Rightangled triangles and trigonometry

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1 Rightangled triangles and trigonometry 5 syllabusref Strand: Applied geometry eferenceence Core topic: Elements of applied geometry In this cha 5A 5B 5C 5D 5E 5F chapter Pythagoras theorem Shadow sticks Calculating trigonometric ratios Finding an unknown side Finding angles Angles of elevation and depression
2 158 Maths Quest Maths A Year 11 for Queensland Introduction Thousands of years before Christ, ancient civilisations were able to build enormous structures like Stonehenge in England and the pyramids in Egypt. How did they do it? Human ingenuity allows us to achieve many things that at first seem impossible. How do we determine the height of a tall object without physically scaling it? Sometimes, even tasks that appear to be very simple (for example, planning and designing a staircase) present us with unexpected problems. How in fact do we go about the practical task of planning and designing a staircase? The theorem of Pythagoras and the geometry of rightangled triangles (trigonometry) can be employed to answer these questions. As you progress through this chapter, solutions to problems such as these will become clear. To put the mathematics of Pythagoras into perspective, let us first consider the times in which he lived.
3 Chapter 5 Rightangled triangles and trigonometry 159 History of mathematics PYTHAGORAS OF SAMOS (circa 580BC 500BC) During his life: Taoism is founded KungFutse (Confucius) is born Buddhism is founded. Pythagoras was a famous Greek mathematician and mystic but is now best known for his theorem about the sides of a triangle. He was born on Samos Island. It is believed that he was born about 580 BC and died about 500 BC, but because of the way dates were recorded then, various dates are given for his life. Not much was known about his personal life but it was known that he had a wife, son and daughter. When Pythagoras was a young man, he travelled to Egypt and Babylonia (Mesopotamia) where he learned much of his mathematics and developed an interest in investigating it further. He founded a cult with the idea that the essence of all things is a number. This group believed that all nature could be expressed in terms of numbers. They found, however, that some numbers could not be expressed as rational numbers, such as 2. They kept this information to themselves and there is a story that they killed one member who told somebody else about this fact. Pythagoras showed that musical notes had a mathematical pattern. He stretched a string tightly and found that it produced a certain sound and then found that if he halved the length of the string, it produced a sound that was in harmony with the first. He also found that if it was not exactly half, or a multiple of a half then it was a clashing sound. This approach is still used in musical instruments today. He is credited with the discovery now known as Pythagoras theorem. This states that For a rightangled triangle, the square of the hypotenuse (long side) is equal to the sum of the squares of the two short sides and can be written as c 2 = a 2 + b 2 using the diagram below. Other people knew of this idea long before Pythagoras. There is a Babylonian tablet known as Plimpton 322 (believed to have been made about 1500 years before Pythagoras was born), c b which has a set of values of the type known as Pythagorean triples. a Some examples of Pythagorean triples are: 3, 4 and 5 8, 15 and 17 and a more difficult example: 20, 99 and 101. For his investigations and demonstrations with right angles, Pythagoras used a string in which he had tied knots. Questions 1. Where was Pythagoras born? 2. Where did he travel to and learn most of his mathematics? 3. What is the formula for his famous theorem? 4. What did he start to investigate about patterns in nature? 5. What is the name of the ancient tablet that contains Pythagorean triples? 6. What is a Pythagorean triple?
4 160 Maths Quest Maths A Year 11 for Queensland 1 Name the hypotenuse in each of the following triangles. a A b D E c C H B d K F G I J L 2 State Pythagoras theorem. 3 Calculate the unknown lengths in the following rightangled triangles. a b c 6 m a 8 m 4 Complete each of the following to form ratios that are equivalent to the ratio 3 : 4 : 5. a 9 : 12 : b 1.5 : : 2.5 c : 1 : 1.25 d 7.5 : 10 : 5 The diagram at right depicts a tree in a field on a sunny day. Copy the diagram. a Draw the position of the shadow cast by the tree. b Mark the angle of elevation of the sun. 6 Find the value of the unknown angles in the following triangles. a a b 27 6 m 4 m b 20 cm 34 7 Express the following angles in degrees, minutes and seconds. a 35.5 b c Express the following angles in degrees, correct to 2 decimal places. a b c Define the trigonometric ratios: a sine b cosine c tangent. b
5 Chapter 5 Rightangled triangles and trigonometry Copy the diagrams below and label the sides as Opposite, Adjacent or Hypotenuse with respect to the marked angle. a b c 11 Using your calculator, determine the following (correct to 4 decimal places). a sin 45 b cos 30 c tan 22.5 d sin e tan Copy the following diagrams and mark the angles of elevation or depression, from the observer to the object, on each. a b Observer Ship Sea c Ground Bird Observer 13 Solve the following for x. 5.2 x a = b = x Ground
6 162 Maths Quest Maths A Year 11 for Queensland Rightangled triangles The properties of rightangled triangles enable us to calculate lengths and angles which are sometimes not able to be measured. The first property we will investigate is Pythagoras theorem. Pythagoras theorem Pythagoras theorem allows us to calculate the length of a side A of a rightangled triangle, if we know the lengths of the hypotenuse other two sides. Consider LABC at right. AB is the hypotenuse (the longest side). It is opposite the right angle. C B Note that the sides of a triangle can be named in either of two ways. 1. A side can be named by the two capital letters given to the vertices at each end. This is what has been done in the figure above to name the hypotenuse AB. 2. We can also name a side by using the lowercase letter of the opposite vertex. In the figure above, we could have named the hypotenuse c. WORKED Example 1 Name the hypotenuse in the triangle at right. P THINK 1 The hypotenuse is opposite the right angle. 2 The vertices at each end or the lowercase letter of the opposite vertex can be used to name the side. WRITE Q The hypotenuse is QR or p. R GC program Pythagoras Pythagoras calculations Consider the rightangled triangle ABC (below) with sides 3 cm, 4 cm and 5 cm. Squares have been constructed on each of the sides. The area of each square has been calculated (A = S 2 ) and indicated. Note that the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. 25 cm 2 = 16 cm cm 2 Alternatively: (5 cm) 2 = (4 cm) 2 + (3 cm) 2 Which means: hypotenuse 2 = base 2 + height 2 A = 9 cm 2 A C 5 cm 3 cm 4 cm A = 16 cm 2 A = 25 cm 2 B
7 Chapter 5 Rightangled triangles and trigonometry 163 This result is known as Pythagoras theorem. Pythagoras theorem states: In any rightangled triangle, the square of the hypotenuse is equal to the sum of the squares of the two shorter sides. That is, Hypotenuse 2 = base 2 + height 2 This is the formula used to find the length of the hypotenuse in a rightangled triangle when we are given the lengths of the two shorter sides. WORKED Example 2 Find the length of the hypotenuse in the triangle at right. 8 cm THINK WRITE 15 cm 1 Write the formula. Hypotenuse 2 = base 2 + height 2 2 Substitute the lengths of the shorter c 2 = sides. 3 Evaluate the expression for c 2. = = Find the value of c by taking the square c = 289 root. = 17 cm c EXCEL Pythagoras Spreadsheet In this example, the answer is a whole number because we are able to find 289 exactly. In most examples this will not be possible. In such cases, we are asked to write the answer correct to a given number of decimal places. By rearranging Pythagoras theorem, we can write the formula to find the length of a shorter side of a triangle. Since hypotenuse 2 = base 2 + height 2 it follows that base 2 = hypotenuse 2 height 2 and height 2 = hypotenuse 2 base 2 The method of solving this type of question is the same as in the previous example, except that here we use subtraction instead of addition. For this reason, it is important to look at each question carefully to determine whether you are finding the length of the hypotenuse or one of the shorter sides. WORKED Example 3 Find the length of side PQ in triangle PQR, correct to 1 decimal place. 16 m R 9 m Pythagoras Mathcad THINK WRITE P r 1 Write the formula. Base 2 = hypotenuse 2 height 2 2 Substitute the lengths of the known r 2 = sides. 3 Evaluate the expression. = = Find the answer by finding the square r = 175 root. = 13.2 m Q
