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Maths A - Chapter 5

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Maths A - Chapter 5

1. 1. 5syllabussyllabusrrefefererenceence Strand: Applied geometry Core topic: Elements of applied geometry In thisIn this chachapterpter 5A Pythagoras’ theorem 5B Shadow sticks 5C Calculating trigonometric ratios 5D Finding an unknown side 5E Finding angles 5F Angles of elevation and depression Right-angled triangles and trigonometry MQ Maths A Yr 11 - 05 Page 157 Wednesday, July 4, 2001 4:39 PM
2. 2. 158 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d Introduction Thousands of years before Christ, ancient civi- lisations 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 right-angled triangles (trigonometry) can be employed to answer these questions. As you pro- gress 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. MQ Maths A Yr 11 - 05 Page 158 Wednesday, July 4, 2001 4:39 PM
4. 4. 160 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 1 Name the hypotenuse in each of the following triangles. a b c d 2 State Pythagoras’ theorem. 3 Calculate the unknown lengths in the following right-angled triangles. a b c 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 b 7 Express the following angles in degrees, minutes and seconds. a 35.5° b 27.23° c 68.125° 8 Express the following angles in degrees, correct to 2 decimal places. a 45°27′ b 84°35′22″ c 64°28″. 9 Deﬁne the trigonometric ratios: a sine b cosine c tangent. C A B ED F H G I K J L 6 m 8 m a 6 m 4 m b 20 cm a 34° b 27° MQ Maths A Yr 11 - 05 Page 160 Wednesday, July 4, 2001 4:39 PM
5. 5. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 161 10 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 67°15′ e tan 38°12′22″ 12 Copy the following diagrams and mark the angles of elevation or depression, from the observer to the object, on each. a b c 13 Solve the following for x. a = b = Ground Observer Ship Sea Ground Observer Bird 5.2 4.5 ------- x 1.8 ------- 5.2 4.5 ------- 1.8 x ------- MQ Maths A Yr 11 - 05 Page 161 Wednesday, July 4, 2001 4:39 PM
6. 6. 162 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d Right-angled triangles The properties of right-angled triangles enable us to calculate lengths and angles which are sometimes not able to be meas- ured. The ﬁrst property we will investigate is Pythagoras’ theorem. Pythagoras’ theorem Pythagoras’ theorem allows us to calculate the length of a side of a right-angled triangle, if we know the lengths of the other two sides. Consider LABC at right. AB is the hypotenuse (the longest side). It is opposite the right angle. 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 lower-case letter of the opposite vertex. In the ﬁgure above, we could have named the hypotenuse ‘c’. Consider the right-angled 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 = S2 ) 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 cm2 = 16 cm2 + 9 cm2 Alternatively: (5 cm)2 = (4 cm)2 + (3 cm)2 Which means: hypotenuse2 = base2 + height2 hypotenuse A C B Name the hypotenuse in the triangle at right. THINK WRITE The hypotenuse is opposite the right angle. The vertices at each end or the lower- case letter of the opposite vertex can be used to name the side. The hypotenuse is QR or p. Q P R1 2 1WORKEDExample A = 9 cm2 A = 25 cm2 C B A 3 cm 4 cm 5 cm A = 16 cm2Pythagoras calculations GCpr ogram Pythagoras MQ Maths A Yr 11 - 05 Page 162 Wednesday, July 4, 2001 4:39 PM
7. 7. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 163 This result is known as Pythagoras’ theorem. Pythagoras’ theorem states: In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the two shorter sides. That is, Hypotenuse2 = base2 + height2 This is the formula used to ﬁnd the length of the hypotenuse in a right-angled tri- angle when we are given the lengths of the two shorter sides. In this example, the answer is a whole number because we are able to ﬁnd 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 hypotenuse2 = base2 + height2 it follows that base2 = hypotenuse2 − height2 and height2 = hypotenuse2 − base2 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. Find the length of the hypotenuse in the triangle at right. THINK WRITE Write the formula. Hypotenuse2 = base2 + height2 Substitute the lengths of the shorter sides. c2 = 152 + 82 Evaluate the expression for c2 . = 225 + 64 = 289 Find the value of c by taking the square root. c = = 17 cm 8 cm 15 cm c 1 2 3 4 289 2WORKEDExample E XCEL Spread sheet Pythagoras 289 Find the length of side PQ in triangle PQR, correct to 1 decimal place. THINK WRITE Write the formula. Base2 = hypotenuse2 − height2 Substitute the lengths of the known sides. r2 = 162 − 92 Evaluate the expression. = 256 − 81 = 175 Find the answer by ﬁnding the square root. r = = 13.2 m QP R 16 m 9 m r 1 2 3 4 175 3WORKEDExample Math cad Pythagoras MQ Maths A Yr 11 - 05 Page 163 Wednesday, July 4, 2001 4:39 PM
