Mathematics Stage 5 MS5.1.2 Trigonometry Part 2 Applying trigonometry
Number: M43684 Title: MS5.1.2 Trigonometry This publication is copyright New South Wales Department of Education and Training (DET), however it may contain material from other sources which is not owned by DET. We would like to acknowledge the following people and organisations whose material has been used: Extracts from Mathematics Syllabus Years 7-10 Board of Studies, NSW 2002 Unit overview pp iiiiv, Part 1 p 3-4, Part 2 p 3-4 COMMONWEALTH OF AUSTRALIA Copyright Regulations 1969 WARNING This material has been reproduced and communicated to you on behalf of the New South Wales Department of Education and Training (Centre for Learning Innovation) pursuant to Part VB of the Copyright Act 1968 (the Act). The material in this communication may be subject to copyright under the Act. Any further reproduction or communication of this material by you may be the subject of copyright protection under the Act. CLI Project Team acknowledgement: Writer(s): James Stamell Editor(s): Dr Ric Morante Illustrator(s): Thomas Brown, Tim Hutchinson Desktop Publisher(s): Gayle Reddy Version date: May 2, 2005 All reasonable efforts have been made to obtain copyright permissions. All claims will be settled in good faith. Published by Centre for Learning Innovation (CLI) 51 Wentworth Rd Strathfield NSW 2135 Copyright of this material is reserved to the Crown in the right of the State of New South Wales. Reproduction or transmittal in whole, or in part, other than in accordance with provisions of the Copyright Act, is prohibited without the written authority of the Centre for Learning Innovation (CLI). State of New South Wales, Department of Education and Training 2005.
Contents Part 2 Introduction Part 2...3 Indicators...3 Preliminary quiz...5 Calculating sides...9 Finding the opposite side...9 Finding adjacent or hypotenuse...14 Calculating the hypotenuse...16 Choosing the ratio...21 General problems...25 Gradient of a line...28 Elevation and depression...31 Further elevation and depression...35 Reviewing trigonometry...39 Suggested answers Part 2...41 Exercises Part 2...45 Part 2 Applying trigonometry 1
2 MS5.1.2 Trigonometry
Introduction Part 2 This second part of trigonometry deals with deciding which of the three trigonometric functions to use in answering a question, and to solve problems where trigonometry can be applied. It is also extended to cover angles of elevation and depression. Throughout these notes diagrams are given. Occasionally incomplete diagrams are provided to encourage students to add relevant information to them that will aid them in answering the question. Indicators By the end of Part 2, you will have been given the opportunity to work towards aspects of knowledge and skills including: selecting and using appropriate trigonometric ratios in right-angled triangles to find unknown sides, including the hypotenuse selecting and using appropriate trigonometric ratios on right-angled triangles to find unknown angles correct to the nearest degree identifying angles of elevation and depression solving problems involving angles of elevation and depression when given a diagram. By the end of Part 2, you will have been given the opportunity to work mathematically by: solving problems in practical situations involving right-angled triangles interpreting diagrams in questions involving angles of elevation and depression Part 2 Applying trigonometry 3
relating the tangent ratio to the gradient of a line. Source: Extracts from outcomes of the Mathematics Years 7 10 syllabus <www.boardofstudies.nsw.edu.au/writing_briefs/mathematics/mathematics_ 710_syllabus.pdf > (accessed 04 November 2003). Board of Studies NSW, 2002. 4 MS5.1.2 Trigonometry
Preliminary quiz Before you start this part, use this preliminary quiz to revise some skills you will need. Activity Preliminary quiz Try these. 1 Use your calculator to find, correct to 3 decimal places, a b c sin 29 tan 10 cos 53 d 4 cos25 5 2 Find angles for which a b sin A = 0.563 tan B = 12 c cos C = 3 20 d tan α = 1 Part 2 Applying trigonometry 5