8 164 Maths Quest Maths A Year 11 for Queensland Because Pythagoras theorem applies only in rightangled triangles, we can use the theorem to test whether a triangle is rightangled, acuteangled or obtuseangled. If we assume, as a starting point, that the triangle is rightangled, we could regard the longest side as the hypotenuse. We could then calculate values for hypotenuse 2, base 2 and height 2. Interpreting the relationships between these three values, we could conclude the following. If hypotenuse 2 = base 2 + height 2 the triangle is rightangled. If hypotenuse 2 > base 2 + height 2 the triangle is obtuseangled (since the hypotenuse is longer than it should be for that particular base and height). If hypotenuse 2 < base 2 + height 2 the triangle is acuteangled (since the hypotenuse is too short for that particular base and height). Determine whether the triangle shown is rightangled, acuteangled or obtuseangled. THINK 1 Assume that the longest side is a hypotenuse. 2 Calculate the hypotenuse 2 and base 2 + height 2 separately. 3 4 WORKED Example 4 Write down an equality or inequality statement. Write a conclusion. WRITE 4 cm 5 cm 7 cm hypotenuse 2 = 7 2 = 49 base 2 + height 2 = = = 41 hypotenuse 2 > base 2 + height 2 Therefore the triangle is obtuseangled. Pythagorean triads (or Pythagorean triples) are sets of 3 numbers which satisfy Pythagoras theorem. The first rightangled triangle we dealt with in this section had side lengths of 3 cm, 4 cm and 5 cm. This satisfied Pythagoras theorem, so the numbers 3, 4 and 5 form a Pythagorean triad or triple. In fact, any multiple of these numbers, for example 1. 6, 8 and , 2 and 2.5 would also form a Pythagorean triad or triple. Some other triads are: 5, 12, 13 8, 15, 17 9, 40, 41
9 Chapter 5 Rightangled triangles and trigonometry 165 WORKED Example 5 Is the set of numbers 4, 6, 7 a Pythagorean triad? EXCEL Triads Spreadsheet THINK 1 Find the sum of the squares of the two smaller numbers. 2 Find the square of the largest number. 3 Compare the two results. The numbers form a Pythagorean triad if the results are the same. Write your answer. 4 WRITE = = = , 6, 7 is not a Pythagorean triad Pythagoras theorem can be used to solve more practical problems. In these cases, it is necessary to draw a diagram that will help you decide the appropriate method for finding a solution. The diagram simply needs to represent the triangle; it does not need to show details of the situation described. WORKED Example 6 The fire brigade attends a blaze in a tall building. They need to rescue a person from the 6th floor of the building, which is 30 metres above ground level. Their ladder is 32 metres long and must be at least 10 metres from the foot of the building. Can the ladder be used to reach the people needing rescue? THINK 1 Draw a diagram and show all given information. WRITE Burning building c 30 m c 30 m Fire engine 10 m 10 m 2 Write the formula after deciding if you Hypotenuse 2 = base 2 + height 2 are finding the hypotenuse or a shorter side. 3 Substitute the lengths of the known c 2 = sides. 4 Evaluate the expression. = = Find the answer by taking the square c = 1000 root. = m 6 Give a written answer. The ladder will be long enough to make the rescue, since it is 32 m long.
10 166 Maths Quest Maths A Year 11 for Queensland remember remember 1. Make sure that you can identify the hypotenuse of a rightangled triangle. It is the side opposite the right angle, and the longest side. 2. If you are finding the length of the hypotenuse use hypotenuse 2 = base 2 + height If you are finding the length of a shorter side use either base 2 = hypotenuse 2 height 2 or height 2 = hypotenuse 2 base Pythagoras theorem can be used to test whether a triangle is rightangled. (a) if hypotenuse 2 = base 2 + height 2, the triangle is rightangled (b) if hypotenuse 2 > base 2 + height 2, the triangle is obtuseangled (c) if hypotenuse 2 < base 2 + height 2, the triangle is acuteangled. 5. Pythagorean triads or Pythagorean triples are sets of three numbers which satisfy Pythagoras theorem for example 3, 4, Read the question carefully to make sure that you give the answer in the correct form. 7. Begin problem questions with a diagram and finish with an answer in words. 5A Pythagoras theorem WORKED Example 1 Name the hypotenuse in each of the following triangles. a b c 1 P X B Y Q R Z A C WORKED Example 2 2 Find the length of the hypotenuse in each of the following triangles. a b c x m z 5 cm 80 mm 12 cm 60 m 150 mm 11 m 3 In each of the following, find the length of the hypotenuse, correct to 2 decimal places. a b c 4.9 m 8.6 km 11.3 km a 9 cm 4.9 m 6 cm
11 Chapter 5 Rightangled triangles and trigonometry 167 WORKED Example 3 4 Find the length of each unknown shorter side in the rightangled triangles below, correct to 1 decimal place. a b c 2.01 m 12 cm p 2.2 m 2.9 m 4.37 m t 6 cm 5 In each of the following rightangled triangles, find the length of the side marked with a pronumeral, correct to 1 decimal place. a b 10.5 cm q 8 cm m 24.5 cm z c 4 cm 33 mm d m p 34 mm a m WORKED Example 4 6 Use the converse of Pythagoras theorem to determine if the following triangles are rightangled, acuteangled or obtuseangled. a b c 31 m 10 cm 8 cm 41 cm 40 cm 38 m 16 m 6 cm 9 cm 7 8 multiple choice The hypotenuse in LXYZ at right is: A XY B XZ C YZ D impossible to tell multiple choice Which of the following triangles is definitely rightangled? A B 5 m 24.5 m Z Y X 24.5 m 32.5 m 25 m C 20.5 m 24.5 m 84 m D 25.4 m 87.5 m 24.5 m 35.3 m
12 168 Maths Quest Maths A Year 11 for Queensland WORKED Example 5 9 Are the following sets of numbers Pythagorean triads? a 9, 12, 15 b 4, 5, 6 c 30, 40, 50 d 3, 6, 9 e 0.6, 0.8, 1.0 f 7, 24, 25 g 6, 13, 14 h 14, 20, 30 i 11, 60, 61 j 10, 24, 26 k 12, 16, 20 l 2, 3, 4 10 Complete the following Pythagorean triads. a 9,, 15 b, 24, 25 c 1.5, 2.0, d 3,, 5 e 11, 60, f 10,, 26 g, 40, 41 h 0.7, 2.4, 11 For each of the sets which were Pythagorean triads in question 9 state which side the right angle is opposite. 12 multiple choice Which of the following is a Pythagorean triad? A 7, 14, 21 B 1.2, 1.5, 3.6 C 3, 6, 9 D 12, 13, 25 E 15, 20, multiple choice Which of the following is not a Pythagorean triad? A 5, 4, 3 B 6, 9, 11 C 13, 84, 85 D 0.9, 4.0, 4.1 E 5, 12, 13 WORKED Example 6 14 A television antenna is 12 m high. To support it, wires are attached to the ground 5 m from the foot of the antenna. Find the length of each wire. 15 Susie needs to clean the guttering on her roof. She places her ladder 1.2 m back from the edge of the guttering that is 3 m above the ground. How long will Susie s ladder need to be (correct to 2 decimal places)? 16 A rectangular gate is 3.5 m long and 1.3 m wide. The gate is to be strengthened by a diagonal brace as shown at right. How long should the brace be (correct to 2 decimal places)? 3.5 m 1.3 m 17 A 2.5m ladder leans against a brick wall. The foot of the ladder is 1.2 m from the foot of the wall. How high up the wall will the ladder reach (correct to 1 decimal place)? 18 Use the measurements in the diagram at right to find the height of the flagpole, correct to 1 decimal place. 7.9 m 2.4 m 19 An isosceles, rightangled triangle has a hypotenuse of 10 cm. Calculate the length of the shorter sides. (Hint: Call both shorter sides x.)
