Engineering Mathematics 4 Episode 6 ppsx

TRIGONOMETRIC WAVEFORMS 189 30° y = 5 sin A y = 5 sin ( A + 30°) 30° 090° 180° 270° 360° A ° 5 −5 y Figure 22.23 In general, y = sin.pt − a/ lags y = sin pt by a=p, hence 7 sin2A /3 lags 7 sin 2A by /3/2, i.e. /6 rad or 30 ° Asketchofy D 7sin2A  /3 is shown in Fig. 22.24. π/6 π/2 3π/2 2ππ π/6 y 7 090° 180° 270° 360° A ° −7 y = 7sin 2 A y = 7sin (2 A − π/3) Figure 22.24 Problem 10. Sketch y D 2cosωt 3/10 over one cycle Amplitude D 2 and period D 2/ω rad. 2cosωt3/10 lags 2 cos ωt by 3/10ω seconds. Asketchofy D 2cosωt  3/10 is shown in Fig. 22.25. Now try the following exercise Exercise 85 Further problems on sine and cosine curves In Problems 1 to 7 state the amplitude and period of the waveform and sketch the curve between 0 ° and 360 ° . 1. y D cos 3A [1, 120 ° ] 3p/10w rads y = 2 cos w t y = 2 cos (w t − 3p/10) 2 0 −2 p/2wp/w 3p/2w 2p/w t y Figure 22.25 2. y D 2sin 5x 2 [2, 144 ° ] 3. y D 3sin4t [3, 90 ° ] 4. y D 3cos  2 [3, 720 ° ] 5. y D 7 2 sin 3x 8  7 2 , 960 °  6. y D 6sint  45 °  [6, 360 ° ] 7. y D 4cos2 C30 °  [4, 180 ° ] 22.5 Sinusoidal form A sin.!t ± a/ In Fig. 22.26, let OR represent a vector that is free to rotate anticlockwise about O at a veloc- ity of ω rad/s. A rotating vector is called a pha- sor. After a time t seconds OR will have turned through an angle ωt radians (shown as angle TOR in Fig. 22.26). If ST is constructed perpendicular to OR,thensinωt D ST/OT,i.e.ST D OT sin ωt. y 1.0 −1.0 00 SR ω t π/2 3π/22πω t π ω rads/s ω t 90° 180° 270° 360° y = sin ω t T Figure 22.26 If all such vertical components are projected on to a graph of y against ωt, a sine wave results of amplitude OR (as shown in Section 22.3). 190 ENGINEERING MATHEMATICS If phasor OR makes one revolution (i.e. 2 radians) in T seconds, then the angular velocity, ω D 2/T rad/s, from which, T = 2p=! seconds T is known as the periodic time. The number of complete cycles occurring per second is called the frequency, f Frequency D number of cycles second D 1 T D ω 2 Hz i.e. f = ! 2p Hz Hence angular velocity, ! = 2pf rad/s Amplitude is the name given to the maximum or peak value of a sine wave, as explained in Section 22.4. The amplitude of the sine wave shown in Fig. 22.26 has an amplitude of 1. A sine or cosine wave may not always start at 0 ° . To show this a periodic function is represented by y D sinωt š ˛ or y D cosωt š ˛,where˛ is a phase displacement compared with y D sin A or y D cos A.Agraphofy D sinωt  ˛ lags y D sin ωt by angle ˛, and a graph of y D sinωt C ˛ leads y D sin ωt by angle ˛. The angle ωt is measured in radians  i.e.  ω rad s  t s D ωt radians  hence angle ˛ should also be in radians. The relationship between degrees and radians is: 360 ° D 2 radians or 180 ° = p radians Hence 1 rad D 180  D 57.30 ° and, for example, 71 ° D 71 ð  180 D 1.239 rad Summarising, given a general sinusoidal function y = A sin.!t ± a/, then: (i) A D amplitude (ii) ω D angular velocity D 2f rad/s (iii) 2 ω D periodic time T seconds (iv) ω 2 D frequency, f hertz (v) ˛ D angle of lead or lag (compared with y D A sin ωt) Problem 11. An alternating current is given by i D 30 sin100t C0.27 amperes. Find the amplitude, periodic time, frequency and phase angle (in degrees and minutes) i D 30 sin100tC0.27A, hence amplitude = 30 A. Angular velocity ω D 100, hence periodic time, T D 2 ω D 2 100 D 1 50 D 0 .02 s or 20 ms Frequency, f D 1 T D 1 0.02 D 50 Hz Phase