3 Use the cosine ratio to calculate the length of the side marked. 126 m G E 36 x F 4 Use the tangent ratio to calculate the length of the side marked. e 7.6 cm 38 5 Use the sine ratio to calculate the size of the angle between north and the diagonal line. N 5 6 E 6 MS5.1.2 Trigonometry
6 Use the tangent ratio to calculate the size of the angle marked. F 9.5 cm G h 15.3 cm H Check your response by going to the suggested answers section. Part 2 Applying trigonometry 7
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Calculating sides You have seen how to use the trigonometric ratios to find the acute angles of a right-angled triangle. Now you will see how they can be used to find the length of different sides in a right-angled triangle. Finding the opposite side The first example shows you how to calculate the side opposite an acute angle in a right-angled triangle. Follow through the steps in this example. Do your own working in the margin if you wish. Calculate the length, AB, correct to one decimal place. A x cm C 27 8 cm B Solution Mark the side you need to calculate, say x cm. AB is opposite the 27 angle, and the other known side, CB, is adjacent to this angle. So the ratio to use for the 27 angle is opposite AB. This is the tangent ratio: tan C = adjacent BC. Part 2 Applying trigonometry 9
Substituting in values, tan 27 = x 8 x = tan 27 8 x = 8 tan 27 x = 4.1 cm, correct to one decimal place. 8 tan 27 = Always make use of a triangle to answer trigonometry questions. Activity Calculating sides Try this. 1 Complete to calculate the value of x. x 20 7 tan 20 = x = tan x = tan = units (correct to one decimal place). Check your response by going to the suggested answers section. No matter which acute angle is given, you can always find the other if necessary so that you can use the angle opposite the wanted side. 10 MS5.1.2 Trigonometry
To find the other shorter side in a right-angled triangle: select the acute angle opposite the wanted side use the tangent ratio for this angle. Follow through the steps in this example. Do your own working in the margin if you wish. Write the equation you would use to find the value of x in the diagram. Find x to two decimal places. D 12 E 24 x F Solution The tangent ratio is used in this example. Since tan θ = opposite, you can make x the opposite side by adjacent referring to D. By doing it this way you have the unknown side (x) in the numerator. D = 90 24 = 66 tan D = opposite adjacent = EF DF x = tan 66 12 x = 12 tan 66 = 26.95 units (correct to two decimal places). In later sessions you will learn alternate ways to answer questions of this type. Part 2 Applying trigonometry 11
Activity Calculating sides Try these. 2 Complete to calculate the value of XY correct to two decimal places. (Side lengths are in centimetres.) Z=90 = tan Z = ZY tan 24 = 66 X z z = tan 24 = cm. Z 6 Y Check your response by going to the suggested answers section. So far the examples have involved the two shorter sides of a right-angled triangle. If you know the length of the hypotenuse (in a right-angled triangle) you can use the sine ratio to find the length of a side opposite a known angle. Follow through the steps in this example. Do your own working in the margin if you wish. In ABC, C = 90, B = 35 and AB = 7 cm (hypotenuse). Find the length of AC correct to 2 decimal places. A 7 B 35 C 12 MS5.1.2 Trigonometry
Solution AC is opposite B, so you use 35. sin 35 = opposite hypotenuse sin 35 = AC 7 AC = 7 sin 35 AC = 4.02 cm (correct to two decimal places) Don t forget to include the correct units in the answer. Leave your answer correct to the number of decimal places required. Activity Calculating sides Try this. 3 Complete to calculate the value of x correct to one decimal place. (Side lengths are in millimetres.) sin 67 = sin 67 = opposite hypotenuse x x x = sin = mm (correct to one decimal place) 67 35.4 Check your response by going to the suggested answers section. Part 2 Applying trigonometry 13