13 iinvestigation iinvestigation Chapter 5 Rightangled triangles and trigonometry 169 nvestigation Pythagoras theorem Resources: Paper, pencil, protractor, compass, calculator. As mentioned previously, Pythagoras theorem states: In a rightangled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. 3 cm 5 cm 4 cm In this investigation we consider replacing the word square/s with other geometric shapes. 1 What would happen if the word square/s were to be replaced by the word semicircle/s? Draw diagrams and investigate to see whether the statement would still be true in all cases. (Make sure that the triangles you draw are indeed rightangled.) 2 What would happen if we were to replace square/s with circle/s or equilateral triangle/s? Remember that it must work for all situations. 3 What are your conclusions? 4 Prepare a formal report on your investigation. Support your conclusions with detailed diagrams and mathematical evidence. nvestigation Constructing a staircase Resources: Pen, paper, calculator. With an understanding of the theorem of Pythagoras, you should now be able to investigate the staircase problem. Building stairs prompts the following questions: 1. How much timber should be ordered? 2. Where should the stairs start if we are to get the correct slope?
14 iinvestigation 170 Maths Quest Maths A Year 11 for Queensland We start by considering the general structure of a staircase. Treads 250 mm Mrs Kelly has contacted a local builder to 170 mm construct an external set of stairs from the ground to an existing deck 2550 mm above the ground. 1 Use the information in the diagram above Stringers 2 to determine: level Ground a the number of treads, 170 mm apart, required to reach a height of 2550 mm Footings (remember that no treads are necessary at the top or base of the staircase) b the horizontal distance from the deck where the footings should be cemented (knowing that each tread is 250 mm wide) c d the length of each stringer the length of timber that must be ordered if the treads and stringers are constructed from 250mm by 50mm hardwood, and the stairs are 1 metre wide. 50 mm 250 mm 2 Draw a diagram for Mrs Kelly, showing the structure of her staircase (with the correct number of treads). Mark on your diagram all known measurements and indicate the total length of timber required for the job. nvestigation Constructing a paper box Resources: Paper, scissors. How is it possible to determine the starting dimensions for constructing a box of a particular size? Pythagoras theorem has the answer. 1 Take a square piece of paper of a reasonable size. a Valley fold along all lines of symmetry. Remember to crease all folds sharply. Open the square out flat. b Fold each of the corners in a valley fold in towards the centre.
15 Chapter 5 Rightangled triangles and trigonometry 171 c Valley fold the top, bottom, left and right sides of the resulting shape in towards the centre, opening each fold before making the next. These folds will form the base of the box. d Open two of the corner folds opposite each other as indicated. e Fold up the edges opposite each other (those which already have centre folds) to make one pair of the box s sides. f Fold in both other ends as shown, completing the other two sides of the box. 2 With the base of the box completed, a lid could then be fashioned from another piece of square paper about 1 cm larger.
16 172 Maths Quest Maths A Year 11 for Queensland 3 Having completed our box, the question now arises: How is it possible to determine the dimensions of square paper required to construct a box of a particular size? a Open your box and look at all the creases formed. It should appear as below. b c The base of the box is outlined with respect to the size of the original square paper. Note that the box s base is 2 units in length, while the whole diagonal of the paper is 8 units in length. This means that the box s base is onequarter the length of the diagonal of the original paper. What is the height of the box compared with the length of the diagonal of the original square? 4 We now have the knowledge to make a box with a side length of, say, 5 cm. a The diagonal of our starting square must be S 4 5 cm = 20 cm b Using Pythagoras theorem, we can now determine the length of the side of the square (S). S Hypotenuse 2 = base 2 + height = S 2 + S = 2S = S = S 2 So S = 200 = 14.1 cm c So, starting with a piece of paper 14.1 cm square, after folding, the resulting box would be 2.5 cm high, with a side length of 5 cm. d e Verify these measurements for yourself by folding such a piece of paper. Try this exercise with other paper sizes. This is yet another example of the importance of geometry in many facets of our daily lives. 20 cm
17 iinvestigation Chapter 5 Rightangled triangles and trigonometry 173 Similar triangles and shadow sticks Sometimes it is not possible to measure heights of objects such as trees, tall buildings and mountains. It is, however, often possible to measure the length of a shadow cast by these objects. If we compare the length of these shadows with the length of a shadow cast by a stick of known height under the same conditions, it is possible to calculate heights which are difficult to measure. Consider the situation below. Stick Building Stick s shadow Building s shadow We need to determine the height of the building. Take a stick whose height we know (or can measure) and place it, vertically, in the sunshine, near the building. The rays from the sun are parallel, so we have two similar rightangled triangles. (The building and the stick are at right angles to the ground.) We can measure both shadows. Stick Stick s shadow The larger rightangled triangle formed by the building and its shadow is some scale factor of the smaller rightangled triangle formed by the stick and the stick s shadow. Scale factor = Building s shadow Building length of building shadow length of stick shadow We can then apply the same scale factor to the height of the stick to determine the height of the building. So, Height of building = height of stick scale factor nvestigation Using shadows to determine the height of a tall object Resources: Tall object, shadow stick, measuring tape, sunshine! We shall apply the theory just developed, to determine the height of a tall object in your school grounds. 1 Choose a tall, accessible object in your school grounds (tree, building, etc.) as the object whose height you wish to determine. Let its height be B metres.
18 174 Maths Quest Maths A Year 11 for Queensland 2 Measure the height of your shadow stick as S metres. 3 Place your shadow stick vertically in the sunshine near your tall object. Measure the length of its shadow on the ground. Let this be s metres. 4 Measure the length of the building s shadow on the ground. Let this be b metres. 5 Draw a diagram representing this situation. Mark all known measurements on the diagram. 6 Calculate the scale factor F (tallobject shadow length divided by stick shadow b length; that is, F =  ). s 7 The height of your tall object can then be determined by multiplying the height of the stick by the scale factor; that is, B = S F 8 Write up your activity, explaining all steps involved. WORKED Example 7 At the same time as a tree cast a shadow of 14 m, a 168cmtall girl cast a shadow of 140 cm. Find the height of the tree. Give the answer to 1 decimal place. Shadow (140 cm) Sun s rays Girl (168 cm) 14 metres THINK Identify the two similar triangles and draw them separately. (We assume that both the girl and the tree are perpendicular to the ground.) Identify the side of the triangle whose length is required. Calculate the scale factor. Note: Measurements must be in the same units. Apply the scale factor to the girl s height. Write the final answer, specifying units. WRITE 168 cm 140 cm 14 m length of tree shadow Scale factor = length of girl s shadow 14 m = m = 10 Height of tree = girl s height scale factor = 1.68 m 10 = 16.8 m Height of tree is 16.8 m x
19 remember remember Chapter 5 Rightangled triangles and trigonometry To determine the height of objects too difficult to measure, we can use the shadow stick method. length of shadow of object 2. Scale factor = length of shadow of stick 3. Height of object = height of stick scale factor. 5B Shadow sticks WORKED Example 7 1 The following diagrams represent the measurements taken from shadows of sticks and objects. Use the figures to determine the heights of the objects. a b a b c 3.6 m 2.5 m 8.2 m d 4.5 m 3.5 m 20 m c d 1.5 m 3 m 1.2 m 4.2 m 5.8 m 15 m 2 At the same time as a building cast a shadow of 14.3 metres, a 2metre stick cast a shadow of 5.3 metres. What is the height of the building? 3 A boulder on the shoreline cast a shadow of 15.8 m on the beach at the same time as a 1.5metre stick cast a shadow of 3.7 m. Determine the height of the boulder. 4 Find the height of the flagpole shown in the diagram at right. Guy wire 5 The shadow of a tree is 4 metres and at the same time the shadow of a 1metre stick is 25 cm. Assuming both the tree and stick are perpendicular to the horizontal ground, what is the height of the tree? 1 m 0.9 m 9 m
20 176 Maths Quest Maths A Year 11 for Queensland Questions 6 and 7 refer to the following information. A young tennis player s serve is shown in the diagram. 0.9 m y 1.1 m x 5 m 10 m Assume the ball travels in a straight line. 6 multiple choice The height of the ball just as it is hit, x, is closest to: A 3.6 m B 2.7 m C 2.5 m D 1.8 m E 1.6 m 7 multiple choice The height of the player, y, as shown is closest to: A 190 cm B 180 cm C 170 cm D 160 cm E 150 cm WorkSHEET 5.1 Calculating trigonometric ratios We have already looked at Pythagoras theorem, which enabled us to find the length of one side of a rightangled triangle given the lengths of the other two. However, to deal with other relationships in rightangled triangles, we need to turn to trigonometry. Trigonometry allows us to work with the angles also; that is, deal with relationships between angles and sides of rightangled triangles. For example, trigonometry enables us to find the length of a side, given the length of another side and the magnitude of an angle. So that we are clear about which lines and angles we are describing, we need to identify the given angle, and name the shorter sides with reference to it. For this reason, we label the sides opposite and acent that is, the sides opposite hypotenuse acent opposite and acent to the given angle. The diagram shows this relationship between the sides and the angle,. The trigonometric ratios are constant for a particular angle, and this is the reason the shadowstick method worked.