8. 8. 164 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d Because Pythagoras’ theorem applies only in right-angled triangles, we can use the theorem to test whether a triangle is right-angled, acute-angled or obtuse-angled. If we assume, as a starting point, that the triangle is right-angled, we could regard the longest side as the hypotenuse. We could then calculate values for hypotenuse2 , base2 and height2 . Interpreting the relationships between these three values, we could conclude the following. If hypotenuse2 = base2 + height2 the triangle is right-angled. If hypotenuse2 > base2 + height2 the triangle is obtuse-angled (since the hypotenuse is longer than it should be for that particular base and height). If hypotenuse2 < base2 + height2 the triangle is acute-angled (since the hypotenuse is too short for that particular base and height). Pythagorean triads (or Pythagorean triples) are sets of 3 numbers which satisfy Pythagoras’ theorem. The ﬁrst right-angled 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 10 1.5, 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 Determine whether the triangle shown is right-angled, acute-angled or obtuse-angled. THINK WRITE Assume that the longest side is a hypotenuse. Calculate the hypotenuse2 and base2 + height2 separately. hypotenuse2 = 72 = 49 base2 + height2 = 52 + 42 = 25 + 16 = 41 Write down an equality or inequality statement. hypotenuse2 > base2 + height2 Write a conclusion. Therefore the triangle is obtuse-angled. 4 cm 5 cm 7 cm 1 2 3 4 4WORKEDExample MQ Maths A Yr 11 - 05 Page 164 Wednesday, July 4, 2001 4:39 PM
10. 10. 166 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d Pythagoras’ theorem 1 Name the hypotenuse in each of the following triangles. a b c 2 Find the length of the hypotenuse in each of the following triangles. a b c 3 In each of the following, ﬁnd the length of the hypotenuse, correct to 2 decimal places. a b c remember 1. Make sure that you can identify the hypotenuse of a right-angled triangle. It is the side opposite the right angle, and the longest side. 2. If you are ﬁnding the length of the hypotenuse use hypotenuse2 = base2 + height2 . 3. If you are ﬁnding the length of a shorter side use either base2 = hypotenuse2 − height2 or height2 = hypotenuse2 − base2 . 4. Pythagoras’ theorem can be used to test whether a triangle is right-angled. (a) if hypotenuse2 = base2 + height2 , the triangle is right-angled (b) if hypotenuse2 > base2 + height2 , the triangle is obtuse-angled (c) if hypotenuse2 < base2 + height2 , the triangle is acute-angled. 5. Pythagorean triads or Pythagorean triples are sets of three numbers which satisfy Pythagoras’ theorem — for example 3, 4, 5. 6. 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. remember 5A WORKED Example 1 P Q R X Y Z B A C WORKED Example 2 12 cm 5 cm x 150 mm 80 mm m 60 m 11 m z 9 cm 6 cm a 4.9 m 4.9 m 8.6 km 11.3 km MQ Maths A Yr 11 - 05 Page 166 Wednesday, July 4, 2001 4:39 PM
11. 11. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 167 4 Find the length of each unknown shorter side in the right-angled triangles below, correct to 1 decimal place. a b c 5 In each of the following right-angled triangles, ﬁnd the length of the side marked with a pronumeral, correct to 1 decimal place. a b c d 6 Use the converse of Pythagoras’ theorem to determine if the following triangles are right-angled, acute-angled or obtuse-angled. a b c 7 The hypotenuse in LXYZ at right is: 8 Which of the following triangles is deﬁnitely right-angled? A XY B XZ C YZ D impossible to tell A B C D WORKED Example 3 12 cm 6 cm p 2.2 m 2.9 m q 4.37 m 2.01 m t 8 cm 4 cm m 10.5 cm 24.5 cm z 33 mm 34 mm a 37.25 m 52.75 m p WORKED Example 4 10 cm 6 cm 8 cm 41 cm 40 cm 9 cm 31 m 38 m 16 m mmultiple choiceultiple choice X Y Z mmultiple choiceultiple choice 24.5 m 32.5 m 20.5 m 5 m 24.5 m 25 m 24.5 m 84 m 87.5 m 25.4 m 24.5 m 35.3 m MQ Maths A Yr 11 - 05 Page 167 Wednesday, July 4, 2001 4:39 PM