Here is a summary of the definitions of three trig functions that you will be using throughout this topic on trigonometry. In any right-angled triangle ABC with C = 90 opposite sin A = hypotenuse a c adjacent cos A = hypotenuse b c opposite tan A = adjacent a b B c a A b C c 2 = a 2 + b 2 Pythagoras theorem opposite sin B = hypotenuse b c adjacent cos B = hypotenuse a c opposite tan B = adjacent b a I remember the trig. functions sin, cos and tan and their ratios by saying SOH CAH TOA S O H C A H T O A Finding adjacent or hypotenuse When you know the length of the hypotenuse and wish to find the length of the side adjacent (or next to) the given angle you must use the cosine ratio. Follow through the steps in this example. Do your own working in the margin if you wish. Calculate the value of x correct to 2 decimal places. 14 MS5.1.2 Trigonometry
Solution cos B = adjacent hypotenuse 10 cos 43 = BC AB cos 43 = x 43 B 10 x x =10 cos 43 = 7.31 units (correct to 2 decimal places) A C Activity Calculating sides Try this. 4 Calculate the value of x correct to two significant figures. 20 cm 37 x cm Check your response by going to the suggested answers section. In this activity you should realise that as you are using the adjacent side and the hypotenuse, the trig function you need is cos. Part 2 Applying trigonometry 15
Calculating the hypotenuse How would you find the length of the hypotenuse in this triangle? (This is not an accurate drawing.) 30 A There are various methods: you may decide to draw a scale drawing and measure, if you are not particularly concerned about accuracy B 3 cm you may calculate the other angle as 60, use the tan ratio to find AC, and use Pythagoras Theorem c 2 = a 2 + b 2 to calculate AB you may use the ratio involving BC the opposite and AB the hypotenuse. C The example shows how you can use the third method, which is the easiest. Follow through the steps in this example. Do your own working in the margin if you wish. Calculate the length of AB in this triangle. A sin 30 = BC AB = opposite hypotenuse x 30 sin 30 = 3 x x sin 30 = 3 (multiply both sides by x) B 3 cm C 16 MS5.1.2 Trigonometry
Solution 3 x = sin 30 x = 6 cm (divide both sides by sin 30 ) Notice that in this problem, when you use sine, the unknown x appears in the denominator. To make x the subject in sin 30 = 3 x you multiplied by x, then divided by sin 30. But an easy short-cut is to swap the x and sin 30 around. sin 30 = 3 x 3 x = sin 30 Activity Calculating sides Try this. 5 Calculate the value of y correct to two significant figures. 18 cm y 60 Part 2 Applying trigonometry 17
Check your response by going to the suggested answers section. In the above activity the sides involved are the opposite and hypotenuse. These are the two sides needed for the sine ratio. Did you use the sine ratio in the activity? In the next activity the sides involved are the adjacent and hypotenuse. Which trig ratio will you use? Activity Calculating sides Try this. 6 Calculate the value of z correct to three significant figures. 12.8 m 25 z Check your response by going to the suggested answers section. 18 MS5.1.2 Trigonometry
Always use the correct trig. ratio and a diagram of a right-angled triangle. The triangle does not need to be drawn to scale. S O H C A H T O A You have been practising calculating one of the sides in a right-angled triangle. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 2.1 Calculating sides. In later sessions you will use the knowledge you have gained in trigonometry to answer questions in practical situations. Part 2 Applying trigonometry 19
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Choosing the ratio Until now you have been directed to the trig function (sine, cosine, or tangent) you needed to use. You have now reached a stage where you should be able to decide which of these three trig ratios is the appropriate one to use in any given situation. sin opp hyp SOH CAH TOA adj cos hyp opp tan adj hypotenuse * adjacent opposite For example, if the opposite side and hypotenuse are being used, the trig. ratio you want is sine. Follow through the steps in this example. Do your own working in the margin if you wish. a Calculate the angle, a, correct to the nearest degree. 12 cm a A 8 cm b B Calculate z correct to one decimal place. C X 17 cm z Z 25 Y Part 2 Applying trigonometry 21
