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SKILLS CHECK for Week 11 Refresh your memory by attempting the questions below. You will need these skills when working through this week’s lessons. Answers are given on the next page. Question 1: Estimating magnitude (size) of angles Estimate the size of angle ABC in each of the diagrams shown. a. b. A C C c. A A A C B B A C B A A B C BA B A Question 2: Drawing angles A A Draw an accurate diagram to represent an angle with a magnitude of: a. 30o b. 75o c. 130o Question 3: Calculations of sizes of angles and lengths G H GH is parallel to DF 400 m 410 D a. i. Which trig ratio would you use to find the distance DE? ii. What is the size of the following angles? 1. DGE 2. HGE 3. GEF iii. GFE = 290. What is the size of: 1. FGH 2. FGE F E b. Using the above diagram, state whether each statement is TRUE or FALSE i. GF2 = GE2 + EF2 ii. distance GF sin139 distance EF sin EGF iii. EF2 GE2 GF2 2 GE GF cos 41 iv. area of triangle EFG = ½ EF 400 Answers to Skills Check / Extra Resources 1 a. 600 400 b. 2 a. b. 1150 c. c. A A 300 750 B A 1300 B A 3 a. i. tan ratio C B at D andAso The triangle has a right angle A 400 tan 41 . Alternatively (since DGE = 490) DE ii. 1. DGE = 490 2. HGE = 410 3. GEF = 1390 C DE A 400 tan 49 iii. 1. FGH = 290 2. FGE = 120 b. i. GF2 = GE2 + EF2 FALSE The triangle is not right-angled and so Pythagoras’ rule cannot be used. ii. distance GF distance EF TRUE sin139 sin EGF This is the correct application of the sine rule. iii. EF2 GE2 GF2 2 GE GF cos 41 FALSE The angle required is FGE = 120, so using the cosine rule will give EF2 GE2 GF2 2 GE GF cos12 iv. area of triangle EFG = ½ EF 400 TRUE EF is the base of the triangle and 400 is the height perpendicular to EF, so the rule Area = ½ base height can be used. More information: http://www.mathsisfun.com/angles.html http://www.mathsisfun.com/algebra/trig-sine-law.html http://www.mathsisfun.com/algebra/trig-cosine-law.html http://www.mathsisfun.com/algebra/trig-solving-triangles.html http://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html More information: https://www.mathsonline.com.au/ There are many lessons on this website covering the geometry/trigonometry module. For some examples (once you have logged on), go to: 11 & 12 General > Measurement > Further Trigonometry > The Sine Rule: Finding a Side 11 & 12 General > Measurement > Further Trigonometry > The Cosine Rule: Finding a Side N.B. If you have further queries, please contact your teacher. A Week 11 Applications of Geometry & Trigonometry (I) Angles of elevation & depression Bearings & Directions Examples of bearings Page 2 Page 5 Page 9 Triangulation Work for submission Answers to exercises Page 14 Page 17 Page 23 Reference: ESSENTIAL Further Mathematics Chapter 14 Key knowledge Angles of elevation and depression. Basic geometric and trigonometric concepts associated with orienteering and navigation. Concepts associated with triangulation. Key skills Use angles of elevation and depression to solve different problems. Basic geometric and trigonometric concepts associated with problem solving. Use the triangulation concepts to solve problems in practical situations. Tasks this week relate to outcomes 1, 2 and 3. Week 11 Page 2 LESSON 1 ANGLES OF ELEVATION AND DEPRESSION ANGLES OF ELEVATION: Looking up from the horizontal gives an angle of elevation. Angle of elevation is measured from the horizontal upwards. ANGLES OF DEPRESSION: Looking down from the horizontal gives an angle of depression. Angle of depression is measured from the horizontal downwards. Angles of elevation and depression are in a vertical plane We can see from the following diagram that the angle of depression given from one location can give us the angle of elevation from the other position using the alternate angle law. Watch a powerpoint! Insert your DECV course CD into your computer and then click: Further Mathematics Unit 4 > Further Resources > Angles of Elevation