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Angles and Triangles Unit (Level IV Graduate Math) Draft (NSSAL) C. David Pilmer ©2014 (Last Update: April 2015) This resource is the intellectual property of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education. The following are permitted to use and reproduce this resource for classroom purposes. Nova Scotia instructors delivering the Nova Scotia Adult Learning Program Canadian public school teachers delivering public school curriculum Canadian non-profit tuition-free adult basic education programs Nova Scotia Community College instructors The following are not permitted to use or reproduce this resource without the written authorization of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education. Upgrading programs at post-secondary institutions (exception: NSCC) Core programs at post-secondary institutions (exception: NSCC) Public or private schools outside of Canada Basic adult education programs outside of Canada Individuals, not including teachers or instructors, are permitted to use this resource for their own learning. They are not permitted to make multiple copies of the resource for distribution. Nor are they permitted to use this resource under the direction of a teacher or instructor at an unauthorized learning institution. Table of Contents Introduction………………………………………………………………………………… ii Negotiated Completion Date………………………………………………………………. ii The Big Picture…………………………………………………………………………….. iii Course Timelines…………………………………………………………………………... iv Angles and Degrees ………………………………………………………………………. 1 Relationships Amongst Angles …………………………………………………………… 9 Investigating Similar Triangles …………………………………………………………… 15 Using Similar Triangles …………………………………………………………………… 17 Similar Triangle Application Problems …………………………………………………… 26 The Pythagorean Theorem ………………………………………………………………… 30 Putting It Together ………………………………………………………………………… 37 Post-Unit Reflections………………………………………………………………………. 43 Soft Skills Rubric …………………………………………………………………………. 44 Answers……………………………………………………………………………………. 45 NSSAL ©2009 i Draft C. D. Pilmer Introduction This unit has three components. 1. The first component focuses on measuring and estimating acute and obtuse angles in degree measure. This is followed by understanding and exploiting relationships amongst angles (e.g. supplementary angles, opposite angles, 180o in a triangle, etc.). 2. The second component of the unit focuses on using similar triangles to solve real world problems. 3. The third and final component looks at real world applications and multi-step problems involving the Pythagorean Theorem. Negotiated Completion Date After working for a few days on this unit, sit down with your instructor and negotiate a completion date for this unit. Start Date: _________________ Completion Date: _________________ Instructor Signature: __________________________ Student Signature: NSSAL ©2009 __________________________ ii Draft C. D. Pilmer The Big Picture The following flow chart shows the six required units and the four optional units (choose two of the four) in Level IV Graduate Math. These have been presented in a suggested order. Instructors and students may choose to alter this order to best serve the needs of the learner. Not all of the units take the same amount of time to complete. For example, the Consumer Finance Unit is quite short. By contrast, the Graphs and Function Unit and the Measurement Units (A and B) tend to take the greatest amount of time. Math in the Real World Unit (Required) Fractions, decimals, percentages, ratios, proportions, and signed numbers in real world applications Math Games and Puzzles Solving Equations Unit (Required) Solve and check equations of the form , and . , Consumer Finance Unit (Required) Simple Interest and Compound Interest TVM Solver (Loans and Investments) Graphs and Functions Unit (Required) Understanding Graphs Linear Functions and Line of Best Fit Measurement Unit (Required) Part A: Imperial and Metric Measures