21 iinvestigation Chapter 5 Rightangled triangles and trigonometry 177 nvestigation Looking at the tangent ratio Resources: Ruler, spreadsheet. I The tangent ratio for a given angle is a ratio of sides in similar rightangled triangles, such as those in the diagram. BAC is common to each triangle and is equal to 30. We are going to look at the ratio of the opposite side length to the acent side length in each triangle. H 1 Complete each of the following measurements. a BC = mm AB = mm b DE = mm AD = mm c FG = mm AF = mm d HI = mm AH = mm 2 Set up the spreadsheet below: G F E D C B A Triangle Opposite side length Adjacent side length Opposite Adjacent ABC ADE AFG AHI 3 Enter values in columns 2 and 3 from the measurements above. 4 Provide a formula in column 4 to calculate the ratio of the opposite side length to the acent side length. This is the tangent ratio. 5 The BAC is common to each triangle. Consider the calculated values in column 4. What is your conclusion? Trigonometry uses the ratio of side lengths to calculate the lengths of sides and the size of angles. The ratio of the opposite side to the acent side is called the tangent ratio. This ratio is fixed for any particular angle. The tangent ratio for any angle,, can be found using the result: opposite side tan = acent side In the investigation above, we found that for a 30 angle the ratio was approximately We can find a more accurate value for the tangent ratio on a calculator by using the tan button. Calculators require a particular sequence of button presses in order to perform this calculation. Investigate the sequence required for your particular calculator. For all calculations in trigonometry you will need to make sure that your calculator is in DEGREES MODE. For most calculators you can check this by looking for a DEG in the display.
22 iinvestigation 178 Maths Quest Maths A Year 11 for Queensland When measuring angles: 1 degree = 60 minutes 1 minute = 60 seconds You need to be able to enter angles using both degrees and minutes into your calculator. Most calculators use a DMS (Degrees, Minutes, Seconds) button or a button. Check with your teacher to see how to do this. WORKED Example 8 Using your calculator, find the following, correct to 3 decimal places. 8 a tan 60 b 15 tan 75 c d tan tan 69 THINK WRITE/DISPLAY a Press tan and enter 60, then press =. a tan 60 = b Enter 15, press and tan, enter 75, then b 15 tan 75 = press =. 8 c Enter 8, press and tan, enter 69, then c = tan 69 press =. d Press tan, enter 49, press DMS, enter 32, d tan = press DMS, then press =. Note: Some calculators require that the angle size be entered before the trigonometric functions. The tangent ratio is used to solve problems involving the opposite side and the acent side of a rightangled triangle. The tangent ratio does not allow us to solve problems that involve the hypotenuse. The sine ratio (abbreviated to sin; pronounced sine) is the name given to the ratio of the opposite side and the hypotenuse. nvestigation Looking at the sine ratio Resources: Ruler, spreadsheet. I The tangent ratio for a given angle is a ratio of G the opposite side and the acent side in a E rightangled triangle. The sine ratio is the ratio C of the opposite side and the hypotenuse. 1 Complete each of the following A H F D B measurements as before. (As we saw earlier, BAC is common to all these similar triangles and so in this exercise, we look at the ratio of the side opposite BAC to the hypotenuse of each triangle.) a BC = mm AC = mm b DE = mm AE = mm c FG = mm AG = mm d HI = mm AI = mm
23 Chapter 5 Rightangled triangles and trigonometry Set up the spreadsheet below: Triangle Opposite side length Hypotenuse length Opposite Hypotenuse ABC ADE AFG AHI 3 Enter values in columns 2 and 3 from the measurements above. 4 Provide a formula in column 4 to calculate the ratio of the length of the opposite side to the length of the hypotenuse. This is the sine ratio. 5 The BAC is common to each triangle. You should notice that the values in column 4 are the same (or very close, allowing for measurement error). In any rightangled triangle with equal angles, the ratio of the length of the opposite side to the length of the hypotenuse will remain the same, regardless of the size of the triangle. The formula for the sine ratio is: sin = opposite side hypotenuse The value of the sine ratio for any angle is found using the sin function on the calculator. sin 30 = 0.5 Check this on your calculator. WORKED Example 9 Find, correct to 3 decimal places: 18 a sin 57 b 9 sin 45 c d 9.6 sin sin 44 THINK WRITE/DISPLAY a Press sin and enter 57, then press =. a sin 57 = b Enter 9, press and sin, enter 45, b 9 sin 45 = then press =. 18 c Enter 18, press and sin, enter 44, c = sin 44 then press =. d Enter 9.6, press and sin, enter 26, d 9.6 sin = press DMS, enter 12, press DMS, then press =. Note: Check the sequence of button presses required by your calculator.
24 iinvestigation 180 Maths Quest Maths A Year 11 for Queensland A third trigonometric ratio is the cosine ratio. This ratio compares the length of the acent side and the hypotenuse. nvestigation Looking at the cosine ratio Resources: Ruler, spreadsheet. I 1 Complete each of the following measurements. G a AB = mm AC = mm E b AD = mm AE = mm C c AF = mm AG = mm d AH = mm AI = mm H F D B A 2 Set up a spreadsheet of similar format to the two previous spreadsheets to calculate the ratio of the length of the acent side to that of the hypotenuse. 3 You should again find this ratio constant (or nearly so). This is the cosine ratio. The cosine ratio is found using the formula: acent side cos = hypotenuse To calculate the cosine ratio for a given angle on your calculator, use the cos function. On your calculator check the calculation: cos 30 = WORKED Example 10 Find, correct to 3 decimal places: a cos 27 b 6 cos 55 c d cos 74 cos THINK WRITE/DISPLAY a Press cos and enter 27, then press =. a cos 27 = b Enter 6, press and cos, enter 55, b 6 cos 55 = then press = c Enter 21.3, press and cos, enter 74, c = cos 74 then press =. 4.5 d Enter 4.5, press and cos, enter 82, d = cos press DMS, enter 46, press DMS, then press =. Note: Check the sequence requirements for your calculator. Similarly, if we are given the sine, cosine or tangent of an angle, we are able to calculate the size of that angle using the calculator. We do this using the inverse functions. On most calculators these are the second function of the sin, cos and tan functions and are denoted sin 1, cos 1 and tan 1.