Solution a The two sides involved are the adjacent (8 cm) and hypotenuse (12 cm). Therefore use cosine. cos a = 8 12 a = 48.189685 = 48 (correct to the nearest degree) use SHIFT cos b This time the two sides involved are the opposite (17 cm) and the hypotenuse (z). Sine is needed. sin 25 = 17 z 17 z = sin 25 = 81.8 cm (correct to one decimal place) In each case, look over your answers to check if they are reasonable. Activity Choosing the ratio Try these. 1 Use the appropriate trigonometric ratio to calculate x in each triangle. a 50 80 mm x 22 MS5.1.2 Trigonometry
b x 60 100 m 2 Calculate the size of the marked angle, correct to the nearest degree. a 3 8 α b 41 α 9 Part 2 Applying trigonometry 23
c 7 5 α β Check your response by going to the suggested answers section. You have been practising choosing and using the correct trig ratio in a right-angled triangle. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 2.2 Choosing the ratio. 24 MS5.1.2 Trigonometry
General problems You have learned how to use the trigonometric ratios of an acute angle in a right-angled triangle in two different ways. To find the size of the acute angles of the triangle To find the length of the side given one of the acute angles. You will now use this knowledge to solve simple problems. In each case you will need to use a labelled diagram to represent the situation. In this set of notes a diagram will always be given, but you may want to add more information to it. Follow through the steps in this example. Do your own working in the margin if you wish. Merill skis 132 metres down an even slope and drops 20 m in height. What angle (to the nearest degree) does the slope make with the horizontal? 132 m 20 m α Part 2 Applying trigonometry 25
Solution The diagram helps you identify that the sine ratio is required for this question. sin α = 20 132 α = 8.715 = 9 (to the nearest degree) sin = opp. hyp. Try this calculation. Did you obtain the same answer? Activity General problems Try these. 1 A ladder has its foot 1.3 metres from the base of a wall on level ground. The ladder makes an angle of 75 with the horizontal. Calculate the length of the ladder. y 75 1.3 m 26 MS5.1.2 Trigonometry
2 A boat is at anchor 500 m from the foot of a cliff 40 m high. Calculate the angle, θ, which the line of sight to the top of the cliff makes with the horizontal. not drawn to scale 40 m 500 m θ Check your response by going to the suggested answers section. In each of these examples you can see that, regardless of the situation, a right-angled triangle is involved. Once you have identified that triangle you then need to decide which of the three trig functions: sine cosine or tangent you will use. Use a diagram as a visual representation of the problem. It does not need to be drawn to scale. Check that the answer you obtain looks reasonable for the problem you are solving. It is very easy to mistakenly choose the wrong trig ratio, or press the wrong key on your calculator. Part 2 Applying trigonometry 27
Gradient of a line In co-ordinate geometry you established that the slope of the line, m, is given by m = rise run. Also from trigonometry, tan θ = opposite and so, from the adjacent triangle shown, tan θ = rise run. y θ run rise This means that m = tan θ. O x The angle in the triangle is the same as the angle between the line and the positive (right-hand) side of the x-axis. Can you see why? (Think of parallel lines, and corresponding angles.) So the gradient of a line is simply the tan of the angle it makes with the positive side of the x-axis. Follow through the steps in this example. Do your own working in the margin if you wish. What angle does the line y = 1 x +1 make with the x-axis? 2 Solution Make a small table of values to draw the line. x 1 0 1 2 y 0.5 1 1.5 2 28 MS5.1.2 Trigonometry