angle, a D 0.27 rad D  0.27 ð 180   ° D 15.47 ° or 15 ° 28  leading i = 30 sin.100pt/ Problem 12. An oscillating mechanism has a maximum displacement of 2.5 m and a frequency of 60 Hz. At time t D 0the displacement is 90 cm. Express the displacement in the general form A sinωt š˛ Amplitude D maximum displacement D 2.5m Angular velocity, ω D 2f D 260 D 120 rad/s Hence displacement D 2.5sin120t C ˛ m When t D 0, displacement D 90 cm D 0.90 m Hence, 0.90 D 2.5sin0 C˛ i.e. sin ˛ D 0.90 2.5 D 0.36 Hence ˛ D sin 1 0.36 D 21.10 ° D 21 ° 6 0 D 0.368 rad Thus, displacement = 2.5sin.120pt Y 0.368/ m TRIGONOMETRIC WAVEFORMS 191 Problem 13. The instantaneous value of voltage in an a.c. circuit at any time t seconds is given by v D 340 sin50t 0.541 volts. Determine the: (a) amplitude, periodic time, frequency and phase angle (in degrees) (b) value of the voltage when t D 0 (c) value of the voltage when t D 10 ms (d) time when the voltage first reaches 200 V, and (e) time when the voltage is a maximum Sketch one cycle of the waveform (a) Amplitude = 340 V Angular velocity, ω D 50 Hence periodic time, T D 2 ω D 2 50 D 1 25 D 0 .04 s or 40 ms Frequency f D 1 T D 1 0.04 D 25 Hz Phase angle D 0.541 rad D  0.541 ð 180   D 31 ° lagging v D 340 sin50t (b) When t D 0, v D 340 sin0 0.541 D 340 sin31 °  D −175.1V (c) When t = 10 ms, then v D 340 sin  50 10 10 3  0.541  D 340 sin1.0298 D 340 sin 59 ° D 291.4 volts (d) When v D 200 volts, then 200 D 340 sin50t 0.541 200 340 D sin50t 0.541 Hence 50t 0.541 D sin 1 200 340 D 36.03 ° or 0.6288 rad 50t D 0.6288 C0.541 D 1.1698 Hence when v D 200 V, time, t D 1.1698 50 D 7 .447 ms (e) When the voltage is a maximum, v D 340 V Hence 340 D 340 sin50t 0.541 1 D sin50t  0.541 50t  0.541 D sin 1 1 D 90 ° or 1.5708 rad 50t D 1.5708 C0.541 D 2.1118 Hence time, t D 2.1118 50 D 13 .44 ms Asketchof v D 340 sin50t 0.541 volts is shown in Fig. 22.27. Voltage v 340 291.4 200 0 −175.1 −340 7.447 13.44 10 20 30 40 t (ms) v = 340 sin (50π t − 0.541) v = 340 sin 50π t Figure 22.27 Now try the following exercise Exercise 86 Further problems on the sinusoidal form A sin .!t ± a/ In Problems 1 to 3 find the amplitude, peri- odic time, frequency and phase angle (stating whether it is leading or lagging sin ωt)ofthe alternating quantities given. 1. i D 40 sin50t C0.29 mA  40, 0.04 s, 25 Hz, 0.29 rad (or 16 ° 37 0 ) leading 40 sin 5t  2. y D 75 sin40t 0.54 cm  75 cm, 0.157 s, 6.37 Hz, 0.54 rad (or 30 ° 56 0 ) lagging 75 sin 40t  192 ENGINEERING MATHEMATICS 3. v D 300 sin200t 0.412 V  300 V, 0.01 s, 100 Hz, 0.412 rad (or 23 ° 36 0 ) lagging 300 sin 200t  4. A sinusoidal voltage has a maximum value of 120 V and a frequency of 50 Hz. At time t D 0, the voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form v D A sinωt š˛.  (a) v D 120 sin 100t volts (b) v D 120 sin100t C0.43 volts  5. An alternating current has a periodic time of 25 ms and a maximum value of 20 A. When time t D 0, current i D10 amperes. Express the current i in the form i D A sinωt š˛.  i D 20 sin  80t   6  amperes  6. An oscillating mechanism has a maximum displacement of 3.2 m and a frequency of 50 Hz. At time t D 0 the displacement is 150 cm. Express the displacement in the general form A sinωt š ˛. [3.2sin100t C0.488) m] 7. The current in an a.c. circuit at any time t seconds is given by: i D 5sin100t 0.432 amperes Determine (a) the amplitude, periodic time, frequency and phase angle (in degrees) (b) the value of current at t D 0, (c) the value of current at t D 8ms, (d) the time when the current is first a maximum, (e) the time when the current first reaches 3A. Sketch one cycle of the waveform showing relevant points.         (a) 5 A, 20 ms, 50 Hz, 24 ° 45 0 lagging (b) 2.093 A (c) 4.363 A (d) 6.375 ms (e) 3.423 ms         22.6 Waveform harmonics Let an instantaneous voltage v be represented by v D V m sin 2ft volts. This is a waveform which varies sinusoidally with time t, has a frequency f, and a maximum value V m . Alternating voltages are usually assumed to have wave-shapes which are sinusoidal where only one frequency is present. If the waveform is not sinusoidal it is called a com- plex wave, and, whatever its shape, it may be split up mathematically into components called the fun- damental and a number of harmonics. This process is called harmonic analysis. The fundamental (or first harmonic) is sinusoidal and has the supply fre- quency, f; the other harmonics are also sine waves having frequencies which are integer multiples of f. Thus, if the supply frequency is 50 Hz, then the third harmonic frequency is 150 Hz, the fifth 250 Hz, and so on. A complex waveform comprising the sum of the fundamental and a third harmonic of about half the amplitude of the fundamental is shown in Fig. 22.28(a), both waveforms being initially in phase with each other. If further odd harmonic waveforms of the appropriate amplitudes are added, Complex waveform Complex waveform Fundamental Fundamental Third harmonic Third harmonic v 0 v 0 v 0 v 0 v 0 v 0 t t t t t t (b)(a) (c) (e) (f) (d) A B Complex waveform Complex waveform Fundamental Fundamental Second harmonic Second harmonic Complex waveform Complex waveform Fundamental Fundamental Second harmonic Second harmonic Third harmonic Third harmonic Figure 22.28 TRIGONOMETRIC WAVEFORMS 193 a good approximation to a square wave results. In Fig. 22.28(b), the third harmonic is shown having an initial phase displacement from the fundamental. The positive and negative half cycles of each of the complex waveforms shown in Figs. 22.28(a) and (b) are identical in shape, and this is a feature of waveforms containing the fundamental and only odd harmonics. A complex waveform comprising the sum of the fundamental and a second harmonic of about half the amplitude of the fundamental is shown in Fig. 22.28(c), each waveform being initially in phase with each other. If further even harmonics of appropriate amplitudes are added a good approxi- mation to a triangular wave results. In Fig. 22.28(c), the negative cycle, if reversed, appears as a mirror image of the positive cycle about point A. In Fig. 22.28(d) the second harmonic is shown with an initial phase displacement from the fundamen- tal and the positive and negative half cycles are dissimilar. A complex waveform comprising the sum of the fundamental, a second harmonic and a third harmonic is shown in Fig. 22.28(e), each waveform being initially ‘in-phase’. The negative half cycle, if reversed, appears as a mirror image of the positive cycle about point B. In Fig. 22.28(f), a complex waveform comprising the sum of the fundamental, a second harmonic and a third harmonic are shown with initial phase displacement. The positive and negative half cycles are seen to be dissimilar. The features mentioned relative to Figs. 22.28(a) to (f) make it possible to recognise the harmonics present in a complex waveform. 