1. AoPS Chapter 2 Angles

2. When 2 rays share an origin, they form an angle. Vertex: O Sides: OA, OB, OC Angle AOB, < BOA Angle BOC, < COB Angle AOC, < COA

3. Problem 2.3 Given that < AOC = 60° and < AOB = 20°, find < BOC.

4. Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle?

5. Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle? What is the measure of a central angle that cuts off ⅓ of a circle?

6. Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle? What is the measure of a central angle that cuts off ⅓ of a circle? What is the measure of a central angle that cuts off ½ of a circle?

7. Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle? What is the measure of a central angle that cuts off ⅓ of a circle? What is the measure of a central angle that cuts off ½ of a circle? What is the measure of a central angle that cuts off ⅛ of a circle?

8. Right Angles An angle that measures exactly 90°. We usually mark right angles with a little box as shown. Two lines, rays, or segments that form a right angle are said to be perpendicular. JK KL J KL

9. Adjacent Angles We call angles that share a side adjacent angles. Ray OB is the common side and O is the common vertex.

10. Straight & Vertical Angles A straight angle measures 180 °.

11. Straight & Vertical Angles A straight angle measures 180 °. Vertical angles are formed by the intersection of 2 lines.

12. Straight & Vertical Angles Vertical angles are formed by the intersection of 2 lines. They are opposite angles and are always congruent. What are the measures of angles 1, 2, and 3?

13. Supplementary Angles

14. Linear Pair

15. Complementary Angles

16. Problem 2.7 If the measure of < 2 = 55°, what is the measure of < 1, <3, <4?

17. Parallel Lines Two lines on the same plane that never intersect are parallel. XY || m m

18. Corresponding Angles When lines are parallel, many special relation- ships exist. Corresponding angles are congruent.

19. Alternate Interior Angles When lines are parallel, many special relation- ships exist. Alternate interior angles are congruent.

20. Alternate Exterior Angles When lines are parallel, many special relation- ships exist. Alternate exterior angles are congruent.

21. Same-side Interior Angles When lines are parallel, many special relation- ships exist. Same-side interior angles are supplemen- tary.

22. Angles in a Triangle The sum of the measures of a triangle is 180°.

23. Problem 2.18 One angle in a triangle is twice another angle, and the third angle is 54°. What is the measure of the smallest angle?

24. Problem 2.18 One angle in a triangle is twice another angle, and the third angle is 54°. What is the measure of the smallest angle? 2x + x + 54 = 180 3x = 126 x = 42

25. Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50°E

26. Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° Start with the parallel lines See that < ABE = 90° because m || n and AB | m E

27. Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° Start with the parallel lines See that < ABE = 90° because m || n and AB | m Then from ABE we have <EAB + <ABE + <AEB = 180°, so <EAB = 180° - <ABE - <AEB = 40°. E

28. Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° We can then use ACD to find <ACD = 180° - <DAC - <ADC = 180° - 40° - 3x = 140° - 3x E

29. Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° We can then use ACD to find <ACD = 180° - <DAC - <ADC = 180° - 40° - 3x = 140° - 3x When angle-chasing, it’s best to write the values you find for angles on your diagram as you find them, even when these values include variables. E 140° - 3x 40°

30. Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° We can then use ACD to find <ACD = 180° - <DAC - <ADC = 180° - 40° - 3x = 140° - 3x This diagram now suggests a way to finish the problem. We have three angles with vertices at C that together make a straight line, so E 140° - 3x 40°

31. Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° so we have <ACD + <DCB + <BCE = Substitution gives 140° - 3x + x + 3x = 180° so x = 40°. E 140° - 3x 40°

32. Problem 2.12 Given AB || CD and AD || BC and given the measures of the four angles as shown in terms of x and y, find x and y. 3y + 15° 3x - 15° AB D C y x

33. Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB D C y x

34. Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB D C y x

35. Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15°

36. Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

37. Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

38. Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: <EAN + <EAB = (3y + 15) + (3x – 15) =180° 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

39. Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: <EAN + <EAB = (3y + 15) + (3x – 15) =180° <FBA + <FBG +< GBH (3y + 15) + x + y =180° 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

40. Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: <EAN + <EAB = (3y + 15) + (3x – 15) =180° <FBA + <FBG +< GBH (3y + 15) + x + y =180° Rearranging these gives x + y = 60° x + 4y = 165° 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

41. Problem 2.12 Subtracting the 1 st from the 2 nd gives us 3y = 105°, so y = 35°. Of course we didn’t have to label every angle – we could have stopped when we had enough information to set up a pair of equations to solve for x and y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

42. Problem 2.12 Subtracting the 1 st from the 2 nd gives us 3y = 105°, so y = 35°. Of course we didn’t have to label every angle – we could have stopped when we had enough information to set up a pair of equations to solve for x and y. Since AB || CD, we have <DAN=<KDM so 3x – 15 = x + y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

