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So, the measure of arc TS is 144. ANSWER: 144 10-6 Secants, Tangents, and Angle Measures Find each measure. Assume that segments that appear to be tangent are tangent. 3. 1. SOLUTION: SOLUTION: ANSWER: 110 ANSWER: 73 4. 2. SOLUTION: SOLUTION: So, the measure of arc TS is 144. ANSWER: 144 3. ANSWER: 31 5. SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero Page 1 So, the measure of arc QTS is 248. So, the measure of arc LP is 150. 10-6ANSWER: Secants, Tangents, and Angle Measures 31 ANSWER: 150 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground? 5. SOLUTION: SOLUTION: Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting tangents to the circle formed by the barrel. One arc has a measure of 165. The other arc is the major arc with the same endpoints, so its measure is 360 – 165 or 195. So, the measure of arc QTS is 248. ANSWER: 248 6. Therefore, the measure of the angle the ramp makes with the ground is 15. SOLUTION: ANSWER: 15 Find each measure. Assume that segments that appear to be tangent are tangent. So, the measure of arc LP is 150. 8. ANSWER: 150 7. STUNTS A ramp is attached to the first of several barrels that have been strapped together for a circus motorcycle stunt as shown. What is the measure of the angle the ramp makes with the ground? SOLUTION: ANSWER: 82 SOLUTION: eSolutions Manual - Powered by Cognero Let x be the measure of the angle the ramp makes with the ground which is formed by two intersecting 9. Page 2 Therefore, the measure of the angle the ramp makes with the ground is 15. 10-6ANSWER: Secants, Tangents, and Angle Measures 15 Find each measure. Assume that segments that appear to be tangent are tangent. ANSWER: 71.5 10. 8. SOLUTION: SOLUTION: ∠JMK and ∠HMJ form a linear pair. ANSWER: 82 9. ANSWER: 102 11. SOLUTION: SOLUTION: Therefore, the measure of arc RQ is 28. ANSWER: 71.5 10. ANSWER: 28 12. SOLUTION: SOLUTION: ∠JMK and ∠HMJ form a linear pair. eSolutions Manual - Powered by Cognero ANSWER: Page 3 Therefore, the measure of arc RQ is 28. So, the measure of arc PM is 144. 10-6ANSWER: Secants, Tangents, and Angle Measures 28 12. ANSWER: 144 14. SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints. SOLUTION: ANSWER: 97 ANSWER: 103 13. 15. SOLUTION: SOLUTION: So, the measure of arc PM is 144. By Theorem 10.13, ANSWER: 144 . 14. ANSWER: 125 SOLUTION: Arc BD and arc BCD are a minor and major arc that share the same endpoints. eSolutions Manual - Powered by Cognero ANSWER: 103 16. SOLUTION: By Theorem 10.13, Page 4 . 10-6ANSWER: Secants, Tangents, and Angle Measures 125 ANSWER: 196 17. SPORTS The multi-sport field shown includes a 16. softball field and a soccer field. If find each measure. , SOLUTION: By Theorem 10.13, Solve for . a. b. . We know that Substitute. SOLUTION: . a. By Theorem 10.13, . Substitute. ANSWER: 196 Simplify. 17. SPORTS The multi-sport field shown includes a softball field and a soccer field. If find each measure. b. By Theorem 10.14, , . Substitute. Simplify. ANSWER: a. 100 b. 20 a. b. SOLUTION: CCSS STRUCTURE Find each measure. a. By Theorem 10.13, . 18. Substitute. Simplify. b. By Theorem 10.14, SOLUTION: . Substitute. eSolutions Manual - Powered by Cognero By Theorem 10.14, . Page 5 Substitute. ANSWER: 100 10-6a.Secants, Tangents, and Angle Measures b. 20 ANSWER: 81 CCSS STRUCTURE Find each measure. 19. 18. SOLUTION: By Theorem 10.14, SOLUTION: . By Theorem 10.14, Substitute. Substitute. Simplify. Simplify. ANSWER: 81 ANSWER: 74 . 20. m(arc JNM) 19. SOLUTION: By Theorem 10.14, . SOLUTION: Substitute. Simplify. So, the measure of arc JNM is 205. ANSWER: 205 ANSWER: 74 eSolutions Manual - Powered by Cognero 20. m(arc JNM) 21. Page 6 So, the measure of arc JNM is 205. 