23 Cartesian and polar co-ordinates 23.1 Introduction There are two ways in which the position of a point in a plane can be represented. These are (a) by Cartesian co-ordinates,i.e.(x, y), and (b) by polar co-ordinates,i.e.(r, Â), where r is a ‘radius’ from a fixed point and  is an angle from a fixed point. 23.2 Changing from Cartesian into polar co-ordinates In Fig. 23.1, if lengths x and y are known, then the length of r can be obtained from Pythagoras’ theorem (see Chapter 21) since OPQ is a right- angled triangle. Hence r 2 D x 2 C y 2  from which, r =  x 2 Y y 2 y OQ P x x q r y Figure 23.1 From trigonometric ratios (see Chapter 21), tan  D y x from which q = tan −1 y x r D  x 2 C y 2 and  D tan 1 y x are the two formulae we need to change from Cartesian to polar co- ordinates. The angle Â, which may be expressed in degrees or radians, must always be measured from the positive x-axis, i.e. measured from the line OQ in Fig. 23.1. It is suggested that when changing from Cartesian to polar co-ordinates a diagram should always be sketched. Problem 1. Change the Cartesian co-ordinates (3, 4) into polar co-ordinates. A diagram representing the point (3, 4) is shown in Fig. 23.2. q Figure 23.2 From Pythagoras’ theorem, r D p 3 2 C 4 2 D 5 (note that  5 has no meaning in this con- text). By trigonometric ratios,  D tan 1 4 3 D 53.13 ° or 0.927 rad [note that 53.13 ° D 53.13 ð/180 rad D 0.927 rad.] Hence (3, 4) in Cartesian co-ordinates corre- sponds to (5, 53.13 ° ) or (5, 0.927 rad) in polar co-ordinates. Problem 2. Express in polar co-ordinates the position (4, 3) A diagram representing the point using the Cartesian co-ordinates (4, 3) is shown in Fig. 23.3. From Pythagoras’ theorem, r D p 4 2 C 3 2 D 5 By trigonometric ratios, ˛ D tan 1 3 4 D 36.87 ° or 0.644 rad. Hence  D 180 °  36.87 ° D 143.13 ° or  D  0.644 D 2.498 rad . CARTESIAN AND POLAR CO-ORDINATES 195 a q q Figure 23.3 Hence the position of point P in polar co-ordinate form is (5, 143.13 ° ) or (5, 2.498 rad). Problem 3. Express (5, 12) in polar co-ordinates. A sketch showing the position (5, 12) is shown in Fig. 23.4. r D  5 2 C 12 2 D 13 and ˛ D tan 1 12 5 D 67.38 ° or 1.176 rad . Hence  D 180 ° C 67.38 ° D 247.38 ° or  D  C1.176 D 4.318 rad . a q Figure 23.4 Thus (−5, −12) in Cartesian co-ordinates corre- sponds to (13, 247.38 ° ) or (13, 4.318 rad) in polar co-ordinates. Problem 4. Express (2, 5) in polar co-ordinates. A sketch showing the position (2, 5) is shown in Fig. 23.5. r D  2 2 C 5 2 D p 29 D 5.385 correct to 3 decimal places ˛ D tan 1 5 2 D 68.20 ° or 1.190 rad a q Figure 23.5 Hence  D 360 °  68.20 ° D 291.80 ° or  D 2 1.190 D 5.093 rad Thus (2, −5) in Cartesian co-ordinates corre- sponds to (5.385, 291.80 ° ) or (5.385, 5.093 rad) in polar co-ordinates. Now try the following exercise Exercise 87 Further problems on chang- ing from Cartesian into polar co-ordinates In Problems 1 to 8, express the given Carte- sian co-ordinates as polar co-ordinates, correct to 2 decimal places, in both degrees and in radians. 