Now draw the line on the x-y plane and use it to determine the gradient. 3 y 2 1 1 θ 2 θ 2 1 0 1 2 3 1 The gradient, m = 1 2, and so tan θ = 1 2. Using your calculator, θ = 27 (to the nearest degree). You can use any two points on the line to establish rise and run. x Activity General problems Try these. 3 Draw the line y = 2x 3 on the co-ordinate plane and use it to determine the angle the line makes with the x-axis. 3 2 1 y 1 0 1 2 3 4 5 6 1 2 3 x Check your response by going to the suggested answers section. Part 2 Applying trigonometry 29
Check that you can now solve general problems using right-angled triangle trigonometry by yourself. Go to the exercises section and complete Exercise 2.3 General problems. 30 MS5.1.2 Trigonometry
Elevation and depression If someone is looking at an object, the straight line from the eye of the observer to the object is the line of sight. line of sight If the object is above the eye level of the observer, then the angle they raise their eyes, from the horizontal to the line of sight of the object, is called the angle of elevation. line of sight angle of elevation horizontal at eye level If the object is below the eye level of the observer, then the angle they look down, from the horizontal to the line of sight of the object, is called the angle of depression. horizontal at eye level angle of depression line of sight Note: the angle of elevation or depression of one point from another is always measured from the horizontal. Part 2 Applying trigonometry 31
Since these horizontal lines are parallel, the angle of elevation must be equal to the angle of depression, because they are alternate angles formed by parallel straight lines. horizontal angle of depression angle of elevation horizontal You can use angle of elevation or angle of depression to answer questions involving trigonometry. Follow through the steps in this example. Do your own working in the margin if you wish. From a plane travelling at a height of 6 km, the angle of depression of an airfield is found to be 14. How far must the plane fly to be directly above the airfield? P 14 angle of depression (directly above the airfield) Q 6 km ground level A (airfield) Solution PQA = 90 because the plane flies horizontally. At Q, it is directly above A. PQA is the right-angled triangle. You want to calculate the length PQ. 32 MS5.1.2 Trigonometry
tan P = AQ PQ 6 tan 14 = PQ 6 PQ = tan 14 = 24.065 (The two shorter sides involve using the tangent ratio.) The plane must fly 24 km (to the nearest kilometre). When answering questions like this, give your answer to an appropriate number of significant figures. Correct to the nearest kilometre for this question is reasonable, but writing 24.06469 km is not. Activity Elevation and depression Try these. 1 From the top of a cliff 20 m high the angle of depression of a ship at sea is 15. Calculate the distance from the ship to the foot of the cliff. 15 75 C 20 m S 15? F Part 2 Applying trigonometry 33
2 A flagpole of height 10 metres casts a shadow of length 16 metres on level ground. Calculate the angle of elevation of the sun at this time. F 10 m D x 16 m E Check your response by going to the suggested answers section. Notice that the angle of depression, in the first question, is the angle outside the triangle. It is the angle measured downwards from the horizontal. You could use alternate angles, and work with CSF inside the triangle. Or you could calculate the complement of 15 (which is 75 ) and work with FCS. The choice is yours. The two angles are equal. 15 75 C The horizontal lines are parallel, and the angles are alternate. The two angles add up to 90. 20 m S 15? F Either way, the answer you will arrive at is the same. Go to the exercises section and complete Exercise 2.4 Elevation and depression. 34 MS5.1.2 Trigonometry