and Depression Week 11 Page 3 Example 1 A pilot flying a plane at altitude 400 m, sees a small boat at an angle of depression of 1.2º (a) Draw a diagram (b) Find the horizontal distance of the boat to the plane. Answers (a) Plane P 1.20 400 m 400 m Opposite side 1.20 adjacent side (d) Boat B (b) We have a right- angled triangle. We are looking for side d (the adjacent side) Notice that 1.2º is the same as 1º 12' Use the tan ratio. tan 1.2º = d = opposite 400 = adjacent d 400 = 400 = 19 096 m to the nearest metre tan 1.20 0.020947 Example 2 A boat sights a lighthouse light 75 metres above sea level at an angle of elevation of 7.1º. How far is the boat from the lighthouse? Answers 75 m 75 m Opposite side 7.10 7.10 B d adjacent side B We have a right-angled triangle. We are looking for side d (the adjacent side) Use the tan ratio: tan 7.1º = d= opposite 75 = adjacent d Notice that 7.1 is the same as 7 6' . 75 75 = = 602 metres to the nearest metre. tan 7.10 0.124556 Week 11 Page 4 Study Essential, Chapter 14.1 and the Examples 1, 2 and 3 Practice Exercise 1 (Answers at the end of this week) 1. 2. Use your calculator to change: (a) 20.6º into degrees and minutes (b) 42º 16' into degrees, to one decimal place (c) 14.8º into degrees and minutes (d) 65º 43' into degrees, to one decimal place. From the top of a vertical cliff 130 m high, the angle of depression of a buoy at sea is 18º. What is the distance of the buoy from the base of the cliff? 120 m T 3. A 120 m tall tower is on top of a hill. From the base of the hill, the angle of B elevation of the base of the tower is 9º, and 12º for the top of the tower. (a) Find TAB, ABC, TBA, 12 0 90 and ATC C A 4. 5. (b) Find distance AB, to the nearest metre. (c) Find distance BC, the height of the hill, to the nearest metre. A 2 m tall man standing at the edge of a 50 metre tall cliff sights two buoys with angles of depression 18º and 20º. (a) Draw a diagram. (b) Work out the distance between the buoys, correct to one decimal place. A surveyor notes a church spire at an angle of elevation of 18º. She walks 150 metres to a fence and notes a second angle of elevation of 33º. (a) (b) (c) 6. Draw a diagram. Find the height of the spire to the nearest half metre. Find the distance from the fence to the church, to the nearest half metre. Try questions 1, 3, 5, 6, 9,10,11 and 18 of Exercise 14A, Essential textbook. (Answers for these questions are at the back of your textbook. Contact your teacher if you need some help with solving these questions) Week 11 Page 5 LESSON 2 BEARINGS AND DIRECTIONS True Bearing Trigonometry is used to determine directions and to calculate distances. The direction of one object from another in a horizontal plane can be determined by using bearings. The bearings are usually quoted in terms of an angle measured clockwise from North (and sometimes a capital T) and are known as True bearings. In diagram 1, A is at 5 km from C on a bearing of 56º. This is a bearing of 056ºT. This bearing represents a direction of 56º (always measured in a clockwise direction) from North. In diagram 2, B is at a distance of 3.3 km from C on a bearing of 310º T. This is a bearing of 310ºT. This bearing represents a direction of 310º (again measured in a clockwise direction) from North. North B 3.3 km B North A 56 0 C 3100 5 km 3 km East C Clockwise direction A Diagram 1 Diagram 2 Compass Bearing Sometimes bearings are also quoted in terms of an angle measured East or West, or North or South. In diagram 1, 056ºT could be expressed as N 56º E or E 34º N. In diagram 2, the bearing 310ºT, can be expressed as W 40º N or N 50º W. A bearing of N56º E means that you are initially pointing in a Northern direction; you rotate 56º towards the East and then head off in that direction. A bearing of E34ºN means that you are initially pointing in a Eastern direction; you rotate 34º towards the North and then head off in that direction. When dealing with problems that involve bearings, always