25 Chapter 5 Rightangled triangles and trigonometry 181 WORKED Example 11 Find, correct to the nearest degree, given that sin = THINK WRITE/DISPLAY 1 Press 2nd F [sin 1 ] and enter.738, then press =. 2 Round your answer to the nearest degree. = 48 Note: Check the sequence requirements for your calculator. Problems sometimes supply angles in degrees, minutes and seconds, or require answers to be written in the form of degrees, minutes and seconds. On most calculators, you will use the DMS (Degrees, Minutes, Seconds) function or the function. If you are using a graphics calculator, the angle function provides this facility. Check with your teacher for the requirements of your particular calculator. WORKED Example 12 Given that tan = 1.647, calculate to the nearest minute. THINK 1 Press 2nd F [tan 1 ] and enter 1.647, then press =. 2 Convert your answer to degrees and minutes by pressing DMS. WRITE/DISPLAY = remember remember 1. The tangent ratio is the ratio of the opposite side and the acent side. opposite side tan = acent side 2. The sine ratio is the ratio of the opposite side and the hypotenuse. opposite side sin = hypotenuse 3. The cosine ratio is the ratio of the acent side and the hypotenuse. acent side cos = hypotenuse 4. The value of the trigonometric ratios can be found using the sin, cos and tan functions on your calculator. 5. The angle can be found when given the trigonometric ratio using the sin 1, cos 1 and tan 1 functions on your calculator.
26 182 Maths Quest Maths A Year 11 for Queensland 5C Calculating trigonometric ratios WORKED Example 8 WORKED Example 9 WORKED Example 10 1 Calculate the value of each of the following, correct to 3 decimal places. 8.6 a tan 57 b 9 tan 63 c d tan tan 12 2 Calculate the value of each of the following, correct to 3 decimal places a sin 37 b 9.3 sin 13 c d sin 72 sin Calculate the value of each of the following, correct to 3 decimal places. 6 a cos 45 b 0.25 cos 9 c d 5.9 cos 2 3 cos 24 4 Calculate the value of each of the following, correct to 3 decimal places, if necessary. a sin 30 b cos 15 c tan 45 d 48 tan 85 e 128 cos 60 f 9.35 sin 8 g h i cos 32 tan 20 sin 72 5 Calculate the value of each of the following, correct to 2 decimal places. a sin b tan c cos d 9 cos e 4.9 sin f 2.39 tan 8 59 g h i tan cos sin 75 5 WORKED Example 11 WORKED Example 12 6 Find, correct to the nearest degree, given that sin = Find, correct to the nearest degree, given that: a sin = b cos = c tan = Find, correct to the nearest minute, given that cos = Find, correct to the nearest minute, given that: a tan = b cos = c sin = Finding an unknown side We can use the trigonometric ratios to find the length of one side of a rightangled triangle if we know the length of another side and an angle. Consider the triangle at right. In this triangle we are asked to find the length of the opposite side and have been given the length of the acent side. hyp cm x opp
27 Chapter 5 Rightangled triangles and trigonometry 183 opposite x We know from the formula that: tan = In this example, tan 30 = From acent 14 our calculator we know that tan 30 = We can set up an equation that will allow us to find the value of x. opp tan = x tan 30 = x = 14 tan 30 = cm Use the tangent ratio to find the value of h in the triangle at right, correct to 2 decimal places. THINK WRITE 1 Label the sides of the triangle opp, and hyp. hyp h opp cm 2 Write the tangent formula. opp tan = h Substitute for (55 ) and the acent tan 55 = side (17 cm). 4 Make h the subject of the equation. h = 17 tan 55 Calculate. = cm 5 WORKED Example cm h In the example above, we were told to use the tangent ratio. In practice, we need to be able to look at a problem and then decide if the solution is found using the sin, cos or tan ratio. To do this we need to examine the three formulas. opposite side tan = acent side We use this formula when we are finding either the opposite or acent side and are given the length of the other. opposite side sin = hypotenuse The sin ratio is used when finding the opposite side or the hypotenuse when given the length of the other. acent side cos = hypotenuse The cos ratio is for problems where we are finding the acent side or the hypotenuse and are given the length of the other. To make the decision we need to label the sides of the triangle and make a decision based on these labels. Cabri Trig ratios Geometry
28 184 Maths Quest Maths A Year 11 for Queensland WORKED Example 14 Find the length of the side marked x, correct to 2 decimal places. THINK WRITE 1 Label the sides of the triangle. hyp 24 m x opp 24 m 50 x x is the opposite side and 24 m is the hypotenuse, therefore use the sin formula. 50 opp sin = hyp Substitute for and the hypotenuse. x sin 50 = Make x the subject of the equation. x = 24 sin 50 Calculate. = m To remember each of the formulas more easily, we can use this acronym: SOHCAHTOA We pronounce this acronym as Sock ca toe her. The initials of the acronym represent the three trigonometric formulas. S O H C T opp sin = A cos = tan = hyp O hyp H A opp Care needs to be taken at the substitution stage. In the previous two examples, the unknown side was the numerator in the fraction, hence we multiplied to find the answer. If after substitution, the unknown side is in the denominator, the final step is done by division. Find the length of the side marked z in the triangle at right. THINK 1 WORKED Example 15 Label the sides opp, and hyp. WRITE 23 15' hyp z opp 23 15' z 12.5 m 2 3 Choose the cosine ratio because we are finding the hypotenuse and have been given the acent side m Write the formula. cos = hyp
29 Chapter 5 Rightangled triangles and trigonometry 185 THINK WRITE 4 Substitute for and the acent side cos = z 5 Make z the subject of the equation. z cos = z = cos Calculate. = m Note: In calculating the length of the hypotenuse, the process will always involve division by the trigonometric function. Trigonometry is used to solve many practical problems. In these cases, it is necessary to draw a diagram to represent the problem and then use trigonometry to solve the problem. With written problems that require you to draw the diagram, it is necessary to give the answer in words. A flying fox is used in an army training camp. The flying fox is supported by a cable that runs from the top of a cliff face to a point 100 m from the base of the cliff. The cable makes a 15 angle with the horizontal. Find the length of the cable used to support the flying fox. THINK Draw a diagram and show information WORKED Example Label the sides of the triangle opp, and hyp. Choose the cosine ratio because we are finding the hypotenuse and have been given the acent side. WRITE Write the formula. cos = hyp Substitute for and the acent side. 100 cos 15 = f Make f the subject of the equation. f cos 15 = f = cos 15 Calculate. Give a written answer. 16 opp f hyp m = m The cable is approximately m long.