10-6ANSWER: Secants, Tangents, and Angle Measures 205 21. ANSWER: 30 23. SOLUTION: By Theorem 10.14, SOLUTION: By Theorem 10.14, . Substitute. . Substitute. Solve for . Simplify. ANSWER: 22 ANSWER: 185 24. JEWELRY In the circular necklace shown, A and B are tangent points. If x = 260, what is y? 22. SOLUTION: By Theorem 10.14, . Substitute. SOLUTION: Simplify. By Theorem 10.14, . Substitute. Simplify. ANSWER: 30 23. ANSWER: 80 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite. Page 7 SOLUTION: eSolutions Manual - Powered by Cognero By Theorem 10.14, . ANSWER: 9 10-6ANSWER: Secants, Tangents, and Angle Measures 80 25. SPACE A satellite orbits above Earth’s equator. Find x, the measure of the planet’s arc, that is visible to the satellite. 27. SOLUTION: By Theorem 10.14, SOLUTION: The measure of the visible arc is x and the measure of the arc that is not visible is 360 – x. Use Theorem 10.14 to find the value of x. Solve for Therefore, the measure of the planet’s arc that is visible to the satellite is 168. . . ANSWER: 20 ANSWER: 168 ALGEBRA Find the value of x. 28. SOLUTION: 26. By Theorem 10.14, SOLUTION: By Theorem 10.14, Solve for Solve for . . . . ANSWER: 9 eSolutions Manual - Powered by Cognero 27. SOLUTION: ANSWER: 19 29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used? Page 8 ANSWER: a. 145 b. 30 10-6ANSWER: Secants, Tangents, and Angle Measures 19 29. PHOTOGRAPHY A photographer frames a carousel in his camera shot as shown so that the lines of sight form tangents to the carousel. a. If the camera’s viewing angle is , what is the arc measure of the carousel that appears in the shot? b. If you want to capture an arc measure of in the photograph, what viewing angle should be used? CCSS ARGUMENTS For each case of Theorem 10.14, write a two-column proof. 30. Case 1 Given: secants and Prove: SOLUTION: Statements (Reasons) 1. and are secants to the circle. (Given) 2. SOLUTION: a. Let x be the measure of the carousel that appears in the shot. By Theorem 10.14, Solve for . . b. Let x be the measure of the camera’s viewing angle. By Theorem 10.14, Solve for . (The , measure of an inscribed the measure of its intercepted arc.) 3. Theorem) (Exterior 4. (Substitution) 5. (Subtraction Prop.) 6. (Distributive Prop.) ANSWER: Statements (Reasons) 1. and are secants to the circle. (Given) 2. (The , . ANSWER: a. 145 b. 30 CCSS ARGUMENTS For each case of Theorem 10.14, write a two-column proof. 30. Case 1 Given: secants and measure of an inscribed the measure of its intercepted arc.) 3. Theorem) (Exterior 4. (Substitution) 5. (Subtraction Prop.) 6. (Distributive Prop.) 31. Case 2 Given: tangent eSolutions Manual - Powered by Cognero Prove: and secant Page 9 Prove: (Subtraction Prop.) 5. (Distributive Prop.) 10-66.Secants, Tangents, and Angle Measures 31. Case 2 Given: tangent 5. (Subtraction Prop.) 6. (Distributive Prop.) 32. Case 3 Given: tangent and secant and Prove: Prove: SOLUTION: Statements (Reasons) 1. is a tangent to the circle and the circle. (Given) SOLUTION: Statements (Reasons) 1. and are tangents to the circle. (Given) 2. is a secant to ( The , the measure of its meas. of an inscribed intercepted arc.) 3. Theorem) (Exterior 2. of an secant-tangent (The meas. the measure of its intercepted arc.) 3. Theorem) (Exterior 4. (Substitution) 4. (Substitution) 5. (Subtraction Prop.) 5. (Subtraction Prop.) 6. (Distributive Prop.) 6. (Distributive Prop.) ANSWER: Statements (Reasons) 1. is a tangent to the circle and the circle. (Given) 2. is a secant to ( The , meas. of an inscribed the measure of its intercepted arc.) 3. Theorem) (Exterior 4. (Substitution) 5. (Subtraction Prop.) 6. (Distributive Prop.) 32. Case 3 Given: tangent 2. of an secant-tangent (The meas. the measure of its intercepted arc.) 3. Theorem) (Exterior 4. (Substitution) 5. (Subtraction Prop.) 6. (Distributive Prop.) 33. PROOF Write a paragraph proof of Theorem 10.13. and eSolutions Manual - Powered by Cognero Prove: ANSWER: Statements (Reasons) 1. and are tangents to the circle. (Given) Page 10 6. (Distributive Prop.) measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by 10-6 Secants, Tangents, and Angle Measures 33. PROOF Write a paragraph proof of Theorem 10.13. a. Given: is a tangent of . is a secant of . ∠CAE is acute. Prove: m∠CAE = m(arc CA) b. Prove that if ∠CAB is obtuse, m∠CAB = m (arc CDA) SOLUTION: a. Proof: By Theorem 10.