draw a set of axes at each reference point Week 11 Page 6 Example 1 N D A N 300 2100 O 3000 1200 O B C The bearing of A from O is 030º T Example 2 of B from O is 120º T The bearing The bearing of C from O is 210º T The bearing of D from O is 300º T Watch two video lessons on How to Find the Bearing! Insert your DECV course CD into your computer and then click: (i) Further Mathematics Unit 4 > Further Resources > Bearings and Angles (I) (ii) Further Mathematics Unit 4 > Further Resources > Bearings and Angles (II) Courtesy: www.mathsonline.com.au Example 2 Simon travels 6 km due North and then 8 km due East. N (a) What is the bearing of the finish from the start? (b) To travel directly back to the start 8 km Finish 6 km (i) (ii) What bearing will Simon need to follow? How far will he have to travel? Start Bearing Answers Draw a diagram. The bearing of the finish point from the starting point is the angle shown. We have the opposite and adjacent sides, so we use the tan ratio tan = opposite 8 = = 1.3333 adjacent 6 = tan–1 1.3333 = 53º 7' 48 = 53º 8’ (to nearest minutes). The bearing of the finish point from the starting point is: 053º T (to nearest degree). 8 ( opposite) 6 (adjacent) Week 11 Page 7 (b) (i) Draw a diagram to show the bearing we are looking for. To travel directly back to the start, Simon will follow the bearing shown. Bearing The bearing is (180º + a) a + b = 90º and we can find angle b from the triangle 1800 b a 538 b = 180º – 90º – 53º8' = 36º 52' 180º = 179º 60' a = 90º – 36º 52' 179º 60' – 143º 8' = 36º 52' = 89º 60' – 56º 52' Subtract degrees and minutes separately = 53º8' The bearing is 180º + 53º8'= 233º 8' Travelling back, Simon has to follow the bearing of 233º T (nearest degree) (ii) d is the distance travelled back to the start. Use Pythagoras’ theorem: 8 km 2338 d2 = 62 + 82 = 36 + 64 = 100 d = 100 = 10 km Simon has to travel 10 km back to his starting position. 6 km d Example 3 Tanya hikes 4 km on a bearing of 56º, then 3 km on a bearing of 35º (a) Draw a diagram showing her journey. (b) How far north is she from her starting point? Answers (a) Draw a diagram Finish 350 350 3 km 560 560 Start 4 km 900 350 = 550 900 560 = 340 5 Week 11 Page 8 (b) We have two right angled triangles The distance north of the starting point is (a + b) a a = 4 × sin 34º = 2.2368 4 b sin 55º = b = 3 × sin 55º = 2.4575 3 sin 34º = a + b = 2.2368 + 2.4575 = 4.6943 4.7 km (one decimal place). 3 km b 550 4 km 340 Tanya is 4.7 km north from the starting point. a Practice Exercise 2 (Answers at the end of this week) 1. Start Costa travelled 5 km due south, then 8 km due west. (a) (i) (ii) To travel directly back to the start what bearing will Costa need to follow? How far will he have to travel on this bearing? 5 km Finish (b) 2. 3. 4. 5. 8 km Tomas decides to take the shortest route from start to finish. What bearing should he follow? Finish Tamara travelled 6 km due west, then 10 km due north. (a) To travel directly back to the start, what bearing will she need to follow? (b) Shawn takes the shortest route from the start to the finish. What bearing did he follow? 10 km Samantha travelled 8 km on a bearing of 40º then 4 km on a bearing of 75º. (a) Draw a diagram of the journey. (b) How far north from the start was she? (c) How far east from the start was she? Trent travelled 2 km on a bearing of 125º then 9 km on a bearing of 30º. (a) Draw a diagram of the journey. (b) How far north of the start is he? Chris travels 5 km on a bearing of 320º, then 10 km on a bearing of 55º. (a) Draw a diagram of the journey (b) How far east of the start is he? Start Week 11 Page 9 MORE EXAMPLES OF BEARING LESSON 3 We’ll need to know some angle rules. There are two angle rules that we use in bearings N 300 300 60 N 0 0 60 0 0 300 30 1800 0 70 0 180 0 0 60 1 2 0 1200 110 Rule 1: Alternate angles are equal 0 1 8 0 0 Rule 2: Angles6 on a straight line add up0 to 1800 0 0 Let’s see some examples of how we use these rules. Example 1 The hiking club hikes 7 km on a bearing of 50° to the mineral springs, then 5 km on a bearing of 140° to the hut. (a) (b) (c) Springs Show that the angle at the springs is 90° How far will the hike back to the camp be? They are going to hike from the hut directly back to camp. What bearing should they follow? 