Part B: Perimeter, Area and Volume Angles and Triangles Unit (Required) Angle and Line Relationships Similar Triangles Pythagorean Theorem Choose two of the four. Linear Functions and Systems of Equations Unit NSSAL ©2009 Trigonometry Unit iii Descriptive Statistics Unit Numeracy Unit Draft C. D. Pilmer Course Timelines Graduate Level IV Math is a two credit course within the Adult Learning Program. As a two credit course, learners are expected to complete 200 hours of course material. Since most ALP math classes meet for 6 hours each week, the course should be completed within 35 weeks. The curriculum developers have worked diligently to ensure that the course can be completed within this time span. Below you will find a chart containing the unit names and suggested completion times. The hours listed are classroom hours. Unit Name Minimum Completion Time in Hours 24 20 15 25 22 14 18 18 Total: 156 hours Math in the Real World Unit Solving Equations Unit Consumer Finance Unit Graphs and Functions Unit Measurement Unit (Parts A & B) Angles and Triangles Unit Selected Unit #1 Selected Unit #2 Maximum Completion Time in Hours 34 28 18 30 30 16 22 22 Total: 200 hours As one can see, this course covers numerous topics and for this reason may seem daunting. You can complete this course in a timely manner if you manage your time wisely, remain focused, and seek assistance from your instructor when needed. NSSAL ©2009 iv Draft C. D. Pilmer Angles and Degrees Aerial tricks done by skiers, figure skaters, and skateboarders are often described using degrees. If the skier does a complete rotation off a jump, she is said to have done a 360, which means a 360 degree rotation. Similarly, if the skier does two complete rotations off a jump, she is said to have done a 720, which means a 720 degree rotation. In this section, we are going to learn how angles (or rotations) are measured in degrees. An angle is: Formed by two line segments that start at the same point, called a vertex. Measured in degrees, whose symbol is o. vertex 30o Angles that measure less than 90o are called acute angles. 70o 18o 40o Angles that measure are between 90o and 180o are called obtuse angles. 107o 140o 153o The degree measure of an angle is often determined using a protractor. The vertex of the angle is placed on the crosshairs of the protractor, and one of the line segments is placed so that it is along the horizontal base line of the protractor. Most protractors have an inner and outer scale; use the one that is most appropriate for the angle you are measuring. 113o 55o 150o 28o NSSAL ©2009 5 Draft C. D. Pilmer Questions 1. Estimate the measure of the following angles. Also state whether this is an acute or obtuse angle. (a) (b) (c) (d) ? ? ? ? 2. Determine the measure of the following angles. Also state whether this is an acute or obtuse angle. (a) (b) (c) (d) (e) (f) NSSAL ©2009 6 Draft C. D. Pilmer 3. Use a protractor to determine the degree measures of these angles. (a) (b) ? ? (c) (d) ? ? (e) (f) ? ? (g) (h) ? NSSAL ©2009 ? 7 Draft C. D. Pilmer (i) (j) ? ? (k) (l) ? ? (m) (n) ? ? NSSAL ©2009 8 Draft C. D. Pilmer Relationships Amongst Angles Before looking at triangles, we need to know a few things about angles. Naming Angles The most common way to name an angle is to use the three letters on the shape that define the angle, with the middle letter representing the vertex of the angle. In the diagram CAB 40 , ABC 50 , and BCA 90 . C 90o 40o With simple geometric figures, as is the case with the diagram provided, the vertex alone can be used to define the angle. A 40 , B 50 , and C 90 50o A Naming Sides of Triangles The side opposite (across from) a specific vertex on a triangle is named using the same letter as the vertex but using a lower case letter. In the diagram, the side opposite vertex A or A is called side a. The side opposite vertex B or B is called side b. The side opposite vertex C or C is called side c. B C b A Complementary Angles Two angles are complementary if they