43. Problem 2.12 Subtracting the 1 st from the 2 nd gives us 3y = 105°, so y = 35°, then x = 25°. Of course we didn’t have to label every angle - we could have stopped when we had enough information to setup a pair of equations to solve for x and y. Since AB || CD, we have <DAN=<KDM so 3x – 15 = x + y. Also <HBC & <BCI are supplementary so (3y + 15) + (3x – 15) = 180°. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

44. Problem 2.12 Solving these 2 equations gives us the same answer as before. Solving a problem with 2 different methods is an excellent way to check your answer. Since AB || CD, we have <DAN=<KDM so 3x – 15 = x + y. Also <HBC & <BCI are supplementary so (3y + 15) + (3x – 15) = 180°. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

45. Problem 2.13 Given that WV || YZ and WZ || VY, find x. Z V Y W P 140° 3x x

46. Problem 2.13 Given that WV || YZ and WZ || VY, find x. W e start by using what we know about parallel lines to find as many angles as we can. We find <V = 180 – 140 = 40° since WZ || VY. Z V Y W P 140° 3x x

47. Problem 2.13 Given that WV || YZ and WZ || VY, find x. W e start by using what we know about parallel lines to find as many angles as we can. We find <V = 180 – 140 = 40° since WZ || VY. Similarly, <Z = 40° and <ZYV = 180° - <Z = 140°. Z V Y W P 140° 3x x

48. Problem 2.13 Given that WV || YZ and WZ || VY, find x. W e start by using what we know about parallel lines to find as many angles as we can. We find <V = 180 – 140 = 40° since WZ || VY. Similarly, <Z = 40° and <ZYV = 180° - <Z = 140°. Since <PYV = 3x, we have <PYZ = <VYZ - <PYV = 140° - 3x. Since this angle, the 90° angle, & the angle with measure x together give us straight line PY, we have Z V Y W P 140° 3x x

49. Problem 2.13 < PYZ + 90° + x = 180° W e then substitute <PYZ = 140° – 3x into this equation, and we have 140° - 3x + 90° + x = 180°. We then solve this equation for x to find that x = 25°. Using information about angles to find information about other angles has 3 important tools: straight angles, vertical angles,and parallel lines. Z V Y W P 140° 3x x

50. Reflex Angle Angle that measures more than 180° is called a reflex angle. Suppose instead of measuring an angle in the ‘regular’ way, we go the ‘long’ way around. The ‘regular’ angle has a measure of 40°. What is the measure of the reflex angle?

51. Reflex Angle Angle that measures more than 180° is called a reflex angle. Suppose instead of measuring an angle in the ‘regular’ way, we go the ‘long’ way around. The ‘regular’ angle has a measure of 40°. What is the measure of the reflex angle? 360° - 40° = 320°.

52. Exterior Angles Find x given the angle measures shown. x is known as an exterior angle and the 35° and the 79° angles are known as its remote interior angles. A ACB 79° 35° x°x°

53. Exterior Angles Find x given the angle measures shown. x is known as an exterior angle and the 35° and the 79° angles are known as its remote interior angles. Important: Any exterior angle of a triangle is equal to the sum of its remote interior angles. A ACB 79° 35° x°x°

54. Exterior Angles Find x given the angle measures shown. So now it is easy to find x: x = 79° + 35° x = 114° Important: Any exterior angle of a triangle is equal to the sum of its remote interior angles. A ACB 79° 35° x°x°

55. Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED.  A B C E D F 30° x + 55° 2x - 20°  G

56. Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. We know we’ll probably need to find x to answer the problem. We could label every angle we know in terms of x, but 1 st we take a minute to look for a faster way to get x.  A B C E D F 30° x + 55° 2x - 20° G 

57. Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. We have <BAC and <ACB of ΔABC so we know the exterior angle <GBC = <BAC + <ACB = 2x + 10.  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

58. Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. From AB || DE, we know that <GBC = <DEF = = x + 55°. Hence we know that 2x + 10 = x + 55, so x = 45°.  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

59. Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. Our desired angle is the supplement of <DEF, so our answer is <CED = 180° - <DEF = 180° - (x + 55) = 80°.  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

60. Problem 2.22 There are many other ways we could have approached this problem. This is almost always true when we have problems involving exterior angles. <CED = 80°.  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

61. Review Problem 2.36 Let Δ ABC have (interior) angles in the ratio of 3:4:5. What is the measure of its smallest exterior angle?

62. Review Problems Let Δ ABC have (interior) angles in the ratio of 3:4:5. What is the measure of its smallest exterior angle? Suppose that ΔABC is labeled in such a way that <A ≤ <B ≤ <C. Then since the angles are in the ratio of 3:4:5, there exists some number x such that <A = 3x, <B = 4x, and <C = 5x. Using the fact that the 3 angles of a triangle add up to 180°, we obtain 12x = <A + <B + <C = 180°, so x = 15°.