500 7 km 1400 5 km Hut Camp Answers (a) Draw a diagram and work out the angles. Angle at the spring 140 500 0 1400 500 500 1800 –1400 = 400 50 (alternate angles) 180 140 40 (supplementary angles) 500 + 400 =900 50 40 90 Week 11 Page 10 (b) d is the distance back to the hut. d is the hypotenuse. 7 km Use Pythagoras’ theorem 5 km d2 = 72 + 5 2 = 49 + 25 = 74 d = 74 8.6 km (one decimal place) d The hike back from the hut to the camp is about 8.6km. (c) 1400 7 500 400 40 alternate 5 500 400 3600 400 = 3200 The bearing back to camp = 320° minus angle We can use trig ratio to find angle a since we have proven that the triangle is a right-angled triangle. We have all three sides, so we can use either sin, cos or tan. I’ll use tan because this uses the sides given in the question. opposite 7 tan = 1.4 adjacent 5 = tan 1 1.4 = 54° 28 The bearing back to camp = 320° 54° 28 = 319° 60 540 28 = 2650 32 Mt. Hederick Example 2 Tod took a bearing from the first hut, of Mt.Hederick and noted it was 250. He then walked 2 km on a bearing of 115° to the next hut. He took a second bearing of Mt Henderick, and noted that it was 330°. (a) (i) (ii) Show that the triangle is a right-angled triangle. Find the angle at Mt Hederick (b) (i) (ii) How far is the first hut from Mt. Hedrick? How far is the second hut from Mt. Hederick? (c) What is the bearing of the first hut from the second hut? 250 Hut 1 Hut 1 1150 2 km 330o Week 11 Page 11 Answers (a) Work out the angles (i) The angle at Mt. Hederick is 25° + 30° = 55° (ii) The angle at the first hut is 90o, hence the triangle is a right-angled triangle. 250 300 250 1150 – 250 = 900 3600 –3300 =300 300 3300 (b) (i) Distance d is from the first hut to Mt. Hedrick d tan 55° = opposite 2 hypotenuse d 1 d 2 tan 55 550 adjacent d is now on the top line 1 2=d tan 55 2 2 tan 55 1.4 km to one decimal place. d= tan 55 (ii) e 2 opposite e2 = d2 + 22 = (1.4)2 + 4 = 1.96 + 4 = 5.96 e = 5.96 = 2.4 km to one decimal place. (c) 550 Hut1 1800 –550 –900 = 35 4500 Hut 1 300 a Hut2 2700 Bearing of first hut from second hut = 2700 + a a Hut 2 2700 The bearing of the first hut from the second hut is (270° + a) Angle a = 90° 35° 30° = 90° 65° = 25° The bearing is 270° + 25° = 295° Week 11 Page 12 Still confused with Bearings? Watch these video lessons Insert your DECV course CD into your computer and then click: (i) Further Mathematics Unit 4 > Further Resources > Compass Question (I) (ii) Further Mathematics Unit 4 > Further Resources > Compass Question (II) Courtesy: www.mathsonline.com.au Practice Exercise 3 (Answers at the end of this week) 1. In the diagrams below work out (i) the angle at B (ii) the bearing of B from C (a) (b) B 110 600 B 800 0 1600 A C A 2. C B In the following diagram work out: (a) The angle at B (b) The bearing of C from B (c) The bearing of A from B (d) The bearing of A from C 3200 500 C A 800 N 3. The bearing of B from A is 120° C is Northeast of A C is due north of B. What is the bearing of (a) A from C? (b) A from B? (c) B from C? C N A 1200 B B M 4. Tom and Tina hike 10 km from their camp on bearing of 040°, then 8 km on a bearing of 130° (a) Work out the angles in the triangle. Show it is a right-angled triangle. (b) From the hut, what is the bearing of the camp? (c) How far is the camp from the hut? 1300 8 km 400 C 10 km H Week 11 Page 13 5. The walking group walks 6 km from P on a bearing of 060° to Q, then 9 km on a bearing of 150° to R. Q 1500 600 (a) Show that the angle at Q is a right angle. P 6 km 9 km (b) They walk directly from R to P: (i) How far is this, correct to one decimal place? (ii) What bearing do they follow? R 6. Carey starts at point A, and notes the