add up to 90o (i.e. form a right angle). In the diagram ABC and CBD are complementary angles. a c B A C B Supplementary Angles Two angles are supplementary if they add up to 180o (i.e. form a straight line). In the diagram DEF and FEG are supplementary angles. D F D NSSAL ©2009 9 E G Draft C. D. Pilmer Opposite Angles When two lines intersect, four angles are formed. The angles that are directly opposite to each other are called opposite angles. These opposite angles are congruent (i.e. equal in measure). In the diagram ABE CBD and ABC DBE . A C B E Right Angle Triangles A right angle triangle has one angle measuring 90o (i.e. a right angle). In the diagram, ABC is a right angle triangle because ABC measures 90o. D A C B Acute Triangles An acute triangle is a triangle where all three angles are less than 90o. In the diagram, FGH is an acute triangle because FGH 70 , HFG 47 , and GHF 63 (All the angles are less than 90o). F G H Obtuse Triangles An obtuse triangle is a triangle where one of the three angles is greater than 90o. In the diagram, PQR is an obtuse triangle because the QPR 107 (It exceeds 90o.) R Q P Interior Angles of a Triangle The interior angles of any triangle add up to 180o. In the diagram: ABC BAC ACB 48 25 107 180 Interior Angles of a Quadrilateral The interior angles of any quadrilateral (i.e. four-sided figure) add up to 360o. In the diagram: HEF EFG FGH GHE 100 128 61 71 360 NSSAL ©2009 10 C A B H G E F Draft C. D. Pilmer Parallel Lines and Transversal Lines Lines are parallel if they are always the same distance apart and never meet. We use arrow symbols to indicate when lines are parallel. When parallel lines are crossed by another line (called a transversal line), angles are formed and special relationships exist among these angles. F E D I H Angle Relationships: G 1. Corresponding angles are equal. FED EIH , DEI HIJ , FEG EIK , and GEI KIJ K J 2. Alternate interior angles are equal. DEI EIK and GEI EIH 3. Alternate exterior angles are equal. FED JIK and FEG JIH 4. Consecutive interior angles are supplementary. DEI EIH 180 GEI EIK 180 Questions Find the identified angles. (Diagrams are not to scale.) 1. 2. A D P N 40o 30 o C B 70o ACD O NPO 3. 4. J F I 75 o P 110 G M H JIG NSSAL ©2009 o 82o O N PMO 11 Draft C. D. Pilmer 5. 6. G F V U o 37 W o D 105 R E T S DEG SVW 7. 8. A D Q P E 150o 50o B C N 95o M AED ONP DEC OPN 9. O 10. N F I G 35o 58o 83o H P M O FGH NMO FIH NOP 11. 12. E o o 32 S D 49 100o 110 V F 65o U T G FDG RUT DGF RUV NSSAL ©2009 R o 12 Draft C. D. Pilmer 13. 14. C M 47o A o N B 81 D Q P H 37o E F R O G CBD ONP FBD OPN BFH NPQ EFG OPR HFG QPR 15. 16. L N o 50 K M L 98o 60o 70o H P 65o I ONP NIH NMP NIJ LMN NMJ LKM MJK LMK LMJ KMP ONI NSSAL ©2009 M N O O 13 J K Draft C. D. Pilmer 17. 18. B C G H H 40o 115o D 95o o 70 o 95 E F I L 58o N J G GHF ILM HFC MLN CFE KLN DCF KNL BCD HIL EFG JIL NSSAL ©2009 K 14 M Draft C. D. Pilmer Investigating Similar Triangles Below we have been provided with two similar triangles. At this point we do not know the formal definition of a similar triangle. We going to examine these two triangles and try to see how they are related and why they are called similar triangles. C A B F D E Step 1: Using a protractor, measure the three interior angles of each triangle. Triangle #1 CAB ABC BCA Triangle #2 FDE DEF EFD What do you notice? NSSAL ©2009 15 Draft C. D. Pilmer Step 2: Using a ruler, measure the length of each side of each triangle to the nearest millimeter. Triangle #1 AB = BC = AC = Triangle #2 DE = EF = DF = Step 3: We are going to use the measurements from Step 2 to work out three ratios. First Ratio: Take the length of AB and divide it by the length of DE. AB DE Second Ratio: Take the length of BC and divide it by the length of EF. BC EF Third Ratio: Take the length of AC and divide it by the length of DF. AC DF What do you notice? Step 4: Based on what you have done, how can you tell if two triangles are similar? NSSAL ©2009 16 Draft C. D. Pilmer Using Similar Triangles Similar