12. 12. 168 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 9 Are the following sets of numbers Pythagorean triads? 10 Complete the following Pythagorean triads. 11 For each of the sets which were Pythagorean triads in question 9 state which side the right angle is opposite. 12 Which of the following is a Pythagorean triad? 13 Which of the following is not a Pythagorean triad? 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)? 17 A 2.5-m 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. 19 An isosceles, right-angled triangle has a hypotenuse of 10 cm. Calculate the length of the shorter sides. (Hint: Call both shorter sides x.) 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 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, __ A 7, 14, 21 B 1.2, 1.5, 3.6 C 3, 6, 9 D 12, 13, 25 E 15, 20, 25 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 5 mmultiple choiceultiple choice mmultiple choiceultiple choice WORKED Example 6 3.5 m 1.3 m 7.9 m 2.4 m MQ Maths A Yr 11 - 05 Page 168 Wednesday, July 4, 2001 4:39 PM
13. 13. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 169 Pythagoras’ theorem Resources: Paper, pencil, protractor, compass, calculator. As mentioned previously, Pythagoras’ theorem states: ‘In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.’ 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 right-angled.) 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. 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? inv estigat ioninv estigat ion 3 cm 4 cm 5 cm inv estigat ioninv estigat ion MQ Maths A Yr 11 - 05 Page 169 Wednesday, July 4, 2001 4:39 PM
14. 14. 170 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d We start by considering the general structure of a staircase. Mrs Kelly has contacted a local builder to 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 to determine: a the number of treads, 170 mm apart, required to reach a height of 2550 mm (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 the length of each stringer d the length of timber that must be ordered if the treads and stringers are constructed from 250-mm by 50-mm hardwood, and the stairs are 1 metre wide. 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. 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. Treads Stringers × 2 Footings level Ground 170 mm 250 mm 250 mm 50 mm inv estigat ioninv estigat ion MQ Maths A Yr 11 - 05 Page 170 Wednesday, July 4, 2001 4:39 PM
15. 15. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 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. MQ Maths A Yr 11 - 05 Page 171 Wednesday, July 4, 2001 4:39 PM
16. 16. 172 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 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 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 one-quarter the length of the diagonal of the original paper. c 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 4 × 5 cm = 20 cm b Using Pythagoras’ theorem, we can now determine the length of the side of the square (S). Hypotenuse2 = base2 + height2 202 = S2 + S2 400 = 2S2 = S2 200 = S2 So S = = 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 Verify these measurements for yourself by folding such a piece of paper. e Try this exercise with other paper sizes. This is yet another example of the importance of geometry in many facets of our daily lives. S S 20 cm 400 2 --------- 200 MQ Maths A Yr 11 - 05 Page 172 Wednesday, July 4, 2001 4:39 PM
20. 20. 176 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d Questions 6 and 7 refer to the following information. A young tennis player’s serve is shown in the diagram. Assume the ball travels in a straight line. 6 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 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 Calculating trigonometric ratios We have already looked at Pythagoras’ theorem, which enabled us to ﬁnd the length of one side of a right-angled triangle given the lengths of the other two. However, to deal with other relationships in right-angled 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 right-angled 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 adjacent — that is, the sides opposite and adjacent to the given angle. The diagram shows this relationship between the sides and the angle, θ. The trigonometric ratios are constant for a particular angle, and this is the reason the shadow-stick method worked. x y1.1m 5 m 10 m 0.9 m mmultiple choiceultiple choice mmultiple choiceultiple choice Work SHEET 5.1 hypotenuse opposite adjacent θ MQ Maths A Yr 11 - 05 Page 176 Wednesday, July 4, 2001 4:39 PM