30 186 Maths Quest Maths A Year 11 for Queensland remember remember 1. Trigonometry can be used to find a side in a rightangled triangle when we are given the length of one side and the size of an angle. 2. The trig formulas are: opp sin = cos = tan = hyp hyp opp Take care to choose the correct trigonometric ratio for each question. 4. Substitute carefully and note the change in the calculation, depending upon whether the unknown side is in the numerator or denominator. 5. Before using your calculator, check that it is in degrees mode. 6. Be sure that you know how to enter degrees and minutes into your calculator. 7. Written problems will require you to draw a diagram and give a written answer. 5D Finding an unknown side 1 Label the sides of each of the following triangles, with respect to the angle marked with the pronumeral. a b c α γ WORKED Example 13 2 Use the tangent ratio to find the length of the side marked x (correct to 1 decimal place). 3 Use the sine ratio to find the length of the side marked a (correct to 2 decimal places). 4 Use the cosine ratio to find the length of the side marked d (correct to nearest whole number). 13 m mm x a 35 cm 31 d
31 Chapter 5 Rightangled triangles and trigonometry 187 WORKED Example 14 5 The following questions use the tan, sin or cos ratios in their solution. Find the size of the side marked with the pronumeral, correct to 1 decimal place. a b c 49 x 13 cm 48 m 68 y 12.5 km 41 z WORKED Example 15 6 Find the length of the side marked with the pronumeral in each of the following (correct to 1 decimal place). a b c p t 87 mm q m 4.8 m 7 Find the length of the side marked with the pronumeral in each of the following (correct to 1 decimal place). a b c 23 b a 0.85 km m x km d 9 e d f m 116 mm cm 11 x m 13 g h i g 44.3 m 83 x km 84 9' m 2.34 m j q k l 60 32' t 75 19' 21.4 m 84.6 km r 26.8 cm 29 32'
32 188 Maths Quest Maths A Year 11 for Queensland 8 9 multiple choice Look at the diagram at right and state which of the following is correct. 9.2 A x = 9.2 sin 69 B x = sin 69 C x = 9.2 cos 69 multiple choice D x = cos 69 Study the triangle at right and state which of the following is correct A tan φ = B tan φ = C sin φ = D cos φ = x φ multiple choice Which of the statements below is not correct? A The value of tan can never be greater than 1. B The value of sin can never be greater than 1. C The value of cos can never be greater than 1. D tan 45 = 1 8 WORKED Example multiple choice Study the diagram at right and state which of the statements is correct. 22 A w = 22 cos 36 B w = sin mm w C w = 22 cos 54 D w = 22 sin A tree casts a 3.6 m shadow when the sun s angle of elevation is 59. Calculate the height of the tree, correct to the nearest metre. 13 A 10m ladder just reaches to the top of a wall when it is leaning at 65 to the ground. How far from the foot of the wall is the ladder (correct to 1 decimal place)? 14 The diagram at right shows the paths of two ships, A and B, after they have left port. If ship B sends a distress signal, how far must ship A sail to give assistance (to the nearest kilometre). Port km 15 A rectangular sign 13.5 m wide has a diagonal brace that makes a 24 angle with the horizontal. a Draw a diagram of this situation. b Calculate the height of the sign, correct to the nearest metre. B 16 A wooden gate has a diagonal brace built in for support. The gate stands 1.4 m high and the diagonal makes a 60 angle with the horizontal. a Draw a diagram of the gate. b Calculate the length that the diagonal brace needs to be. 17 The wire support for a flagpole makes a 70 angle with the ground. If the support is 3.3 m from the base of the flagpole, calculate the length of the wire support (correct to 2 decimal places). A
33 Chapter 5 Rightangled triangles and trigonometry A ship drops anchor vertically with an anchor line 60 m long. After one hour the anchor line makes a 15 angle with the vertical. a Draw a diagram of this situation. b Calculate the depth of water, correct to the nearest metre. c Calculate the distance that the ship has drifted, correct to 1 decimal place. 1 Find the size of the side marked with the pronumeral in each of the following. Where necessary, give your answer correct to 1 decimal place m 20 m a 15 cm b 8 cm 3 30 km 4 c 21 d 40 km 23 m 5 6 f e cm 42.1 m mm mm g h m i j 17 m
34 190 Maths Quest Maths A Year 11 for Queensland Finding angles So far, we have concerned ourselves with finding side lengths. We are also able to use trigonometry to find the sizes of angles when we have been given side lengths. We need to reverse our previous processes. Consider the triangle at right. We want to find the size of the angle marked. 10 cm 5 cm opp Using the formula sin = we know that in this triangle hyp sin = = = 0.5 We then calculate sin 1 (0.5) to find that = 30. As with all trigonometry it is important that you have your calculator set to degrees mode for this work. WORKED Example 17 Find the size of angle, correct to the nearest degree, in the triangle at right. THINK 1 Label the sides of the triangle and choose the tan ratio. WRITE hyp 6.5 m 4.3 m opp 6.5 m 4.3 m Substitute for the opposite and acent sides in the triangle. tan = Make the subject of the equation. = tan Calculate. = 33 = opp In many cases we will need to calculate the size of an angle, correct to the nearest minute. The same method for finding the solution is used; however, you will need to use your calculator to convert to degrees and minutes.
35 Chapter 5 Rightangled triangles and trigonometry 191 WORKED Example 18 Find the size of the angle at right, correct to the nearest minute. THINK WRITE 1 Label the sides of the triangle and choose the opp sin ratio. 4.6 cm 4.6 cm 7.1 cm hyp 7.1 cm Substitute for the opposite side and the hypotenuse in the triangle. Make the subject of the equation. Calculate and convert your answer to degrees and minutes. opp sin = hyp 4.6 = = sin = The same methods can be used to solve problems. As with finding sides, we set the question up by drawing a diagram of the situation. WORKED Example 19 A ladder is leant against a wall. The foot of the ladder is 4 m from the base of the wall and the ladder reaches 10 m up the wall. Calculate the angle that the ladder makes with the ground. THINK Draw a diagram and label the sides. 1 WRITE opp 10 m hyp Choose the tangent ratio and write the formula. 4 m opp tan = Substitute for the opposite and acent sides. = Make the subject of the equation. = tan Calculate. = Give a written answer. The ladder makes an angle of with the ground.
36 192 Maths Quest Maths A Year 11 for Queensland remember remember 1. Make sure that the calculator is in degrees mode. 2. To find an angle given the trig ratio, press 2nd F or SHIFT and then the appropriate ratio button. 3. Be sure to know how to get your calculator to display an answer in degrees and minutes. When rounding off minutes, check if the number of seconds is greater than When solving triangles remember the SOHCAHTOA rule to choose the correct formula. 5. In written problems draw a diagram and give an answer in words. 5E Finding angles WORKED Example 17 WORKED Example 18 1 Use the tangent ratio to find the size of the angle marked with the pronumeral in each of the following, correct to the nearest degree. a b c 7 m 12 m 11 m φ 3 m 2 Use the sine ratio to find the size of the angle marked with the pronumeral in each of the following, correct to the nearest minute. a b 4.6 m c 24 m 13 m 6.5 m 25 mm γ α 9.7 km 5.6 km 162 mm 3 Use the cosine ratio to find the size of the angle marked with the pronumeral in each of the following, correct to the nearest minute. a b c 15 cm 9 cm 4.6 m 2.6 m α 27.8 cm β 19.5 cm
37 Chapter 5 Rightangled triangles and trigonometry In the following triangles, you will need to use all three trig ratios. Find the size of the angle marked, correct to the nearest degree. a b c 11 cm 15 cm 9 cm 14 cm 7 cm 8 cm d 3.6 m e f 32 mm 196 mm 9.2 m 26.8 m 14.9 m 5 In each of the following find the size of the angle marked, correct to the nearest minute. a b c 30 m 63 cm 0.6 m 2.5 m 19.2 m 10 cm d 3.5 m e f 16.3 m 8.3 m 18.5 m 18.9 m 6.3 m 6 multiple choice Look at the triangle drawn at right. Which of the statements below is correct? A ABC = 30 B ABC = 60 C CAB = 30 D ABC = 45 5 cm A C 10 cm B 7 multiple choice Consider the triangle drawn at right. is closest to: A B C 48 4 D m 12.5 m 8 multiple choice 3 The exact value of sin = The angle = 2 A 30 B 45 C 60 D 90
38 194 Maths Quest Maths A Year 11 for Queensland WORKED Example 19 9 A 10m ladder leans against a wall 6 m high. Find the angle that the ladder makes with the horizontal, correct to the nearest degree. kite 10 A kite is flying on a 40m string. The kite is flying 10 m away from the vertical as shown in the figure at right. Find the angle the string makes with the horizontal, correct to 40 m the nearest minute. 10 m 11 A ship s compass shows a course due east of the port from which it sails. After sailing 10 nautical miles, it is found that the ship is 1.5 nautical miles off course as shown in the figure below. Port 1.5 nm 10 nm Find the error in the compass reading, correct to the nearest minute. 12 The diagram below shows a footballer s shot at goal. 7 m 30 m By dividing the isosceles triangle in half in order to form a rightangled triangle, calculate, to the nearest degree, the angle within which the footballer must kick to get the ball to go between the posts. 13 A golfer hits the ball 250 m, but 20 m off centre. Calculate the angle at which the ball deviated from a straight line, correct to the nearest minute.