Triangle Definition In any pair of similar triangles the corresponding angles are equal, and the ratio of the corresponding sides are equal. For our diagram below, ABC and DEF are similar triangles because CAB FDE ABC DEF BCA EFD and AB BC AC DE EF DF F C D A B E Identifying Similar Triangles Sometimes it is little more difficult to identify similar triangles. We often have to recognize relationships between angles, before we can conclude that we are dealing with similar triangles. Consider the following diagrams. Diagram 1 A and B E NSSAL ©2009 D C are similar triangles. Why? 1. because they are the same angle. 2. because they are both labeled as right angles. 3. because there are only 180o in a triangle and if the two triangles have two sets of equal angles, then the third set of angles must be equal. 4. Since all the corresponding angles are equal, then we are dealing with similar triangles. 17 Draft C. D. Pilmer Diagram 2 and are similar triangles. S R Q T P Why? 1. because they are opposite angles. 2. because they are both labeled as right angles. 3. because there are only 180o in a triangle and if the two triangles have two sets of equal angles, then the third set of angles must be equal. 4. Since all the corresponding angles are equal, then we are dealing with similar triangles. Diagram 3 J I and F G H are similar triangles. Why? 1. because they are the same angle. 2. because when dealing with parallel lines cut by a transversal line, corresponding angles are equal. 3. because when dealing with parallel lines cut by a transversal line, corresponding angles are equal. 4. Since all the corresponding angles are equal, then we are dealing with similar triangles. Example 1 If ABC and DEF are similar triangles, find AB given the information supplied in the diagram. D A B NSSAL ©2009 31.7 cm 22.3 cm C E 43.5 cm 18 F Draft C. D. Pilmer Answer: Since we are dealing with similar triangles, we know the following. AB BC AC DE EF DF For the information provided, we are only going to use the following. AB BC DE EF Now we will substitute the known values into our equation and solve for the unknown value (i.e. the length of side AB). Rather than calling our unknown side AB, we can call it side c (note we are using a lowercase c) because it is opposite (or across from) C . AB BC DE EF c 22.3 31.7 43.5 43.5c 22.3 31.7 Procedure: 1. Find the cross products. 2. Set the cross products equal to each other. 3. Divide the number not multiplied by variable by the number multiplied by the variable (in this case the variable is c). 43.5c 706.91 706.91 43.5 c = 16.3 cm c Example 2 If JKL and MNO are similar triangles, find NO given the information supplied in the diagram. O L 12.3 m 18.9 m 20.1 m J K Answer: JL KL MO NO 12.3 20.1 18.9 m 12.3m 20.1 18.9 12.3m 379.89 NSSAL ©2009 M N Rather than calling our unknown side NO, we can call it side m because it is opposite (or across from) . Often people will use the variable x to represent the unknown side. Feel free to do that if you find it easier. 19 Draft C. D. Pilmer 379.89 12.3 m 30.9 m m Example 3 In the diagram below, QS equals 62 mm, QP equals 21 mm, and RT equals 17 mm. Find RS. P T S 57o 57o R Q Answer: We suspect that we are dealing with similar triangles. We have to start by proving this suspicion. PQS TRS because they are both 57o. PSQ TSR because they are the same angle. SPQ STR because there are 180o in any triangle. Since all corresponding angles are equal, we now know that we are dealing with similar triangles. RS RT QS QP t 17 62 21 21t 17 62 21t 1054 1054 t 21 t 50.2 mm NSSAL ©2009 20 Draft C. D. Pilmer Example 4 In the diagram below, EF equals 51.2 cm, FH equals 71.8 cm, and DE equals 168.3 cm. Determine GH. D E F G H Answer: We suspect that we are dealing with similar triangles. We have to start by proving this suspicion. EFD HFG because they are opposite angles. DEH EHG because when dealing with parallel lines cut by a transversal line (EH), alternate interior angles are equal. EDG HGD because when dealing