 2 0

undeﬁned

7π/2

−1

undeﬁned

5π/4 11π/6 2π/3

542

Tan(x) 1

√

3 3 √ − 3 −

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Chapter 9. Right Triangle Trigonometry

9.10 References 1. 2. 3. 4. 5. 6. 7.

. . . . . . .

Trig01-03. Trig01-04. Trig01-05. Trig01-06. Trig01-07. Trig01-08. Trig01-21.

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Chapter 9 right triangles and trigonometry answer key

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