63. Review Problems Let Δ ABC have (interior) angles in the ratio of 3:4:5. What is the measure of its smallest exterior angle? x = 15° Since the exterior angles of the triangle are equal to the sums of their remote interior angles, the smallest exterior angle has measure equal to the least of 3x + 4x = 7x, 3x + 5x = 8x, and 4x + 5x = 9x. Since the smallest of these is 7x, the measure of the smallest exterior angle of the triangle is 7x = 105°.

64. Review Problem 2.40 Three straight lines intersect at O and <COD = <DOE in the diagram. The ratio of <COB to <BOF is 7 : 2. What is the number of degrees in <COD? (Source: MATHCOUNTS) B A F O C D E

65. Review Problem 2.40 Three straight lines intersect at O and <COD = <DOE in the diagram. The ratio of <COB to <BOF is 7 : 2. What is the number of degrees in <COD? (Source: MATHCOUNTS) B A F O C D E Since the ratio of <COB = <BOF is 7 : 2, there is some x such that <COB = 7x and <BOF = 2x. Since <COF is a straight angle, 9x = <COB + <BOF = <COF = 180 so x = 20°.

66. Review Problem 2.40 Three straight lines intersect at O and <COD = <DOE in the diagram. The ratio of <COB to <BOF is 7 : 2. What is the number of degrees in <COD? (Source: MATHCOUNTS) B A F O C D E Since the ratio of <COB = <BOF is 7 : 2, there is some x such that <COB = 7x and <BOF = 2x. Since <COF is a straight angle, 9x = <COB + <BOF = <COF = 180 so x = 20°. Since <BOF & <COE are vertical angles, <COE = <BOF = 2x = 40° Finally, since <COD = <DOE, we have <COD = ½ <COE = 20°

67. Review Problem 2.43 The measures of the angles in the diagram are marked. Find <C. AC || BE. C A B E D 4x4x3x3x 2x2x

68. Review Problem 2.43 The measures of the angles in the diagram are marked. Find <C. AC || BE. C A B E D 4x4x3x3x 2x2x Since AD cuts the parallel segments AC and EB, we have 3x = <CAB = <ABE. Angle ABD is a straight line, so 5x = <ABE + <EBD = 180°, giving x = 36°.

69. Review Problem 2.43 The measures of the angles in the diagram are marked. Find <C. AC || BE. C A B E D 4x4x3x3x 2x2x Since AD cuts the parallel segments AC and EB, we have 3x = <CAB = <ABE. Angle ABD is a straight line, so 5x = <ABE + <EBD = 180°, giving x = 36°. Also, <CBD is an exterior angle of ΔABC, so 4x = <CBD = <C + <CAB = <C + 3x. Therefore, <C = x = 36°.

70. Review Problem 2.44 Three angles of a triangle have measures <A = x – 2y, <B = 3x + 5y, <C = 5x – 3y. Find x.

71. Review Problem 2.44 Three angles of a triangle have measures <A = x – 2y, <B = 3x + 5y, <C = 5x – 3y. Find x. Use the fact that the sum of the measures of the angles of a triangle is 180° to determine 180° = <A + <B + <C = (x – 2y) + (3x + 5y) + (5x – 3y) = 9x. Solving for x, we find that x = 20°.

72. Challenge 2.48 What is the number of degrees of the angle formed by the minute and hour hands of a clock at 11:10 PM? (Source: MATHCOUNTS)

73. Challenge 2.48 What is the number of degrees of the angle formed by the minute and hour hands of a clock at 11:10 PM? (Source: MATHCOUNTS) 1 st, find the position of the minute hand. The minute hand makes a 0° angle with 12 o’clock on the hour, and it travels a total of 360° in an hour, so in the 10 minutes after 11 o’clock, it moves (10/60) * 360° = 60° clockwise from the 12.

74. Challenge 2.48 What is the number of degrees of the angle formed by the minute and hour hands of a clock at 11:10 PM? (Source: MATHCOUNTS) Next, we consider the hour hand. It travels all the way around the clock in 12 hours, so it travels at a rate of (1/12) * 360° = 30° per hour. Since it will be pointing at the 12 at 12 o’clock, and it will travel (5/6 hour)(30° per hour) = 25° in the time between 11:10 PM and 12:00 midnight, it currently rests 25° counterclockwise from the 12. Therefore since the angle made by the 2 hands is the sum of the 2 angles made between the hands & the ray pointing from the center of the clock towards the 12, the angle between the hands is 60° + 25° = 85°

75. These are just a few examples of what can be found in Chapter 2, Introduction to Geometry. If you are interested in the series of books, go to http://www.artofproblemsolving.com/Books/AoPS_B_About.php