bearing of the hill at B to be 50°. He hikes 5 km on a bearing of 075° to the hut at C. He takes a second bearing of the hill to be 320°. (a) Find the angles in the triangle, and show that it is a right angled triangle. B C (b) From the hill at B: (i) What is the bearing of A? (ii) What is the bearing of C? 0 50 3200 5 km A 750 (c) What is the bearing of A from C? (d) (i) How far is B from C (ii) How far is A from B? 7. Chris starts at point S and takes a bearing of 125 to point V The bearing of the tower T from S is 035 . He moves 8 km to V and takes a second bearing of the tower T from V. It is 330°. (a) Find the angles in the triangle and show that it is a right angled triangle. (b) From the tower at T: (i) what is the bearing of S? (ii) what is the bearing of V? (c) What is the bearing of S from V? (d) (i) How far is T from S? (ii) How far is V from T? T 350 S 1250 8 km V 3300 Week 11 Page 14 LESSON 4 TRIANGULATION Have you wondered how the surveyors can measure the distance between two mountains in the countryside or how we can determine how far is that sail ship in the sea. These two cases are examples of distances which cannot be measured directly because of physical restrictions. It is not possible to walk straight from one mountain to the next line nor to walk over the water to measure the distance to the ship. Triangulation is a mathematical technique used to measure such distances. This method involves the use of the theory of solving triangles. When we can use triangulation Triangulation can be used in the following situations: A single point can be observed but can’t be reached or measured due to physical restrictions. Two points can be observed and neither can be reached or measured directly due to physical restrictions. Surveying using triangulation involves using a base line of known length and measuring the angles of points of interest (which may be inaccessible) from both ends of the base line How we can use the triangulation method In both situations we need to have: Two references points from which we can record the bearing of our inaccessible point or points. The distance between our two chosen references points. The bearing of one of these references points from the other must also be measured. In this diagram point B is the inaccessible point, but we know the following: The distance between A and C. The bearing of A from C and C from A. The bearings of B from A and C With the information above, we can calculate the Distance AB &AC or even calculate the area of Triangle ABC In this diagram, we know the following: The distance between A and B. The bearing of A from B or of B from A. The bearings of C and D from A. The bearings of C and D from B Points D & C are inaccessible points, but with the information above, we can calculate distances AD, Example AC, BD and1BC B A 380 600 C b=8 Week 11 Page 15 1. Points A and B mark the position of two lighthouses. The lighthouse at A is 20 km north of the light house at B. The light keeper at A sees a ship on a bearing of 1450. The light keeper at B sees the same ship on a bearing of 600 N 1450 A 20 km (a) How far is the ship from each light house? S (b) The ship’s captain takes bearings of each light house. 600 (i) What is the bearing of the lighthouse at A from the ship? B (ii) What is the bearing of the lighthouse at B from the ship? Answers N A 1450 A 1800 1450 =350 350 20 b 1800 600 350 =850 850 S 600 B B a c sin A sin C (a) a sin 35 a= 20 sin 85 b c sin B sin C 20 sin 85 b sin 60 sin 35 11.5 km b= (b) (i) 20 sin 85 20 sin 85 sin 60 17.4 km (b) (ii) A 1450 From S, the bearing of 360° 35° = 325° The ship is 17.4 km from lighthouse A. The ship is 11.5 km from lighthouse B. the lighthouse A is: S 600 From S, the bearing of 350 350 350 lighthouse Bis: 350 S 360° 35° 85° = 240° 850 600 3600 350 =3250 B S Week 11 Page 16 Study Essential Chapter 14.1 ( Triangulation) and Example 6 Practice Exercise 4 (Answers for question one at the end of this week and answers for question 2 are at the back of your textbook. 