with parallel lines cut by a transversal line (DG), alternate interior angles are equal. Since all corresponding angles are equal, we now know that we are dealing with similar triangles. DE EF GH HF 168.3 51.2 GH 71.8 51.2GH 168.3 71.8 51.2GH 12083.94 12083.94 51.2 GH = 236.0 cm GH NSSAL ©2009 In this question we did not attempt to rename side GH. The reason was that vertex F is shared by both triangles so if we attempted to name a side f, we would not know if we were referring to side GH or side DE. Often people will use the variable x to represent the unknown side. Feel free to do that if you find it easier. 21 Draft C. D. Pilmer Questions: 1. If these two triangles are similar, find PQ. 5.6 m E G 8.2 m P R 10.8 m F Q 2. If the following triangles are similar, determine the measure of CD. S C B 5.2 km R 7.8 km 7.3 km D T NSSAL ©2009 22 Draft C. D. Pilmer 3. If these two triangles are similar, find TV and TU. C T 8.1 km U 19.4 km E 8.8 km 14.1 km V D 4. If the following triangles are similar, determine the measures of FG and EF. 8.7 m J G K 10.8 m 5.1 m 7.4 m L E NSSAL ©2009 23 F Draft C. D. Pilmer 5. Find DE. C E 96 cm 97o D 141 cm 97o 179 cm F B 6. Given that JK = 77 m, KN = 36 m, JN = 85 m, and JL = 116 m, find LM and JM. K L J N M NSSAL ©2009 24 Draft C. D. Pilmer 7. If PN = 78 cm, PQ = 43 cm, NO = 48 cm, and OP = 86 cm, find QR and PR. O P Q N R 8. If WV = 4.6 m, UW = 6.8 m, ST = 13.7 m, and UV = 6.5 m, find SV and TV. S T W U V NSSAL ©2009 25 Draft C. D. Pilmer Similar Triangle Application Questions We will now look at real world problems that can be solved using similar triangles. Example 1 George is standing in a flat field. A 1.6 m height pole located in that field casts a shadow that is 2.3 m long. At the same time, a tree in the same field casts a 11.7 m long shadow. What is the height of the tree? Answer: We will start by drawing a diagram. 1.6 m 11.7 m 2.3 m Since we are dealing with two vertical objects (the tree and pole) and shadows cast at the same time of day, we can conclude that all corresponding angles are equal. If this is the case then we are dealing with similar triangles. h 11.7 1.6 2.3 2.3h 11.7 1.6 2.3h 18.72 18.72 h 2. 3 h 8.1 m The height of the tree is 8.1 m. NSSAL ©2009 26 Draft C. D. Pilmer Example 2 A mirror is placed on the floor between a wall and a small laser. The beam of the laser is pointed at a mirror. The beam is reflected off the mirror and strikes the top of a wall. The distance along the floor between the mirror and the wall is 3.5 metres. The distance along the floor between the mirror and the vertical stand holding the laser is 0.9 m. If the laser is 1.5 m off the floor, how tall is the wall. Please note that the angle the laser beam strikes the mirror is equal to the angle the laser beam reflects to the top of the wall. Answer: Start by drawing a diagram. wall laser 1.5 m mirror 3.5 m 0.9 m If we just look at the similar triangles, we end up with the following. h 1.5 m 3.5 m NSSAL ©2009 0.9 m 27 Draft C. D. Pilmer Questions: 1. A company, that manufactures extension ladders, recommends for safety reasons that the base of the ladder should be 1 metre away from the wall for every 4 metres the ladder rises vertically. If the ladder must reach the top of a 7 metre wall, how far should the base of the ladder be away from the base of this wall? 2. Ryan wants to know the width of a river but he has no means to take that measurement directly. He decides to make some measurements along the shoreline to accomplish this task. The distance from A to C is 161 m. The distance from B to C is 63 m. The distance from C to D is 37 m. Determine the width of the river. River A B C D NSSAL ©2009 28 Draft C. D. Pilmer 3. At a certain time of day, the shadow cast by a flag pole is 10.42 m long. At the same time, the shadow cast by 1.85 m high individual is 1.27 m. How tall is the flag pole? 4. A wheel chair ramp has been built for the new community centre. If someone travels 2 metres horizontally along the ramp, they rise 