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc CF) = 90. By substituting for 90, m∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). Then by subtraction, m∠CAB = m(arc CDA). ANSWER: a. Proof: By Theorem 10.10, . So, ∠FAE is a right ∠ with measure of 90 and arc FCA is a semicircle with measure of 180. Since ∠CAE is acute, C is in the interior of ∠FAE, so by the Angle and Arc Addition Postulates, m∠FAE = m∠FAC + m∠CAE and m(arc FCA) = m(arc FC) + m(arc CA). By substitution, 90 = m∠FAC + m∠CAE and 180 = m(arc FC) + m(arc CA). So, 90 = m(arc FC) + (arc CA) by Division Prop., and m∠FAC + m∠CAE = m(arc FC) + m(arc CA) by substitution. m∠FAC = m(arc FC) since ∠FAC is inscribed, so m(arc FC) + m(arc CA). So, 90 = m(arc FC) + substitution yields m(arc FC) + m∠CAE = m (arc CA) by Division Prop., and m∠FAC + m∠CAE (arc FC) + m(arc CA). By Subtraction Prop., = m(arc FC) + m(arc CA) by substitution. m∠CAE = m(arc CA). m∠FAC = m(arc FC) since ∠FAC is inscribed, so substitution yields m(arc FC) + m∠CAE = m b. Proof: Using the Angle and Arc Addition Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since (arc FC) + m(arc CA). By Subtraction Prop., m∠CAE = m(arc CA). b. Proof: Using the Angle and Arc Addition Postulates, m∠CAB = m∠CAF + m∠FAB and m(ar CDA) = m(arc CF) + m(arc FDA). Since and is a diameter, ∠FAB is a right angle with a measure of 90 and arc FDA is a semicircle with a measure of 180. By substitution, m∠CAB = m∠CAF + 90 and m(arc CDA) = m(arc CF) + 180. Since ∠CAF is inscribed, m∠CAF = m(arc CF) and by substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc ∠CAF is inscribed, m∠CAF = m(arc CF) and by CF) = 90. By substituting for 90, m∠CAB = m(arc substitution, m∠CAB = m(arc CF) + 90. Using the Division and Subtraction Properties on the Arc CF) + m(arc CDA) – m(arc CF). By eSolutions Manual - Powered by Cognero Addition equation yields m(arc CDA) – m(arc subtraction, m∠CAB = m(arc CDA). Page 11 CF) = 90. By substituting for 90, m∠CAB = m(arc 34. OPTICAL ILLUSION The design shown is an Division and Subtraction Properties on the Arc Addition equation yields m(arc CDA) – m(arc 10-6CF) = 90. By substituting for 90, m Secants, Tangents, and Angle Measures ∠CAB = m(arc CF) + m(arc CDA) – m(arc CF). By subtraction, m∠CAB = m(arc CDA). 34. OPTICAL ILLUSION The design shown is an example of optical wallpaper. is a diameter of If m∠A = 26 and SOLUTION: a. Redraw the circle with points A, B, and C in the same place but place point D closer to C each time. Then draw chords and . = 67, what is Refer to the image on page 748. SOLUTION: First, find the measure of arc BD. b. Since is a diameter, arc BDE is a semicircle and has a measure of 180. c. As the measure of ANSWER: 98 35. MULTIPLE REPRESENTATIONS In this problem, you will explore the relationship between Theorems 10.12 and 10.6. a. GEOMETRIC Copy the figure shown. Then draw three successive figures in which the position of point D moves closer to point C, but points A, B, and C remain fixed. gets closer to 0, the measure of x approaches half of becomes an inscribed angle. d. ANSWER: a. b. TABULAR Estimate the measure of for each successive circle, recording the measures of b. and in a table. Then calculate and record the value of x for each