39 Chapter 5 Rightangled triangles and trigonometry Find the length of the hypotenuse in the triangle at right. 5 cm 2 Find x, correct to 1 decimal place. 3 Find y, correct to 1 decimal place. 20 cm 4 Find z, correct to 3 decimal places. 30 cm 12 cm y x 12 cm 3.7 m 12 cm 7.4 m z 5 Calculate sin 56, correct to 4 decimal places. 6 Calculate 9.2 tan 50, correct to 2 decimal places Calculate , cos 8 45 correct to 2 decimal places. 8 Find, given that sin = Find to the nearest degree, given that cos = Find to the nearest minute, given that tan = 2.
40 196 Maths Quest Maths A Year 11 for Queensland Angles of elevation and depression The angle of elevation is measured upwards from a horizontal and refers to the angle at which we need to look up to see an object. Similarly, the angle of depression is the angle at which we need to look down from the horizontal to see an object. We are able to use the angles of elevation and depression to calculate the heights and distances of objects that would otherwise be difficult to measure. Angle of elevation Horizontal Horizontal Angle of depression From a point 50 m from the foot of a building, the angle of elevation to the top of the building is measured as 40. Calculate the height, h, of the building, correct to the nearest metre. THINK Label the sides of the triangle opp, and hyp. Choose the tangent ratio because we are finding the opposite side and have been given the acent side. WRITE hyp m m opp Write the formula. tan = Substitute for and the acent tan 40 = side. 50 Make h the subject of the equation. h = 50 tan 40 Calculate. = 42 m Give a written answer. The height of the building is approximately 42 m. 4 h WORKED Example 20 opp h h In practical situations, the angle of elevation is measured using a clinometer. Therefore, the angle of elevation is measured from a person s height at eye level. For this reason, the height at eye level must be added to the calculated answer. WORKED Example 21 Bryan measures the angle of elevation to the top of a tree as 64, from a point 10 m from the foot of the tree. If the height of Bryan s eyes is 1.6 m, calculate the height of the tree, correct to 1 decimal place m 1.6 m
41 Chapter 5 Rightangled triangles and trigonometry 197 THINK 1 2 Label the sides opp, and hyp. Choose the tangent ratio because we are finding the opposite side and have been given the acent side. WRITE opp h hyp m opp Write the formula. tan = h Substitute for and the acent side. tan 64 = Make h the subject of the equation. h = 10 tan 64 Calculate h. = 20.5 m Add the eye height. Height of tree = Height of tree = 22.1 Give a written answer. The height of the tree is approximately 22.1 m. A similar method for finding the solution is used for problems that involve an angle of depression. WORKED Example 22 When an aeroplane in flight is 2 km from a runway, the angle of depression to the runway is 10. Calculate the altitude of the aeroplane, correct to the nearest metre. h 2 km 10 THINK WRITE 1 2 Label the sides of the triangle opp, and hyp. Choose the tan ratio, because we are finding the opposite side given the acent side. opp h 2 km hyp Write the formula. tan = Substitute for and the acent side, converting 2 km to metres. tan 10 = opp h Make h the subject of the equation. h = 2000 tan 10 Calculate. = 353 m Give a written answer. The altitude of the aeroplane is approximately 353 m.
42 198 Maths Quest Maths A Year 11 for Queensland Angles of elevation and depression can also be calculated by using known measurements. This is done by drawing a rightangled triangle to represent a situation. A 5.2m building casts a 3.6m shadow. Calculate the angle of elevation of the sun, correct to the nearest degree. THINK 1 Label the sides opp, and hyp. 2 Choose the tan ratio because we are given the opposite and acent sides WORKED Example 23 WRITE opp 5.2 m 3.6 m 5.2 m 3.6 m Write the formula. opp tan = Substitute for opposite and acent. 5.2 tan = Make the subject of the equation. = tan Calculate. = 55 Give a written answer. The angle of elevation of the sun is approximately 55. remember remember hyp 1. The angle of elevation is the angle at which you look up to see an object. 2. The angle of depression is the angle at which you look down to see an object. 3. Problems can be solved by using angles of elevation and depression with the aid of a diagram. 4. Written problems should be given an answer in words. 5F Angles of elevation and depression WORKED Example 20 1 From a point 100 m from the foot of a building, the angle of elevation to the top of the building is 15. Calculate the height of the building, correct to 1 decimal place m 2 From a ship the angle of elevation to an aeroplane is 60. The aeroplane is 2300 m due north of the ship. Calculate the altitude of the aeroplane, correct to the nearest metre m
43 Chapter 5 Rightangled triangles and trigonometry From a point out to sea, a ship sights the top of a lighthouse at an angle of elevation of 12. It is known that the top of the lighthouse is 40 m above sea level. Calculate the distance of the ship from the lighthouse, correct to the nearest 10 m. 4 From a point 50 m from the foot of a building, Rod sights the top of a building at an angle of elevation of 37. Given that Rod s eyes are at a height of 1.5 m, calculate the height of the building, correct to 1 decimal place. 1.5 m 12 x 40 m WORKED Example m 5 Richard is flying a kite and sights the kite at an angle of elevation of 65. The altitude of the kite is 40 m and Richard s eyes are at a height of 1.8 m. Calculate the length of string the kite is flying on, correct to 1 decimal place. 1.8 m x m WORKED Example 22 WORKED Example 23 6 Bettina is standing on top of a cliff, 70 m above sea level. She looks directly out to sea and sights a ship at an angle of depression of m Calculate the distance of the ship from shore. 7 From an aeroplane flying at an altitude of 4000 m, the runway is sighted at an angle of depression of 15. Calculate the distance of the aeroplane from the runway, correct to the nearest kilometre. 8 There is a fire on the fifth floor of a building. The closest a fire truck can get to the building is 10 m. The angle of elevation from this point to where people need to be rescued is 69. If the fire truck has a 30m ladder, can the ladder be used to make the rescue? 9 From a navy vessel, a beacon which is 80 m D above sea level is sighted at an angle of elevation of 5. The vessel sailed towards the beacon and thirty minutes later the beacon is at an angle of 80 m elevation of A C Use the diagram at right to complete the following. B a Calculate the distance that the vessel was from the beacon, when the angle of elevation to the beacon was 5 (the distance AC). b Calculate the distance that the vessel sailed in the 30 minutes between the two readings. 10 A 12 m high building casts a shadow 15 m long. Calculate the angle of elevation of the sun, to the nearest degree m m 15 m 4000 m
44 iinvestigation 200 Maths Quest Maths A Year 11 for Queensland 11 An aeroplane which is at an altitude of 1500 m is 4000 m from a ship in a horizontal direction, as shown at right. Calculate the angle of depression from the aeroplane to the ship m 1500 m WorkSHEET The angle of elevation to the top of a tower is 12 from a point 400 m from the foot of the tower. a Draw a diagram of this situation. b Calculate the height of the tower, correct to 1 decimal place. c Calculate the angle of elevation to the top of the tower, from a point 100 m from the foot of the tower. nvestigation Alternative method to determine the height of a tall object Resources: Measuring tape, clinometer or protractor, calculator. In a previous investigation, you determined the height of a tall object using a shadow stick, observing the lengths of shadows. An alternative method could be employed using a clinometer. This is a device which measures angles. Your school may have one. If not, a protractor could be used as a substitute (although the accuracy of the reading would not be as great). 1 Locate a tall building or tree whose height you wish to determine. 2 Take a position some distance from its base, and measure your horizontal distance from the base. 3 Use the measuring tape to measure your height at eye level. 4 Take the clinometer (or protractor) and measure the angle of elevation from your eye level to the top of the tall object. 5 Draw a diagram representing the situation, labelling known lengths and angles. 6 Use a trigonometric ratio to determine the height of your tall building (or tree). (Remember to allow for your height at eye level in your calculations.) 7 Write up this activity, showing detailed diagrams and procedures used to obtain the height of your tall object.