1. A surveyor has measured the angles to An inaccessible point, from both ends of a base line AB Angle CAB equals 68 and angle ABC equals 82 . If AB =112 metres, Find: (a) The length of AC (b) The length of BC. (c) The area of triangle ABC. 2. Try the following questions from Exercise 14A, of your textbook: (i) Question 7, page 375. (ii) Question 12, page 376. (iii) Question 13, page 376. (iv) Question 15, page 376. (v) Question 17, page 376. By now, your supervisor should have received the materials for the second School-Assessed Coursework (SAC 2) for Unit 4. Please check with your supervisor. If he/she has not yet received them, he/she must contact your teacher as soon as possible. Week 11 Page 17 SEND: Work for Submission Show all your workings clearly. 1. Two ships are observed from point O. At a particular time their positions A and B are as shown on the right. The distance between the ships at this time is A. 3.0 km B. 3.2 km C. 4.5 km D. 9.7 km E. 2. 10.4 km [VCAA 2005 Further Maths Exam 1] The bearing of an aeroplane, X, from a control tower, T, is 055°. Another aeroplane, Y, is due east of control tower T. The bearing of aeroplane X from aeroplane Y is 302°. The size of the angle TXY is A. 32° [VCAA 2005 Further Maths Exam 1] B. 35° C. 55° D. 58° E. 113° 3. A hiker walks 4 km from A on a bearing of 30º to a point B, then 6 km on a bearing of 330º to a point C. The distance AC in km is A N C 4 sin 30 B 62 42 48 cos 120 C 62 42 48 cos 120 6 km B D 6 sin 60º E 52 4 km A Week 11 Page 18 4. Ship A and Ship B can both be seen from the lighthouse. Ship A is 5 km from the lighthouse, on a bearing of 028o. Ship B is 5 km from Ship A, on a bearing of 130o. (a) Two angles, x and y, are shown in the diagram. (i) Determine the size of the angle x in degrees. (ii) Determine the size of the angle y in degrees. (b) Determine the bearing of the lighthouse from Ship A. (c) Determine the bearing of the lighthouse from Ship B. 5. Starting from the camp at C, Tim takes a bearing of a mountain at M and notes it to be 25°. He then walks 5 km to the hut at H and takes a second bearingof the same mountain and it is 345°. (a) Work out the angles in the triangle CHM. Prove that it is a right angled triangle. (b) From the mountain at M: (i) what is the bearing of the camp? (ii) what is the bearing of the hut? (c) How far is it (i) from the camp to the mountain (ii) from the hut to the mountain? (d) Tim walks back to camp from the hut. What bearing does he follow? M 250 750 C 5 km H 3450 Week 11 Page 19 6. The base of a lighthouse D, is at the top of a cliff 168 metres above sea level. The angle of depression from D to a boat at C is 28o. The boat heads towards the base of the cliff, A, and stops at B. The distance AB is 128 metres. (a) What is the angle of depression from D to B, correct to the nearest minute? (b) How far did the boat travel from C to B, correct to the nearest metre? 7. Genie Construction is building a new shopping plaza on a plot of land that is a trapezium with the two parallel sides pointing north. The following is a diagram, which is not drawn to scale, of the plot: D N A 160 m 120 m B 230 m m C 2 3 0 Part 1 m (a) Find the area and the perimeter of the site ABCD. (b) Prove that the bearing of D from A is (to the nearest degree) 80 1 6 0 Week 11 Page 20 Part 2 A car park is to be made by running a straight line on a bearing of 25 until it meets the edge of the plot at E. D E A N 25 2 5 120 m B Find the area and the perimeter of the car park ABE. C Week 11 Page 21 SEND: Work for Submission – Exam Practice In Further Mathematics there are two end-of-year examinations. Examination 1 is a set of multiple-choice questions covering the core and the three modules. Geometry and Measurement is one of the modules. In Exam 1, there are a total of 40 questions to be completed in 90 minutes with nine of these questions covering the Geometry and Measurement module. Each question should take, on average, 2 minutes. One mark is given for each correct answer. In order to obtain practice working under such conditions, we suggest you complete the five multiple-choice questions below. Tear out this sheet and include it with the rest of your submission for this week. Restrict your time to 10 minutes only. It is not necessary to show your working as credit is given for correct answers only. Circle the letter beside the correct answer. 