Further elevation and depression When looking up at an object, the angle is an angle of elevation. When looking down at an object, you have an angle of depression. horizontal at eye level of the man angle of depression line of sight angle of elevation horizontal at eye level of the dog Angles of elevation and depression are always measured between the line of sight and the horizontal. Sometimes you are not given a labelled diagram to assist you. So you need to visualise the situation yourself and draw a diagram to help you get the picture clear. Then you can see what triangle to use. So you have three main steps in solving trigonometry questions: draw a diagram to represent the situation locate a right-angled triangle in the diagram use trigonometry to calculate the side or angle needed. Part 2 Applying trigonometry 35
Follow through the steps in this example. Do your own working in the margin if you wish. The angle of elevation of the top of a flagpole from an observer is 39 when measured 12 m away from the flagpole. The distance from the ground to the observer s eyes is 1.8 m. Calculate the total height of the flagpole. Solution A B 39 C E 12 m D You need to calculate the total height of the flag above the ground (that is, AD). To find AC, you use the tangent ratio in ABC. To find the actual height of the flagpole, you add CD to AC. CD = BE (the height of the observer) = 1.8 m AD = AC + BE In ABC, tan B = opposite adjacent tan 39 = AC 12 AC = 12 tan 39 = 9.7 m (correct to 1 decimal place) AD = 9.7 + 1.8 = 11.5 m 36 MS5.1.2 Trigonometry
The actual height of the flagpole is 11.5 m (correct to one decimal place). Sometimes it may be necessary to add, or subtract, values from the ones you calculate using trigonometry to arrive at the answer. Even when a diagram is provided, it may not include all the information. Feel free to write on the diagram and include other information that may help you answer the question. Finally, look over your answer and ask yourself whether it looks reasonable for the problem. In the next activity you will be provided with an incomplete diagram. Mark on it necessary values so you can answer the questions. Activity Further elevation and depression Try these. 1 From the top of a building 60 m high the angle of depression of a parked car is found to be 28. Complete the diagram and calculate the distance the car is parked from the building. Part 2 Applying trigonometry 37
2 Complete the diagram and use it to calculate the angle of elevation (to the nearest degree) of the top of a wall 15 m high from a point on the ground 5 m from the bottom of the wall. Check your response by going to the suggested answers section. You have been practising further examples of elevation and depression. Now check that you can solve these kinds of problems by yourself. Go to the exercises section and complete Exercise 2.5 Further elevation and depression. 38 MS5.1.2 Trigonometry
Reviewing trigonometry There is an enormous number of applications for trigonometry. Of particular interest is the technique of triangulation that is used in astronomy to measure the distance to nearby stars, in geography to measure distances between landmarks, and in satellite navigation systems. But regardless of where it is used, the basic definitions of sine, cosine and tangent are the same. sin opp hyp SOH CAH TOA adj cos hyp opp tan adj hypotenuse adjacent opposite In this session you will practice the ideas you learned in using trigonometry in right-angled triangles. The following exercises will help you to consolidate trigonometry applications. Go to the exercises section and complete Exercise 2.6 Reviewing trigonometry. Part 2 Applying trigonometry 39
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Suggested answers Part 2 Check your responses to the preliminary quiz and activities against these suggested answers. Your answers should be similar. If your answers are very different or if you do not understand an answer, contact your teacher. Activity Preliminary quiz 1 a 0.485 b 0.176 c 0.602 d 0.725 2 a 34 b 85 c 81 d 45 3 cos 36 = x 126 x = 126 cos 36 = 101.9 m 4 tan 38 = e 7.6 e = 7.6 tan 38 = 5.94 cm 5 sin x = 5 6 x = 56 6 tan h = 9.5 15.3 h = 32 Activity Calculating sides 1 tan 20 = x 7 x = tan 20 7 x = 7 tan 20 = 2.55 units Part 2 Applying trigonometry 41