22. 22. 178 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 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 calcu- lator. Most calculators use a (Degrees, Minutes, Seconds) button or a button. Check with your teacher to see how to do this. The tangent ratio is used to solve problems involving the opposite side and the adjacent side of a right-angled 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. Looking at the sine ratio Resources: Ruler, spreadsheet. The tangent ratio for a given angle is a ratio of the opposite side and the adjacent side in a right-angled triangle. The sine ratio is the ratio of the opposite side and the hypotenuse. 1 Complete each of the following 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 DMS ° ’ ” Using your calculator, ﬁnd the following, correct to 3 decimal places. a tan 60° b 15 tan 75° c d tan 49°32′ THINK WRITE/DISPLAY a Press and enter 60, then press . a tan 60° = 1.732 b Enter 15, press and , enter 75, then press . b 15 tan 75° = 55.981 c Enter 8, press and , enter 69, then press . c = 3.071 d Press , enter 49, press , enter 32, press , then press . d tan 49°32′ = 1.172 Note: Some calculators require that the angle size be entered before the trigonometric functions. 8 tan 69° ------------------ tan = × tan = ÷ tan = 8 tan 69° ----------------- tan DMS DMS = 8WORKEDExample inv estigat ioninv estigat ion A B C E G I DFH MQ Maths A Yr 11 - 05 Page 178 Wednesday, July 4, 2001 4:39 PM
23. 23. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 179 In any right-angled 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: The value of the sine ratio for any angle is found using the sin function on the calcu- lator. sin 30° = 0.5 Check this on your calculator. 2 Set up the spreadsheet below: 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). Triangle Opposite side length Hypotenuse length ABC ADE AFG AHI Opposite Hypotenuse ------------------------------ sin θ opposite side hypotenuse -------------------------------= Find, correct to 3 decimal places: a sin 57° b 9 sin 45° c d 9.6 sin 26°12′. THINK WRITE/DISPLAY a Press and enter 57, then press . a sin 57° = 0.839 b Enter 9, press and , enter 45, then press . b 9 sin 45° = 6.364 c Enter 18, press and , enter 44, then press . c = 25.912 d Enter 9.6, press and , enter 26, press , enter 12, press , then press . d 9.6 sin 26°12′ = 4.238 Note: Check the sequence of button presses required by your calculator. 18 sin 44° ----------------- sin = × sin = ÷ sin = 18 sin 44° ----------------- × sin DMS DMS = 9WORKEDExample MQ Maths A Yr 11 - 05 Page 179 Wednesday, July 4, 2001 4:39 PM
24. 24. 180 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d A third trigonometric ratio is the cosine ratio. This ratio compares the length of the adjacent side and the hypotenuse. The cosine ratio is found using the formula: To calculate the cosine ratio for a given angle on your calculator, use the cos func- tion. On your calculator check the calculation: cos 30° = 0.866 Similarly, if we are given the sine, cosine or tangent of an angle, we are able to calcu- late 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 . Looking at the cosine ratio Resources: Ruler, spreadsheet. 1 Complete each of the following measurements. a AB = mm AC = mm b AD = mm AE = mm c AF = mm AG = mm d AH = mm AI = mm 2 Set up a spreadsheet of similar format to the two previous spreadsheets to calculate the ratio of the length of the adjacent side to that of the hypotenuse. 3 You should again ﬁnd this ratio constant (or nearly so). This is the cosine ratio. inv estigationinv estigat ion A B C E G I DFH cos θ adjacent side hypotenuse -------------------------------= Find, correct to 3 decimal places: a cos 27° b 6 cos 55° c d . THINK WRITE/DISPLAY a Press and enter 27, then press . a cos 27° = 0.891 b Enter 6, press and , enter 55, then press . b 6 cos 55° = 3.441 c Enter 21.3, press and , enter 74, then press . c = 77.275 d Enter 4.5, press and , enter 82, press , enter 46, press , then press . d = 35.740 Note: Check the sequence requirements for your calculator. 21.3 cos 74° ------------------ 4.5 cos 82°46′ -------------------------- cos = × cos = ÷ cos = 21.3 cos 74° ------------------ × cos DMS DMS = 4.5 cos 82°46′ -------------------------- 10WORKEDExample MQ Maths A Yr 11 - 05 Page 180 Wednesday, July 4, 2001 4:39 PM