1 A man walks 4 km due east followed by 6 km due south. The bearing he must take to return to the start is closest to: A 034o B 056o C 304o D 326o E 346o 2 A boat sails at a bearing of 265o from A to B. return to A? A 005o B 085o C 090o D 355o E 275o What bearing would be taken from B to Week 11 Page 22 3 From a point on a cliff 200 m above sea level, the angle of depression to a boat is 40o. The distance from the foot of the cliff to the boat to the nearest metre is: A 238 m B 168 m C 153 m D 261 m E 311 m 4 A boat sails from a harbour on a bearing of 045o for 100 km. It then takes a bearing of 190o for 50 km. How far from the harbour is it, correct to the nearest km? A 51 km B 82 km C 66 km D 74 km E 3437 km 5 A hiker walks 3.2 km on a bearing of 120 and then takes a bearing of 055 and walks 6 km. His bearing from the start is: A 013 B 077 C 235 D 257 E 330 Week 11 Page 23 Answers to Exercises PRACTICE EXERCISE 1 1. (a) 20.6º = 20º 36' (c) 14.8º = 14º 48' (b) 42º 16' = 42.3º (d) 65º 43' = 65.7º 2. The distance of the buoy from shore = AB 130 tan 18º = AB 130 0.3249 = AB 130 AB = 0.3249 = 400.123 400 m. C 180 130 m 180 A B 3. (a) T 0 0 0 180 – 99 – 3 = 78 0 1800 – 810 = 990 B 1800 – 900 – 90 = 810 3 0 T 78 0 120 m 90 C (b) B A 120 = AB Sin A Sin 78 AB = 120 × sin 78º = 2242.77 2243 m sin 3 99 0 30 A B (c) BC sin 9º = 2243 BC = sin 9º × 2243 = 350.88 351 m Height of the hill is 351m. 2243 m 90 C A 4. (a) 1800 – 180 – 1600 = 20 D D 50 + 2 = 52 m A 52 m 20 0 20 0 18 0 B 1800 – 200 = 1600 C B C Week 11 Page 24 D (b) Find BD before finding AB, the distance between the buoys 52 52 BD = Sin 20º = BD = 152.0378 Sin 20 AB = BD Sin 2 Sin 18 AB = 152 .0378 × sin 2º = 17.17 17.2 metres Sin 18 Distance between the buyys is 17.2km 20 152.0378 m 18 0 A 160 0 B D D Top of Spire 5. (a) 1800 – 180 – 1470 = 150 18 0 18 0 A A 33 0 150 m B Brick fence (b) Find BD before finding DC, the height of the spire. BD = 150 Sin 15 Sin 18 BD = 150 × sin 18º = 179.0924 Sin 15 Referring to triangle BDC, sin 33º = DC BD DC sin 33º = 179 .0924 A DC = 179.0924 × sin 33º = 97.54 97.5 metres The height of the spire is 97.5 m 1800 – 330 = 1470 B 150 m C Church D 15 0 b a 18 0 b =150 147 0 B D (c) cos 33º = 179BC .0924 BC = 179.0924 × cos 33º = 150.199 150 metres to the nearest half metre 179.0929 B 33 0 C PRACTICE EXERCISE 2 1. (a) (i) The bearing required is angle b we find the angle in the triangle, then = 90º – tan = 85 = 0.625 = tan–1 0.625 = 32º 0' = 90º – 32º = 58º To travel back to the start, Costa needs to follow the bearing of 058º T (ii) Let d = distance travelled directly back to the start d2 = 82 + 52 = 89 d = 89 = 9.433 9.4 km Start d N 5 km Finish 8 km Week 11 Page 25 (b) Start Angle = 180º – 90º – 32º = 58º Bearing Bearing = 180º + 58º = 238º The bearing that Tomas has to follow from start to finish 320 is 238ºT 2 (a) To travel straight back to the start, follow a bearing shown by angle . Finish + = 180º = 180º – Find angle 10 km tan = 6 = 0.6 10 = tan–1 0.6 = 30º 58' Start Bearing = 180º – 30º 58' 6 km = 179º 60' – 30º 58' = 149º 2' The required bearing is 149º 2' (b) To travel directly from start to finish, Shawn has to follow the 30º 58' to follow the bearing shown Bearing = 270º + angle = 180º – 90º – 30º 58' = 59º 2' Bearing = 270º + 59º 2' = 329º 2' 3. (a) 750 4 km Finish 4 8 400 8 km 90º – 75º = 15º Start 90º – 40º = 50º (b) Distance travelled north = x + y 4 y 0 15 8 x 500 sin 50º = x 8 x = 8 × sin 50º = 6.12836 y sin 15º = 4 y = 4 × sin 15º = 1.03527 x + y =47.1636 7.2 km. Samantha is 7.2 km north from the starting point. Week 11 Page 26 (c) Distance travelled east = e + d 150 cos15º = e 4 e e = 4 × cos15º = 3.8637 8 d = 8 × cos50º = 5.1423 500 cos50º = d 8 d e + d = 9.006 9 km. Samantha is 9 km east from the starting point. 