2 Z = 90 66 = 24 tan Z = XY ZY tan 24 = z 6 z = 6 tan 24 = 2.67 cm. 3 sin 67 = opposite hypotenuse sin 67 = x 35.4 x = 35.4 sin 67 = 32.6 mm 4 cos 37 = x 20 x = 20 cos 37 = 16 cm 5 sin 60 = 18 y 18 y = sin 60 = 21 cm 6 cos 25 = 12.8 z 12.8 z = cos 25 = 14.1 m Activity Choosing the ratio 1 a tan = opposite adjacent x tan 50 = 80 x = 80 tan 50 =95.3 mm 42 MS5.1.2 Trigonometry
b cos = adjacent hypotenuse x cos 60 = 100 x = 100 cos 60 = 50 m 2 a sin α = 3 8 α = 22 b cos α = 9 41 α = 77 c tan α = 5 7 α = 36 You can use trig ratios to calculate β. Alternatively, β = 90 36 = 54 Activity General problems 1 cos = adjacent hypotenuse cos 75 = 1.3 y 1.3 y = cos 75 = 5.02 The ladder is 5 m long. 2 tan = opposite adjacent tan θ = 40 500 θ = 85.41 = 85 (correct to the nearest degree) 3 The gradient, m = 2 (so tanθ = 2 ) and the angle the line makes is 63. Part 2 Applying trigonometry 43
Activity Elevation and depression 1 tan 15 = 20 SF 20 SF = tan 15 = 74.6 m (correct to 1 dec. pl.) 2 Let the angle be θ. tan θ = 10 16 θ = 32 (to the nearest degree) Activity Further elevation and depression 1 tan 28 = 60 d 60 d = tan 28 = 112.8 m 28 60 m (correct to 1 dec. pl.) The car is parked 112.8 m from the building. 2 Let the angle be θ. 28 d tan θ = 15 5 θ = 72 15 m (to the nearest degree) The angle of elevation is 72. θ 5 m 44 MS5.1.2 Trigonometry
Exercises Part 2 Exercises 2.1 to 2.6 Name Teacher Exercise 2.1 Finding the opposite side 1 Calculate the value of the pronumeral, correct to one decimal place. (All lengths are in metres.) a a 4 18 b c 30 4 Part 2 Applying trigonometry 45
c d 2.7 70 d 8.4 e 70 e f 42 4 46 MS5.1.2 Trigonometry
2 Use the sine ratio to calculate the value of the pronumeral in each of these. Show all working. Give your answers correct to 3 significant figures. (All lengths are in centimetres.) a y 12 60 b m 20 9.7 c 52 20.69 b Part 2 Applying trigonometry 47
d 22 99.6 n e R h cm P 64 6 cm Q 48 MS5.1.2 Trigonometry
3 Find the value of x in each of the following correct to 2 decimal places. (All lengths are in centimetres.) a 35 x 12 b x 55 4.7 c x 45 11 Part 2 Applying trigonometry 49
d 3.4 74 x 2 Calculate the length of the hypotenuse in the following triangles. (All lengths are in metres.) a 15 x 38 b a 40 30 50 MS5.1.2 Trigonometry
c 25.6 y 55 d 260 x 41 Part 2 Applying trigonometry 51
Exercise 2.2 Choosing the ratio 1 Choose a suitable trigonometric ratio and then calculate x (measured in centimetres), correct to two decimal places. a 30 400 x b x 40 75 c 8 x 25 52 MS5.1.2 Trigonometry
d B 80 A 48 x C e 28 x 59 2 Calculate the value of α (alpha) in degrees in the following triangles. a 7 α 10 Part 2 Applying trigonometry 53
b 6 α 2 c α 36 9 54 MS5.1.2 Trigonometry
Exercise 2.3 General problems 1 A ladder reaches 2.5 m up a vertical wall, and has its foot on level ground 1 m from the base of the wall. Find the angle the ladder makes with the ground. 2.5 m θ 2 A 30 60 90 set square has the side opposite the 60 angle 12 cm long. Find the length of the longest side. (Give your answer correct to the nearest millimetre. Hint: label the longest side with a pronumeral.) 30 12 cm 60 Part 2 Applying trigonometry 55
3 The diagonal BD of a rectangle ABCD makes an angle of 27 with the side AB, which is 7.0 cm long. Calculate the length of the other side, correct to two significant figures. A 7.0 cm 27 B D C 4 To calculate the width of a river, a man stands at a point A directly opposite a tree T on the edge of the opposite bank. He walks 120 m along his bank to another point B and measures ABT = 27. Calculate the width of the river. (The banks are straight and parallel along this section of the river.) A 120 m 27 B T 56 MS5.1.2 Trigonometry
5 O is the centre of the circle. OD = 3.28 cm, and AOD = 49. Calculate the radius of the circle. O r 49 3.28 cm A D B 6 The pitch of a roof is a measure of its steepness. It is often expressed as the ratio of the rise to the run. Drawn to scale rise run θ Use a ruler to measure the rise and run on this scale diagram. Use this information to calculate the pitch angle, θ. Part 2 Applying trigonometry 57