circle. c. VERBAL Describe the relationship between and the value of x as approaches zero. What type of angle does ∠AEB become when d. ANALYTICAL Write an algebraic proof to show the relationship between Theorems 10.12 and 10.6 described in part c. eSolutions Manual - Powered by Cognero SOLUTION: a. Redraw the circle with points A, B, and C in the c. As the measure of gets closer to 0, the measure of x approaches half of becomes an inscribed angle. ∠AEB d. 36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle. SOLUTION: Find the difference of the two intercepted arcs and divide by 2. Page 12 ANSWER: Find the difference of the two intercepted arcs and measure of x approaches half of becomes an inscribed angle. ∠AEB 10-6d.Secants, Tangents, and Angle Measures 36. WRITING IN MATH Explain how to find the measure of an angle formed by a secant and a tangent that intersect outside a circle. ANSWER: 15 38. REASONS Isosceles What can you conclude about 37. CHALLENGE The circles below are concentric. What is x? and Explain. SOLUTION: Find the difference of the two intercepted arcs and divide by 2. ANSWER: Find the difference of the two intercepted arcs and divide by 2. is inscribed in SOLUTION: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by theorem 10.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. So, m(arc AB) = m(arc BC). ANSWER: Sample answer: m∠BAC = m∠BCA because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by theorem 10.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. SOLUTION: So, m(arc AB) = m(arc BC). Use the arcs intercepted on the smaller circle to find m∠A. 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain. Use the arcs intercepted on the larger circle and m∠A to find the value of x. SOLUTION: a. ; for all values except when . at G, then b. Because a diameter is involved the intercepted arcs measure (180 – x) and ANSWER: 15 38. REASONS Isosceles is inscribed in What can you conclude about Explain. eSolutions Manual - Powered by Cognero and x degrees. Hence solving leads to the answer. ANSWER: a. ; at G, then b. for all values except when Page 13 . Because a diameter is because the triangle is isosceles. Since ∠BAC and ∠BCA are inscribed angles, by theorem 10.6, m (arc AB) = 2m∠BCA and m(arc BC) = 2m∠BAC. m(arcTangents, AB) = m(arc 10-6So, Secants, andBC). Angle Measures b. Because a diameter is involved the intercepted arcs measure (180 – x) and leads to x degrees. Hence solving the answer. 39. CCSS ARGUMENTS In the figure, is a diameter and is a tangent. a. Describe the range of possible values for m∠G. Explain. b. If m∠G = 34, find the measures of minor arcs HJ and KH. Explain. SOLUTION: a. ; for all values except when . at G, then b. Because a diameter is involved the intercepted arcs measure (180 – x) and leads to x degrees. Hence solving the answer. ANSWER: a. ; at G, then 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer: By Theorem 10.13, So, 50° = Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°. ANSWER: Sample answer: for all values except when . b. Because a diameter is involved the intercepted arcs measure (180 – x) and leads to x degrees. Hence solving By Theorem 10.13, the answer. So, 50 = Therefore, x (minor arc) = 130, 40. OPEN ENDED Draw a circle and two tangents that intersect outside the circle. Use a protractor to measure the angle that is formed. Find the measures of the minor and major arcs formed. Explain your reasoning. SOLUTION: Sample answer: and y (major arc) = 360 – 130 or 230. 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. SOLUTION: By Theorem 10.13, So, 50° = Sample answer: Use Theorem 10.14 to find each minor arc. Therefore, x (minor arc) = 130, and y (major arc) = 360° – 130° or 230°. eSolutions Manual - Powered by Cognero ANSWER: Sample answer: Page 14 By Theorem 10.13, So, 50 = Therefore, x (minor arc) = 130, 10-6 Secants, Tangents, and Angle Measures and y (major arc) = 360 – 130 or 230. 