0.18 m. How much would one rise, if they traveled 4.5 metres along the same ramp? 5. Montez is building a cottage and has to attach two small pieces of wood to the gable end (see shaded triangles in diagram). The triangular pieces that are 18 inches long must have the same pitch as his roof. The roof’s pitch is 8:12 which means that for every 12 feet one travels horizontally, the roof goes up by 8 feet. How tall should the triangular piece of wood be? NSSAL ©2009 29 Draft C. D. Pilmer The Pythagorean Theorem The Pythagorean Theorem, a formula that describes the relationship among the three sides of a right-angle triangle, is probably one of most famous and useful theorems used in mathematics. The theorem is named after the Greek mathematician, Pythagoras (570 BC to 495 BC), who is credited for the proof, even though many argue that the actual theorem was used previously by the Egyptians in their surveying. B The theorem states that for a right-angle triangle ABC where c is the hypotenuse (i.e. the longest side or the side opposite the c right-angle), the relationship between the three sides, a, b, and c, a can be described by the formula a 2 b 2 c 2 . In words we are saying that in a right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides (referred A b C to as the legs of the triangle). Notice the relationship between the naming of the angles of the triangle (in capital letters) and the naming of the sides (in lower case). The side opposite angle A (or A ) is called a. The side opposite angle B is called b. Does this formula actually work? Consider the triangle that has been drawn below. Al three sides have been measured accurately to two decimal points. When these values are entered into the formula a 2 b 2 c 2 , everything works out (i.e. both sides of the equation are equal to each other. c = 7.66 cm a = 4.58 cm b = 6.14 cm The example is not a formal proof. We don’t have enough time to go over Pythagoras’ proof, or other proofs, however; if you search the internet, you can find these proofs. Example 1 Determine the missing side of the triangle. P Answer: In this case, we are given the two legs of the triangle and must find the hypotenuse. a2 b2 c2 - Use the variables provided. r 2 p2 q2 NSSAL ©2009 30 3.9 m Q 2.3 m q R Draft C. D. Pilmer 3.9 2 2.32 q 2 15.21 5.29 q 2 20.5 q 2 20.5 q q 4.5 m Example 2 Determine the missing side of the triangle. G Answer: In this case, we are given the one leg and the hypotenuse, and must find the other leg. a2 b2 c2 e2 f 2 g 2 6.9 2 f 2 9.2 2 6.9 cm E F 9.2 cm 47.61 f 2 84.64 f 2 84.64 47.61 f 2 37.03 f 37.03 f 6.1 cm 4.7 m Example 3 Determine the length of x. 9.3 m Answer: This question requires that we use the Pythagorean Theorem twice. We use it first to find the side shared by both right-angle triangles. Then the theorem is used to find x. a2 b2 c2 a2 b2 c2 8.5 2 b 2 9.3 2 4 .7 2 3 .8 2 x 2 72.25 b 2 86.49 22.09 14.44 x 2 b 2 86.49 72.25 36.53 x 2 b 2 14.24 36.53 x x 6.0 b 14.24 3.8 x 8.5 m The length of x is 6.0 m. NSSAL ©2009 31 Draft C. D. Pilmer Example 4 Tom traveled directly north for 2.4 km and then headed directly west for 1.9 km. In the end, how far was Tom from his starting point? Answer: In this case we have to create a diagram. Once we have the diagram, we can see that we have the two legs and we must find the hypotenuse. a2 b2 c2 1.9 km N W ? 2.4 km E S 1.9 2 2.4 2 c 2 3.61 5.76 c 2 Starting Point 9.37 c 2 9.37 c c 3.1 km Tom is 3.1 km from his starting point. Questions 1. Solve for the missing side. (a) 12.7 cm 5.9 cm (b) 9.8 m NSSAL ©2009 16.7 m 32 Draft C. D. Pilmer (c) R 9.1 cm 11.7 cm P Q (d) 46 ft S T 52 ft U 2. Determine whether the following triangle is right-angled. D 8m 17 m E 15 m F NSSAL ©2009 33 Draft C. D. Pilmer 3. For each of the following diagrams, solve for x. (a) 5 13 x 16 (b) 15.7 x 8.3 20.3 (c) 43 x 35 26 NSSAL ©2009 34 Draft C. D. Pilmer 4. Maurita leans a 7 metre ladder against a vertical brick wall. If the base of the ladder is 1.5 metres from the brick wall, how far up the wall is the other end of the ladder? 