4. (a) 1250 1250 30 9 km 0 300 90º – 30º = 60º 180º – 125º = 55º 2 km (b) Distance travelled north from the start is (b – a) Finish 9 km Start a 550 sin 60º = b 9 ba b 2 km 600 a b = 9 × sin 60º = 7.794229 cos55º = a 2 a = 2 × cos55º = 1.147152 b – a = 6.647 6.6km Trent is 6.6 km north from the starting point 5 (a) Finish 10 km 550 Finish 5 km 550 3200 90º – 55º = 35º 320º – 270º = 50º Start 2700 Start (b) 10 km f cos35º = 10 f = 10 × cos35º = 8.19152 0 f cos50º = e 5 e = 5 × cos 50º= 3.213938 35 50 0 The distance travelled east from the start equals (f – e) f – e = 4.977582 5.0 km ( Chris is 5 km east from the e 5 km starting point) Week 11 Page 27 PRACTICE EXERCISE 3 1. (a) (i) (ii) 1800 1100 = 700 1100 B 1100 B 600 600 C C 1100 1800 A 600 Angle B = 60° + 70° = 130° (b) From C, bearing of B is 180° + 110° = 290° (i) (ii) 80 B B 0 160 1600 0 800 A 1800 1600 = 200 1600 1800 C Angle B = 80° + 20° = 100° 2. (a) From C, the bearing of B is : 180° + 160° = 340° Angle at B = 50° + 40° = 90° B B 400 500 + 400 = 900 360 320 = 40 0 500 500 0 0 C 3200 C 80 50 = 30 0 A 0 0 30 0 1800 900 300 = 600 A (b) B (c) 1800 B 500 180 40 = 140 0 0 0 500 400 C From B, the bearing of C = 140° 500 A From B, the bearing of A is: 180° + 50° = 230° Week 11 Page 28 B (d) From C, the bearing of A 360° 40° 60° = 260° 900 400 600 3. C 300 N A C N 450 450 1800 450 750 = 600 A 1200 450 = 750 B North east is 45° (b) From B, the bearing of A is: 360° 60° = 300°) (a) From C, the bearing of A is 180° + 45° = 225° C 1800 A 450 600 B A 4. (a) M 1800 C (c) From C, the bearing of B is 180° (due South) B 1300 H 400 H 400 C Angle at M = 40° + 50° = 90° (b) From Hut H, the bearing of camp C is: 360° 50° = 310° , Find angle . 10 tan = = 1.25 8 = tan 1 1.25 = 51° 20 Bearing = 310° 51° 20 = 258° 40 1300 M 500 500 H C Week 11 Page 29 M (c) HC2 = 102 + 82 = 164 HC = 164 = 12.8 km 8 km 10 km H C 5. (a) Q 600 1500 Q 1800 1500 = 300 600 6 km P 9 km P R Angle at Q = 60° + 30° = 90° R (b) (ii) Q 6 km 900 1500 Q 9 km 300 300 P P R R PR2 = 92 + 62 = 117 ( Pythagoras’ theorem) Bearing is 360° 30° = 330° PR = Find 117 = 10.8 km (one decimal place) 6 = 0.666 9 = tan 1 0.666 . . . = 33° 41 Bearing = 329° 60 33° 41 = 296° 19 They have to follow a bearing of 296° 19 tan = 6. (a) 500 500 400 900 C 320 A 5000 75 500= 250 B 3600 3200 = 400 B C 0 250 A 1800 900 250= 650 Week 11 Page 30 (b) (i) (ii) B 180 1800 400 = 1400 0 B 500 500 400 400 C A From B the bearing of A is 180°+ 50°= 230° (c) From B the bearing of C is 140° B 400 650 C 3600 400 650 = 2550 A From C the bearing of A is 255° (d) B 650 C 250 5 km A (i) cos 65° = BC 5 BC = 5 cos 65 = 2.1km 7. (a) (ii) sin 65° = AB 5 AB = 5 sin 65° = 4.5 km. T T 350 300 650 350 S 3600 3300 = 300 1250 350 = 900 S 900 1800 900 650= 250 8 km V 3300 V Week 11 Page 31 (b) (i) T 1800 300 = 1500 T (ii) 1800 350 350 S From T, the bearing of S is 180° + 35° = 215° From T, the bearing of V 180° 30° = 150° V (c) T 350 300 900 S 300 250 250 V From V, the bearing of S is 360° 25° 30° = 305° T (d) 8 tan 65° = ST 8 ST = = 3.7 km tan 65 8 sin 65° = TV 8 TV = = 8.8 km sin 65 650 S 8 km V Week 11 Page 32 PRACTICE EXERCISE 4 C 1. (a) Angle ACB = 180 68 82 30 To find AC, we use the sine rule: 30 AC AB sin B sin C AC 112 sin 82 sin 30 AC sin 30 112 sin 82 AC A 68 82 112m 112 sin 82 221.82 222m sin 30 (b) To find BC we can use the cosine rule: BC 2 AB 2 AC 2 2 AB AC cos 68 BC 2 112 2 222 2 2 112 222 0.3746 BC 2 43199.563 BC 43199.56 208m (c) Area of triangle ABC = = 1 2 1 2 AB AC sin 68 112 222 0.927 = 11526.7 m 2 2. Answers for these questions are at the back of your textbook. Checklist This week you should have submitted this work to me. 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