41. WRITING IN MATH A circle is inscribed within If m∠P = 50 and m∠Q = 60, describe how to find the measures of the three minor arcs formed by the points of tangency. ANSWER: Sample answer: Using Theorem 10.14, 60° = 1/2 ((360 – x) – x) or 120°; repeat for 50° to get 130°. The third arc can be found by adding 50° and 60° and subtracting from 360 to get 110° 42. What is the value of x if and SOLUTION: A 23° B 31° C 64° D 128° Sample answer: Use Theorem 10.14 to find each minor arc. SOLUTION: By Theorem 10.14, . Substitute. Simplify. The sum of the angles of a triangle is 180, so m∠R = 180 – (50 + 60) or 70. Therefore, the measures of the three minor arcs are 130, 120, and 110. ANSWER: Sample answer: Using Theorem 10.14, 60° = 1/2 ((360 – x) – x) or 120°; repeat for 50° to get 130°. The third arc can be found by adding 50° and 60° and subtracting from 360 to get 110° 42. What is the value of x if and So, the correct choice is C. ANSWER: C 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5) SOLUTION: A 23° B 31° C 64° D 128° Here, C is the midpoint of Use the midpoint formula, SOLUTION: eSolutions Manual - Powered By Theorem 10.14, by Cognero . , to find the coordinates of C. . Page 15 Substitute. So, the correct choice is J. So, the correct choice is C. 10-6ANSWER: Secants, Tangents, and Angle Measures C 43. ALGEBRA Points A(–4, 8) and B(6, 2) are both on circle C, and is a diameter. What are the coordinates of C? F (2, 10) G (10, –6) H (5, –3) J (1, 5) . ANSWER: 35 45. SAT/ACT If the circumference of the circle below is 16π, what is the total area of the shaded regions? 2 A 64π units B 32π units2 SOLUTION: Here, C is the midpoint of Use the midpoint formula, , Since 2 C 12π units D 8π units2 . 2 , to find the coordinates of D 2π units C. SOLUTION: First, use the circumference to find the radius of the circle. So, the correct choice is J. ANSWER: J The shaded regions of the circle comprise half of the circle, so its area is half the area of the circle. 44. SHORT RESPONSE If m∠AED = 95 and what is m∠BAC? The area of the shaded regions is 32π. SOLUTION: So, the correct choice is B. Since . ANSWER: B We know that vertical angles are congruent. So, . Since , . We know that the sum of the measures of all interior angles of a triangle is 180. Substitute. Since Find x. Assume that segments that appear to be tangent are tangent. 46. , . ANSWER: 35 SOLUTION: By theorem 10.10, . So, triangle. Use the Pythagorean Theorem. is a right Substitute. 45. SAT/ACT If the circumference of the circle below eSolutions Manual - Powered by Cognero is 16π, what is the total area of the shaded regions? Page 16 The area of the shaded regions is 32π. So, the correct choice is B. 10-6ANSWER: Secants, Tangents, and Angle Measures B ANSWER: 8 Find x. Assume that segments that appear to be tangent are tangent. 48. SOLUTION: Use Theorem 10.10 and the Pythagorean Theorem. 46. SOLUTION: By theorem 10.10, . So, triangle. Use the Pythagorean Theorem. is a right Solve for x. Substitute. ANSWER: ANSWER: 3 49. PROOF Write a two-column proof. Given: is a semicircle; Prove: 47. SOLUTION: If two segments from the same exterior point are tangent to a circle, then they are congruent. SOLUTION: Given: is a semicircle. Prove: ANSWER: 8 48. SOLUTION: Use Theorem 10.10 and the Pythagorean Theorem. Solve for x. eSolutions Manual - Powered by Cognero ANSWER: Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.) 7. (Def. of Page 17 intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) .) (All rt. angles are 10-64.Secants, Tangents, and Angle Measures 5. (Reflexive Prop.) 6. (AA Sim.) 7. (Def. of ANSWER: Given: is a semicircle. 