5. The local community centre has decided to install a new wheelchair ramp. They know that the ramp will rise 1.4 metres for a run of 11.2 metres. How long will the ramp be? rise run 6. Nancy traveled directly south for 1.7 km and then headed directly east for 0.8 km. In the end, how far was Nancy from her starting point? NSSAL ©2009 35 Draft C. D. Pilmer 7. A wire is stretched from the top of a 4.2 metre antenna to a bracket 3.5 metres from the base of the antenna. How long is the wire? Assume that the antenna is vertical and situated on a horizontal flat piece of property. 8. Tylena has let out 50 metres of kite string when she observes that her kite is directly above Candice. If Candice is 34 metres from Tylena, how high is the kite? Assume the kite string is taunt and that the two women are standing on flat horizontal ground. 9. Two planes left the airport. One travelled north at 220 kilometres per hour. The other travelled west at 260 kilometres per hour. How far apart will the planes be in a half hour? 10. The formula for the area of a parallelogram is A bh , where b is the base and h is the height. Determine the area of the parallelogram shown on the right. 35 cm 26 cm 15 cm NSSAL ©2009 36 Draft C. D. Pilmer Putting It Together 1. Determine the measure of the following angles. Also state whether this is an acute or obtuse angle. (a) (b) (c) (d) 2. Use a protractor to determine the degree measures of these angles. (a) (b) ? ? (c) (d) ? ? NSSAL ©2009 37 Draft C. D. Pilmer (e) (f) ? ? (g) (h) ? ? 3. In each case, find the indicated angles. (a) (b) B M N C A D o 58 I H G R o 142 F o 85 E RPM ACH MPO GFC PMN GCF MNO DCF OPQ 38 P 60 o Q O IGC NSSAL ©2009 L 45o Draft C. D. Pilmer (c) (d) P C A B 65 Q o 80 G 55o N D H F M 80o O o L K I 43o J E FEG NOI BGE QOI ABG MNK DBG LIJ DEF HKJ DEG IJK 4. Given the measures of DF and EF , find DE, CD, and BD. F 53.7 cm 44.1 cm B D 21.7 cm E C NSSAL ©2009 39 Draft C. D. Pilmer 5. Calculate the length of GF. G 7.6 m D F E 3.3 m 5.7 m 6. If AB = 18.6 cm, EC = 10.4 cm, AD = 12.4 cm, find ED, BD, CD, and BC. A E B C D NSSAL ©2009 40 Draft C. D. Pilmer 7. Hinto initially walks diagonally across a rectangular property that measures 35 metres by 20 metres. After reaching the corner of the property, he decides to return to his starting position by walking round the edge of the property. How much further does he walk on his return trip? 8. How high is a tower that casts a 13.6 m shadow at the same time as a 1.6 metre high individual casts a 1.1 metre shadow? 9. Calculate the height of an equilateral triangle whose side length is 8 cm? (Note: For an equilateral triangle, all sides are of equal length.) h NSSAL ©2009 41 Draft C. D. Pilmer 10. Jorell places a small mirror on the ground, 6.4 metres from the base of a tree. He then walks backwards until he can see the top of the tree in the mirror. At this point, Jorell is 1.2 metres away from the mirror. If his eye level is 1.5 metres, how tall is the tree? Assume this taking place on flat horizontal ground. 11. A 6.8 metre rope is tied to the top of a 5.2 metre pole. The rope is pulled tight, extended its full length, and stacked into the ground. How far is the staking point from the base of the pole? Assume the pole is vertical and situated on flat horizonatl ground. NSSAL ©2009 42 Draft C. D. Pilmer Post-Unit Reflections What is the most valuable or important thing you learned in this unit? What part did you find most interesting or enjoyable? What was the most challenging part, and how did you respond to this challenge? How did you feel about this math topic when you started this unit? How do you feel about this math topic now? Of the skills you used in this unit, which is your strongest skill? What skill(s) do you feel you need to improve, and how will you improve them? How does what you learned in this unit fit with your personal goals? NSSAL ©2009 43 Draft C. D. Pilmer Soft Skills Rubric Look back over the module you