Exercise 2.4 Elevation and depression 1 From a boat out at sea, Chandra measured the angle of elevation of the top of a cliff 65 m high to be 27. How far is the boat from the foot of the cliff (to the nearest metre)? 65 m 27 2 Kimberley found the angle of depression of the base of a statue in a nearby park to be 65 from a 35 m high window. How far from the building is the statue? (Answer correct to the nearest metre.) 65 35 m 58 MS5.1.2 Trigonometry
3 The altitude of the sun is the angle of elevation of the sun. It is the angle between the horizontal and the sun s rays. A building casts a shadow 40 m long when the altitude of the sun is 50. Find the height of the building, correct to the nearest metre. 50 40 m 4 Calculate the angle of depression of Mrs Lee s cottage which is 1.6 km away down a slope from the top of a hill 225 m higher. 225 m 1. 6 km Part 2 Applying trigonometry 59
5 Find the elevation of the sun when Olav, who is 1.9 m tall, casts a shadow 2.6 m long. (Answer to the nearest degree.) 1.9 m α 2.6 m 6 (Harder) You will need to draw a diagram for this one. The angle of elevation of an aircraft flying at 800 metres above ground level is 64 from a point on the ground. How far is the aircraft from this point? 60 MS5.1.2 Trigonometry
Exercise 2.5 Further elevation and depression In each of these only a partial diagram is provided. Label the diagram appropriately to help you answer the question. 1 From the top of a vertical cliff 40 m above sea level the angle of depression of a boat is measured to be 18. Find the distance from the boat to the bottom of the cliff. 18 40 m 2 An upright stick casts a shadow of length 2 m on level ground. If the stick is 1 m long, find the angle of elevation of the sun at this time. Part 2 Applying trigonometry 61
3 From a plane flying at a height of 4 km above a town A, the angle of depression of a town B is found to be 20. Find the distance between the two towns, correct to one tenth of a kilometre. Town A Town B 4 A tent pole is 2.2 metres high. Calculate the angle of elevation of the top of the pole given the top of the pole is secured to the ground with a rope 3.5 m long. 62 MS5.1.2 Trigonometry
5 A boat is at anchor where the top of the 25 m anchor chain is 16 m above the seabed. Calculate the angle of depression of the anchor chain. 6 The Eiffel Tower was built for the Universal Exhibition in celebration of the French Revolution and opened in 1889. Write a trigonometry question you could answer using the information on this diagram. (You do not need to answer the question.) 324 m 200 m θ Part 2 Applying trigonometry 63
Exercise 2.6 Reviewing trigonometry 1 Calculate the lengths or angles marked. a 4 cm x 30 b F d cm D 50 cm 35 E c g 14.6 9.5 64 MS5.1.2 Trigonometry
d A α 7 mm B 12 mm C e 5.7 mm j 38 2 A boy is flying a kite on 45 metres of line which makes an angle of 44 with the horizontal. How much higher is the kite than the boy s hand? (Answer to the nearest metre.) 45 m h 44 Part 2 Applying trigonometry 65
3 A fishing boat is held at anchor in the water by 23 metres of chain. The depth of the water is 18 metres. Find the angle the chain makes with the sea floor. 23 18 4 Calculate the angle the line makes with the x-axis. 3 y 2 1 2 1 0 1 2 3 1 x 66 MS5.1.2 Trigonometry
4 m 5 One end of a 10-metre rope is tied to the bow of a yacht and the other end to a point on the edge of a jetty. The rope is taut (stretched tightly) and makes an angle of 21 with the horizontal. How far out is the bow from the wharf? 10 m 21 6 A 14-metre fire engine ladder has its foot 4 m away from the side of a building. Calculate the angle the ladder makes with the wall. 14 m Part 2 Applying trigonometry 67
7 A fish notices a bug on the surface of the water 55 cm in front of it, and 65 cm above it. 65 cm 55 cm a Calculate the angle of elevation of the bug from the fish. b How far away is the bug from the fish? 68 MS5.1.2 Trigonometry