4. 5. 6. 7. (All rt. angles are .) (Reflexive Prop.) (AA Sim.) (Def. of 50. REMODELING The diagram at the right shows the floor plan of Trent’s kitchen. Each square on the diagram represents a 3-foot by 3-foot area. While remodeling his kitchen, Trent moved his refrigerator from square A to square B. Describe one possible combination of transformations that could be used to make this move. Prove: Proof: Statements (Reasons) 1. MHT is a semicircle; (Given) 2. is a right angle. (If an inscribed angle intercepts a semicircle, the angle is a right angle.) 3. is a right angle (Def. of ⊥ lines) 4. (All rt. angles are .) 5. (Reflexive Prop.) 6. (AA Sim.) 7. (Def. of 50. REMODELING The diagram at the right shows the floor plan of Trent’s kitchen. Each square on the diagram represents a 3-foot by 3-foot area. While remodeling his kitchen, Trent moved his refrigerator from square A to square B. Describe one possible combination of transformations that could be used to make this move. SOLUTION: The move is a translation 15 feet out from the wall and 21 feet to the left, then a rotation of 90° counterclockwise. ANSWER: The move is a translation 15 feet out from the wall and 21 feet to the left, then a rotation of 90° counterclockwise. SOLUTION: The move is a translation 15 feet out from the wall and 21 feet to the left, then a rotation of 90° counterclockwise. ANSWER: The move is a translation 15 feet out from the wall and 21 feet to the left, then a rotation of 90° counterclockwise. COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 51. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) SOLUTION: Use the Distance Formula to find the length of each side. Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio. If Use a calculator. eSolutions Manual - Powered by Cognero COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an Page 18 ANSWER: ANSWER: The move is a translation 15 feet out from the wall 10-6and 21 feet to the left, then a rotation of 90° Secants, Tangents, and Angle Measures counterclockwise. COORDINATE GEOMETRY Find the measure of each angle to the nearest tenth of a degree by using the Distance Formula and an inverse trigonometric ratio. 51. ∠C in triangle BCD with vertices B(–1, –5), C(–6, – 5), and D(–1, 2) ANSWER: 54.5 52. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side. SOLUTION: Use the Distance Formula to find the length of each side. In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite Write an equation using the sine ratio. Since , triangle BCD is a right triangle. So, CD is the length of the hypotenuse and BC is the length of the leg adjacent to Write an equation using the cosine ratio. If Use a calculator. ANSWER: 54.5 If Use a calculator. ANSWER: 51.3 Solve each equation. 2 53. x + 13x = –36 SOLUTION: 52. ∠X in right triangle XYZ with vertices X(2, 2), Y(2, – 2), and Z(7, –2) SOLUTION: Use the Distance Formula to find the length of each side. Therefore, the solution is –9, –4. ANSWER: –4, –9 2 54. x – 6x = –9 SOLUTION: In right triangle XYZ, ZX is the length of the hypotenuse and YZ is the length of the leg opposite Write an equation using the sine ratio. eSolutions Manual - Powered by Cognero If Therefore, the solution is 3. ANSWER: 3 2 55. 3x + 15x = 0 Page 19 Therefore, the solution is –9, –4. Therefore, the solution is –6. ANSWER: 10-6–4, Secants, Tangents, and Angle Measures –9 ANSWER: –6 2 54. x – 6x = –9 58. SOLUTION: SOLUTION: Therefore, the solution is 3. ANSWER: 3 2 55. 3x + 15x = 0 Therefore, the solution is . ANSWER: SOLUTION: Therefore, the solution is –5, 0. ANSWER: 0, –5 2 56. 28 = x + 3x SOLUTION: Therefore, the solution is –7, 4. ANSWER: –7, 4 2 57. x + 12x + 36 = 0 SOLUTION: Therefore, the solution is –6. ANSWER: –6 58. eSolutions Manual - Powered by Cognero SOLUTION: Page 20