26. 26. 182 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d Calculating trigonometric ratios 1 Calculate the value of each of the following, correct to 3 decimal places. 2 Calculate the value of each of the following, correct to 3 decimal places. 3 Calculate the value of each of the following, correct to 3 decimal places. 4 Calculate the value of each of the following, correct to 3 decimal places, if necessary. 5 Calculate the value of each of the following, correct to 2 decimal places. 6 Find θ, correct to the nearest degree, given that sin θ = 0.167. 7 Find θ, correct to the nearest degree, given that: 8 Find θ, correct to the nearest minute, given that cos θ = 0.058. 9 Find θ, correct to the nearest minute, given that: Finding an unknown side We can use the trigonometric ratios to ﬁnd the length of one side of a right-angled 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 adjacent side. a tan 57° b 9 tan 63° c d tan 33°19′ a sin 37° b 9.3 sin 13° c d a cos 45° b 0.25 cos 9° c d 5.9 cos 2°3′ 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 a sin 24°38′ b tan 57°21′ c cos 84°40′ d 9 cos 55°30′ e 4.9 sin 35°50′ f 2.39 tan 8°59′ g h i a sin θ = 0.698 b cos θ = 0.173 c tan θ = 1.517. a tan θ = 0.931 b cos θ = 0.854 c sin θ = 0.277. 5C WORKED Example 8 8.6 tan 12° ----------------- WORKED Example 9 14.5 sin 72° ----------------- 48 sin 67°40′ ------------------------- WORKED Example 10 6 cos 24° ------------------ 4.5 cos 32° ------------------ 0.5 tan 20° ----------------- 15 sin 72° ----------------- 19 tan 67°45′ ------------------------- 49.6 cos 47°25′ -------------------------- 0.84 sin 75°5′ ---------------------- WORKED Example 11 WORKED Example 12 hyp opp adj 14 cm 30° x MQ Maths A Yr 11 - 05 Page 182 Wednesday, July 4, 2001 4:39 PM
27. 27. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 183 We know from the formula that: tan θ = . In this example, tan 30° = . From our calculator we know that tan 30° = 0.577. We can set up an equation that will allow us to ﬁnd the value of x. tan θ = tan 30° = x = 14 tan 30° = 8.083 cm 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. tan θ = We use this formula when we are ﬁnding either the opposite or adjacent side and are given the length of the other. sin θ = The sin ratio is used when ﬁnding the opposite side or the hypotenuse when given the length of the other. cos θ = The cos ratio is for problems where we are ﬁnding the adjacent side or the hypot- enuse 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. opposite adjacent -------------------- x 14 ------ opp adj --------- x 14 ------ Use the tangent ratio to ﬁnd the value of h in the triangle at right, correct to 2 decimal places. THINK WRITE Label the sides of the triangle opp, adj and hyp. Write the tangent formula. tan θ = Substitute for θ (55°) and the adjacent side (17 cm). tan 55° = Make h the subject of the equation. h = 17 tan 55° Calculate. = 24.28 cm 17 cm 55° h 1 hyp opp adj 17 cm 55° h 2 opp adj --------- 3 h 17 ------ 4 5 13WORKEDExample Cabri Geo metry Trig ratios opposite side adjacent side ------------------------------- opposite side hypotenuse ------------------------------- adjacent side hypotenuse ------------------------------- MQ Maths A Yr 11 - 05 Page 183 Wednesday, July 4, 2001 4:39 PM
28. 28. 184 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 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. sin θ = cos θ = tan θ = 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 x, correct to 2 decimal places. THINK WRITE Label the sides of the triangle. x is the opposite side and 24 m is the hypotenuse, therefore use the sin formula. sin θ = Substitute for θ and the hypotenuse. sin 50° = Make x the subject of the equation. x = 24 sin 50° Calculate. = 18.39 m 24 m 50° x 1 hyp opp adj 24 m 50° x 2 opp hyp --------- 3 x 24 ------ 4 5 14WORKEDExample S O H        opp hyp --------- C A H        adj hyp --------- T O A        opp adj --------- Find the length of the side marked z in the triangle at right. THINK WRITE Label the sides opp, adj and hyp. Choose the cosine ratio because we are ﬁnding the hypotenuse and have been given the adjacent side. Write the formula. cos θ = 12.5 m 23°15' z 1 opp hyp adj 12.5 m 23°15' z 2 3 adj hyp --------- 15WORKEDExample MQ Maths A Yr 11 - 05 Page 184 Wednesday, July 4, 2001 4:39 PM