45 iinvestigation Chapter 5 Rightangled triangles and trigonometry 201 nvestigation Revisiting staircases Resources: Pen, paper (graph paper if possible), calculator. The mathematics used in constructing a staircase is more complex than you may at first suspect. Earlier in this chapter we looked at a simple staircase. Let us now examine the mathematics in more detail. The Home Building Code lays down construction standards for staircases. As a starting point, it is necessary to understand some key terms. Rise and going Rise and going refer to the staircase treads, and their arrangement and construction. Rise The rise is the vertical measurement from the upper surface of one tread to the upper surface of the next tread. Rise of flight The rise of flight is the vertical measurement taken from a floor or landing to the top of the next floor or landing; that is, the height of the staircase. Going The term going is the horizontal measurement from the front edge of one tread to the front edge of the next. Going of flight The going of flight is the horizontal measurement for one flight, taken from the face of the first or bottom step to the face of the last or top step in that flight. The figure below shows rise (R) and going (G) labelled in a staircase. Going of flight R R G R G Rise of flight R G Rise R Going G The home building code sets the following maximums and minimums: Max. (mm) Min. (mm) Rise Going It also states that the sum of two times the rise plus the going must be a maximum of 630 and a minimum of 585. Using R to represent the rise and G to represent going, this could be written as: 2R + G = 585 to 630 mm The building code also states that rise and going must be constant for any one staircase. (Imagine walking up a staircase that had stair heights of varying sizes!)
46 202 Maths Quest Maths A Year 11 for Queensland The following figure shows a staircase with 9 rises. The number of treads used is 8, one less than the number of rises, because the top floor acts as the final step. This staircase has 8 equal goings. Existing top floor Rise of flight Going Task 1 A staircase is to have a rise of flight of 2844 mm. The builder wants to use a rise of approximately 160 mm. 1 Within the building code regulations, determine: a the number of rises b an appropriate rise (R) measurement c a suitable going (G) measurement. 2 Draw a diagram showing the design of the staircase. Include all measurements. Provide calculations to confirm that your staircase conforms to the home building code. Task 2 Often, there are restrictions on the space available for the location of a staircase, and the going must be kept to a minimum. Tony and Mary are renovating their home and wish to install a staircase. The rise of flight is 2608 mm and the restricted going of flight is 3945 mm. 1 Find a suitable rise, if a rise of approximately 160 mm is desired. 2 State the number of treads. 3 Determine the going. Rise 4 Calculate the number of rises. Going of flight 5 Check that the dimensions are in accordance with the building code. 6 Draw a diagram of your completed staircase, indicating all measurements.
47 Chapter 5 Rightangled triangles and trigonometry 203 summary Pythagoras theorem When using Pythagoras theorem to calculate the length of the hypotenuse of a rightangled triangle, the formula is: hypotenuse 2 = base 2 + height 2 To find one of the shorter sides of a rightangled triangle, the formula is: base 2 = hypotenuse 2 height 2 or height 2 = hypotenuse 2 base 2 Pythagoras theorem can be used to test whether a triangle is rightangled. If hypotenuse 2 = base 2 + height 2, the triangle is rightangled. If hypotenuse 2 > base 2 + height 2, the triangle is obtuseangled. If hypotenuse 2 < base 2 + height 2, the triangle is acuteangled. Pythagorean triads or Pythagorean triples are sets of three numbers that satisfy Pythagoras theorem. Shadow sticks Shadow sticks can be used to determine the height of objects too difficult to measure. Height of the object = height of shadow stick scale factor Trigonometry formulas for rightangled triangles opp tan = opp sin = hyp cos = hyp SOHCAHTOA this acronym will help you remember trig formulas. Steps to find a side of a rightangled triangle Label the sides of the triangle opposite, acent and hypotenuse. Choose the correct ratio. Substitute given information. Make the unknown side the subject of the equation. Calculate. Steps to find an angle in a rightangled triangle Label the sides of the triangle opposite, acent and hypotenuse. Choose the correct ratio. Substitute given information. Make the unknown angle the subject of the equation. Calculate by using an inverse trig function. Angles of elevation and depression The angle of elevation is the angle we look up from the horizontal to see an object. The angle of depression is the angle we look down from the horizontal to see an object. Problems are solved using angles of elevation and depression by the same methods as for all rightangled triangles.
48 204 Maths Quest Maths A Year 11 for Queensland CHAPTER review 5A 1 Find the length of the side marked with a pronumeral, in each case writing your answer correct to 2 decimal places. a b c 32 cm n m 9.2 m 4.8 m p 26 cm 9.2 m d q e 1.9 km f 0.6 m 3.2 m 7.25 cm cm r 1.3 km 2.4 m t 5A 5A 5A 5A 5B 5C 5C 2 To travel between the towns of Bolong and Molong, you need to travel west along a road for 45 km, then north along another road for another 87 km. Calculate the straightline distance between the two towns. 3 A rope is 80 m long and runs from a cliff top to the ground, 45 m from the base of the cliff. Calculate the height of the cliff, to the nearest metre. 4 Classify the following triangles as acute, obtuse or rightangled. a 13 cm b 5 cm 5 cm 3 cm c 7 m 12 cm 19 m 9 m 5 Which of the following are Pythagorean triads? a 9, 39, 40 b 3, 4, 5 c 0.3, 0.4, 0.5 d 6, 7, 10 6 A flagpole casts a shadow of 15 metres at the same time as a 1.5metre stick casts a shadow of 2 metres. How high is the flagpole? 7 Calculate each of the following, correct to 4 decimal places. a sin 46 b tan c 4.9 cos d 8.9 sin 67 3 e f cos 75 tan Calculate, correct to the nearest degree, given that: a cos = b tan = 1.23 c sin = 0.8. d 8 cm 4 cm 15 cm 16 cm
49 Chapter 5 Rightangled triangles and trigonometry Calculate, correct to the nearest minute, given that: a cos = b tan = 0.5 c sin = C 10 Find the length of each side marked with a pronumeral, correct to 1 decimal place. a 6 cm b c 3.9 m m 5D 9 q 78 x cm d e f 12.6 cm n q t 6.8 m cm g g h i h 2.9 m 4.8 cm 26 42' 77 18' j 4.32 m k l 38.5 m k j 16 8' 29 51' z 83 30' 63 km 85 12' m 138 mm 11 A rope that is used to support a flagpole makes an angle of 70 with the ground. If the rope is tied down 3.1 m from the foot of the flagpole, find the height of the flagpole, correct to 1 decimal place. 5D
50 206 Maths Quest Maths A Year 11 for Queensland 5D 5D 5E 12 A dirt track runs off a road at an angle of 34 to the road. If I travel for 4.5 km along the dirt track, what is the shortest direct distance back to the road (correct to 1 decimal place)? 13 A fire is burning in a building and people need to be rescued. The fire brigade s ladder must reach a height of 60 m and must be angled at 70 to the horizontal. How long must the ladder be to complete the rescue? 14 Find the size of the angle marked in each of the following, giving your answer correct to the nearest degree. a b 2.3 m c 116 cm 43 cm 16 m 19 m 4.6 m 5E 15 Find the size of the angle marked in each of the following, giving your answer correct to the nearest minute. a 10.8 m b 2.9 m c 11.9 cm 4.6 m 6.1 m 13.8 cm 5E 5E 16 A kite on an 80m string reaches a height of 50 m in a strong wind. Calculate the angle the string makes with the horizontal. 17 There is 50 m of line on a fishing reel. When all the line is out, the bait sits on the bed of a lake and has drifted 20 m from the boat. Calculate the angle that the fishing line makes with the vertical. 5F 18 The top of a building is sighted at an angle of elevation of 40, when an observer is 27 m back from the base. Calculate the height of the building, correct to the nearest metre. h test yourself CHAPTER 5F 5 19 Hakam stands 50 m back from the foot of an 80m telephone tower. Hakam s eyes are at a height of 1.57 m. Calculate the angle of elevation that Hakam must look to see the top of the tower m 50 m m 80 m
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