have just completed and assess yourself using the following rubric. Use pencil or pen and put a checkmark in the column that you think best describes your competency for each description. I will look at how accurately you have done this and will discuss with you any areas for improvement. You will be better prepared for your next step, whether it is work or further education, if you are competent in these areas by the end of the course. Keep all of these rubrics in one place and check for improvement as you progress through the course. Date: Competent demonstrates concept fully consistently Throughout this module, I… the and Approaching Competency demonstrates the concept most of the time Developing Competency demonstrates the concept some of the time Attended every class Let my instructor know if not able to attend class Arrived on time for class Took necessary materials to class Used appropriate language for class Used class time effectively Sustained commitment throughout the module Persevered with tasks despite difficulties Asked for help when needed Offered support/help to others Helped to maintain a positive classroom environment Completed the module according to negotiated timeline Worked effectively without close supervision Comments: (Created by Alice Veenema, Kingstec Campus) NSSAL ©2009 44 Draft C. D. Pilmer Answers Angles and Degrees (pages 1 to 8) 1. (a) (b) (c) (d) acute; any value between or equal to 70o and 80o obtuse; any value between or equal to 130o and 150o acute; any value between or equal to 15o and 30o obtuse; any value between or equal to 150o and 170o 2. (a) 100o, obtuse (c) 145o, obtuse (e) 14o, acute (b) 60o, acute (d) 42o, acute (f) 168o, obtuse 3. If you are off by 2o or less, don't worry about it. (a) 50o (b) (c) 150o (d) o (e) 28 (f) (g) 75o (h) (i) 95o (j) o (k) 118 (l) (m) 53o (n) 34o 123o 113o 148o 12o 98o 85o Relationships Amongst Angles (pages 9 to 14) 1. ACD 140 2. NPO 80 3. JIG 75 4. PMO 78 5. DEG 53 6. SVW 105 7. AED 150 DEC 30 8. ONP 50 OPN 35 9. FGH 62 FIH 145 10. NMO 32 NOP 90 11. FDG 41 DGF 58 12. RUT 85 RUV 95 13. CBD 133 FBD 47 BFH 47 EFG 47 HFG 133 14. ONP 99 OPN 44 NPQ 136 OPR 136 QPR 44 NSSAL ©2009 45 Draft C. D. Pilmer 15. ONP 120 NMP 50 LMN 130 LKM 80 LMK 50 KMP 130 16. NIH 98 NIJ 82 NMJ 125 MJK 125 LMJ 65 ONI 82 17. GHF 52 HFC 110 CFE 70 DCF 100 BCD 80 EFG 110 18. ILM 115 MLN 65 KLN 115 KNL 25 HIL 115 JIL 65 Investigating Similar Triangles (pages 15 and 16) Step 1: Triangle #1 CAB 61 ABC 38 BCA 81 Triangle #2 FDE 61 DEF 38 EFD 81 AB = 80 mm BC = 70 mm AC = 49 mm DE = 122 mm EF = 108 mm DF = 75 mm Corresponding angles are equal. Step 2: Triangle #1 Triangle #2 Step 3: AB 80 0.66 DE 122 BC 70 0.65 EF 108 AC 49 0.65 DF 75 All the ratios are equal. Using Similar Triangles (pages 17 to 25) 1. PQ = 15.8 m 2. CD = 5.6 km 3. TU = 5.1 km TV = 12.1 km 4. EF = 12.7 m FG = 7.4 m NSSAL ©2009 46 Draft C. D. Pilmer 5. DE = 122 cm 6. LM = 54 m JM = 128 m 7. RP = 47 cm RQ = 26 cm 8. TV = 13.1 m SV = 9.3 m Similar Triangle Application Questions (pages 26 to 29) 1. 1.75 m 2. 57.6 m 3. 15.18 m 4. 0.41 m 5. 12 inches The Pythagorean Theorem (pages 30 to 36) 1. (a) 14.0 cm (c) 14.8 cm (b) 13.5 m (d) 24.2 ft 2. It is right-angled because 82 152 17 2 3. (a) 10.6 (c) 10.8 (b) 9.8 4. 6.8 m 5. 11.3 m 6. 1.9 km 7. 5.5 m 8. 36.6 m 9. 170.3 km 10. 743 cm2 NSSAL ©2009 47 Draft C. D. Pilmer Putting It Together (pages 37 to 42) 1. (a) 143o, obtuse (c) 107o, obtuse 2. (a) (c) (e) (g) 49o 155o 78o 65o (b) 17o, acute (d) 55o, acute (b) (d) (f) (h) 3. (a) IGC 122 30o 115o 120o 156o (b) RPM 120 ACH 58 MPO 60 GFC 38 PMN 135 GCF 84 MNO 80 DCF 38 OPQ 120 (c) FEG 47 (d) NOI 80 BGE 90 QOI 100 ABG 65 MNK 125 DBG 115 LIJ 80 DEF 122 HKJ 55 DEG 75 IJK 45 4. Hint: Involves both the Pythagorean Theorem and similar triangles. DE = 30.6 cm, DC = 26.4 cm, BD = 15.0 cm 5. Hint: First find GE. GF = 6.0 m 6. ED = 6.9 cm, DC = 12.5 cm, BD = 22.4 cm, BC = 9.9 cm 7. 55 – 40.3 = 14.7 m 8. 19.8 m 9. 6.9 cm 10. 8 m 11. 4.4 m NSSAL ©2009 48 Draft C. D. Pilmer