29. 29. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 185 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. THINK WRITE Substitute for θ and the adjacent side. cos 23°15′ = Make z the subject of the equation. z cos 23°15′ = 12.5 z = Calculate. = 13.60 m Note: In calculating the length of the hypotenuse, the process will always involve division by the trigonometric function. 4 12.5 z ---------- 5 12.5 cos 23°15′ -------------------------- 6 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 WRITE Draw a diagram and show information. Label the sides of the triangle opp, adj and hyp. Choose the cosine ratio because we are ﬁnding the hypotenuse and have been given the adjacent side. Write the formula. cos θ = Substitute for θ and the adjacent side. cos 15° = Make f the subject of the equation. f cos 15° = 100 f = Calculate. = 103.5 m Give a written answer. The cable is approximately 103.5 m long. 1 15° 100 m opp hyp adj f 2 3 4 adj hyp --------- 5 100 f --------- 6 100 cos 15° ------------------ 7 8 16WORKEDExample MQ Maths A Yr 11 - 05 Page 185 Wednesday, July 4, 2001 4:39 PM
30. 30. 186 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 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 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). remember 1. Trigonometry can be used to ﬁnd a side in a right-angled triangle when we are given the length of one side and the size of an angle. 2. The trig formulas are: sin θ = cos θ = tan θ = 3. 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. opp hyp --------- adj hyp --------- opp adj --------- remember 5D θ α γ WORKED Example 13 51 mm 71° x 13 m 23° a 35 cm 31° d MQ Maths A Yr 11 - 05 Page 186 Wednesday, July 4, 2001 4:39 PM
31. 31. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 187 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 6 Find the length of the side marked with the pronumeral in each of the following (correct to 1 decimal place). a b c 7 Find the length of the side marked with the pronumeral in each of the following (correct to 1 decimal place). a b c d e f g h i j k l WORKED Example 14 68° 13 cmx 49° 48 m y 41° 12.5 km z WORKED Example 15 21° 4.8 m t 77° 87 mm p 36° 8.2 m q 23° 2.3 m a 39° 0.85 kmb 76° 8.5 km x 116 mm 9° m 16.75 cm 11° d 13° 64.75 m x 83° 44.3 m x 20° 15.75 km g 2.34 m 84°9' m 84.6 km 60°32' q 21.4 m 75°19't 26.8 cm 29°32' r MQ Maths A Yr 11 - 05 Page 187 Wednesday, July 4, 2001 4:39 PM
32. 32. 188 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 8 Look at the diagram at right and state which of the following is correct. 9 Study the triangle at right and state which of the following is correct. 10 Which of the statements below is not correct? A The value of tan θ can never be greater than 1. B The value of sin θ can never be greater than 1. C The value of cos θ can never be greater than 1. D tan 45° = 1 11 Study the diagram at right and state which of the statements is correct. 12 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 10-m 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). 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. 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 x = 9.2 sin 69° B C x = 9.2 cos 69° D A tan φ = B tan φ = C sin φ = D cos φ = A w = 22 cos 36° B C w = 22 cos 54° D w = 22 sin 54° mmultiple choiceultiple choice 69° 9.2 x x 9.2 sin 69° -----------------= x 9.2 cos 69° ------------------= mmultiple choiceultiple choice 17 15 8 φ 8 15 ------ 15 8 ------ 15 17 ------ 8 17 ------ mmultiple choiceultiple choice mmultiple choiceultiple choice 36° 22 mm w w 22 sin 36° -----------------= WORKED Example 16 60° Port 23 km A B MQ Maths A Yr 11 - 05 Page 188 Wednesday, July 4, 2001 4:39 PM
33. 33. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 189 18 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. 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. 1 2 3 4 5 6 7 8 9 10 1 15 cm 8 cm a 10 m 20 m b 30 km 40 km c 21° 23 md 40° 39.2 cm e 37° 42.1 m f 50° 14.9 mm g 42° 119 mm h 17° 93.2 m i 45° 17 m j MQ Maths A Yr 11 - 05 Page 189 Wednesday, July 4, 2001 4:39 PM
34. 34. 190 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 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 θ. Using the formula sin θ = we know that in this triangle sin θ = = = 0.5 We then calculate sin−1 (0.5) to ﬁnd that θ = 30°. As with all trigonometry it is important that you have your calculator set to degrees mode for this work. 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. 10 cm 5 cm θopp hyp --------- 5 10 ------ 1 2 --- Find the size of angle θ, correct to the nearest degree, in the triangle at right. THINK WRITE Label the sides of the triangle and choose the tan ratio. tan θ = Substitute for the opposite and adjacent sides in the triangle. = Make θ the subject of the equation. θ = tan−1 Calculate. = 33° 6.5 m 4.3 m θ 1 4.3 m 6.5 m adj hyp opp θ opp adj --------- 2 4.3 6.5 ------- 3 4.3 6.5 -------     4 17WORKEDExample MQ Maths A Yr 11 - 05 Page 190 Wednesday, July 4, 2001 4:39 PM
36. 36. 192 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d Finding angles 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 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 c 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 remember 1. Make sure that the calculator is in degrees mode. 2. To ﬁnd an angle given the trig ratio, press or 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 30. 4. 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. 2nd F SHIFT remember 5E WORKED Example 17 7 m 12 m θ 11 m 3 m φ 25 mm 162 mm γ WORKED Example 18 24 m 13 m θ 4.6 m 6.5 m θ 9.7 km 5.6 km α 15 cm 9 cm θ 2.6 m4.6 m α 27.8 cm 19.5 cm β MQ Maths A Yr 11 - 05 Page 192 Wednesday, July 4, 2001 4:39 PM
37. 37. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 193 4 In the following triangles, you will need to use all three trig ratios. Find the size of the angle marked θ, correct to the nearest degree. a b c d e f 5 In each of the following ﬁnd the size of the angle marked θ, correct to the nearest minute. a b c d e f 6 Look at the triangle drawn at right. Which of the statements below is correct? 7 Consider the triangle drawn at right. θ is closest to: 8 The exact value of sin . The angle θ = A ∠ABC = 30° B ∠ABC = 60° C ∠CAB = 30° D ∠ABC = 45° A 41°55′ B 41°56′ C 48°4′ D 48°5′ A 30° B 45° C 60° D 90° 7 cm 11 cm θ 15 cm 8 cm θ 14 cm 9 cm θ 3.6 m 9.2 m θ 196 mm 32 mm θ 14.9 m26.8 m θ 30 m 19.2 m θ 10 cm 63 cm θ 2.5 m 0.6 m θ 3.5 m 18.5 m θ 16.3 m 8.3 m θ 6.3 m 18.9 m θ mmultiple choiceultiple choice 10 cm 5 cm A BC θ mmultiple choiceultiple choice 12.5 m 9.3 m θ mmultiple choiceultiple choice θ 3 2 -------= MQ Maths A Yr 11 - 05 Page 193 Wednesday, July 4, 2001 4:39 PM
38. 38. 194 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 9 A 10-m ladder leans against a wall 6 m high. Find the angle that the ladder makes with the horizontal, correct to the nearest degree. 10 A kite is ﬂying on a 40-m 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 the nearest minute. 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. Find the error in the compass reading, correct to the nearest minute. 12 The diagram below shows a footballer’s shot at goal. By dividing the isosceles triangle in half in order to form a right-angled 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. WORKED Example 19 40 m 10 m kite 10 nm Port 1.5 nm 7 m 30 m MQ Maths A Yr 11 - 05 Page 194 Wednesday, July 4, 2001 4:39 PM
39. 39. C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 195 1 Find the length of the hypotenuse in the triangle at right. 2 Find x, correct to 1 decimal place. 3 Find y, correct to 1 decimal place. 4 Find z, correct to 3 decimal places. 5 Calculate sin 56°, correct to 4 decimal places. 6 Calculate 9.2 tan 50°, correct to 2 decimal places. 7 Calculate , correct to 2 decimal places. 8 Find θ, given that sin θ = 0.5. 9 Find θ to the nearest degree, given that cos θ = 0.299. 10 Find θ to the nearest minute, given that tan θ = 2. 2 12 cm 5 cm 12 cm 12 cm x 30 cm 20 cm y 3.7 m 7.4 m z 132 cos 8°45′ ----------------------- MQ Maths A Yr 11 - 05 Page 195 Wednesday, July 4, 2001 4:39 PM
40. 40. 196 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d 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. 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. Angle of elevation Horizontal Angle of depression Horizontal 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 WRITE Label the sides of the triangle opp, adj and hyp. Choose the tangent ratio because we are ﬁnding the opposite side and have been given the adjacent side. Write the formula. tan θ = Substitute for θ and the adjacent side. tan 40° = 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. 50 m 40° h 1 50 m 40° hyp opp h adj 2 3 opp adj --------- 4 h 50 ------ 5 6 7 20WORKEDExample 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. 10 m 1.6 m 64° 21WORKEDExample MQ Maths A Yr 11 - 05 Page 196 Wednesday, July 4, 2001 4:39 PM