1. (3, 5) [(5.83, 59.04 ° or (5.83, 1.03 rad)] 2. (6.18, 2.35) [(6.61, 20.82 ° or (6.61, 0.36 rad)] 3. (2, 4) [(4.47, 116.57 ° or (4.47, 2.03 rad)] 4. (5.4, 3.7) [(6.55, 145.58 ° or (6.55, 2.54 rad)] 5. (7, 3) [(7.62, 203.20 ° or (7.62, 3.55 rad)] 6. (2.4, 3.6) [(4.33, 236.31 ° ) or (4.33, 4.12 rad)] 7. (5, 3) [(5.83, 329.04 ° ) or (5.83, 5.74 rad)] 8. (9.6, 12.4) [(15.68, 307.75 ° ) or (15.68, 5.37 rad)] 196 ENGINEERING MATHEMATICS 23.3 Changing from polar into Cartesian co-ordinates From the right-angled triangle OPQ in Fig. 23.6. cos  D x r and sin  D y r , from trigonometric ratios Hence x = r cos q and y = r sin q q Figure 23.6 If length r and angle  are known then x D r cos  and y D r sin  are the two formulae we need to change from polar to Cartesian co-ordinates. Problem 5. Change (4, 32 ° )intoCartesian co-ordinates. A sketch showing the position (4, 32 ° ) is shown in Fig. 23.7. Now x D r cos  D 4 cos 32 ° D 3.39 and y D r sin  D 4sin32 ° D 2.12 q q Figure 23.7 Hence (4, 32 ° ) in polar co-ordinates corresponds to (3.39, 2.12) in Cartesian co-ordinates. Problem 6. Express (6, 137 ° )inCartesian co-ordinates. A sketch showing the position (6, 137 ° ) is shown in Fig. 23.8. x D r cos  D 6 cos 137 ° D4.388 which corresponds to length OA in Fig. 23.8. y D r sin  D 6 sin 137 ° D 4.092 which corresponds to length AB in Fig. 23.8. q q q Figure 23.8 Thus (6, 137 ° ) in polar co-ordinates corresponds to ( −4.388, 4.092) in Cartesian co-ordinates. (Note that when changing from polar to Cartesian co-ordinates it is not quite so essential to draw a sketch. Use of x D r cos  and y D r sin  automatically produces the correct signs.) Problem 7. Express (4.5, 5.16 rad) in Cartesian co-ordinates. A sketch showing the position (4.5, 5.16 rad) is shown in Fig. 23.9. x D r cos  D 4.5cos5.16 D 1.948 which corresponds to length OA in Fig. 23.9. qqqqq Figure 23.9 CARTESIAN AND POLAR CO-ORDINATES 197 y D r sin  D 4. 5sin5. 16 D4.057 which corresponds to length AB in Fig. 23.9. Thus (1.948, −4.057) in Cartesian co-ordinates corresponds to (4.5, 5.16 rad) in polar co- ordinates. 23.4 Use of R → P and P → R functions on calculators Another name for Cartesian co-ordinates is rectan- gular co-ordinates. Many scientific notation calcu- lators possess R ! P and P ! R functions. The R is the first letter of the word rectangular and the P is the first letter of the word polar. Check the operation manual for your particular calculator to determine how to use these two functions. They make changing from Cartesian to polar co-ordinates, and vice-versa, so much quicker and easier. Now try the following exercise Exercise 88 Further problems on chang- ing polar into Cartesian co- ordinates In Problems 1 to 8, express the given polar co-ordinates as Cartesian co-ordinates, correct to3decimalplaces. 1. (5, 75 ° ) [(1.294, 4.830)] 2. (4.4, 1.12 rad) [(1.917, 3.960)] 3. (7, 140 ° )[(5,362, 4.500)] 4. (3.6, 2.5 rad) [(2.884, 2.154)] 5. (10.8, 210 ° )[(9.353, 5.400)] 6. (4, 4 rad) [(2.615, 3.207)] 7. (1.5, 300 ° ) [(0.750, 1.299)] 8. (6, 5.5 rad) [(4.252, 4.233)] 198 ENGINEERING MATHEMATICS Assignment 6 This assignment covers the material in Chapters 21 to 23. The marks for each question are shown in brackets at the end of each question. 1. Fig. A6.1 shows a plan view of a kite design. Calculate the lengths of the dimensions shown as a and b. (4) 2. In Fig. A6.1, evaluate (a) angle  (b) angle ˛ (5) 3. Determine the area of the plan view of a kite shown in Fig. A6.1. (4) 20.0 cm 42.0 cm 60.0 cm b a a q Figure A6.1 4. If the angle of elevation of the top of a 25 m perpendicular building from point A is measured as 27 ° ,determinethe distance to the building. Calculate also the angle of elevation at a point B,20m closer to the building than point A.(5) 5. Evaluate, each correct to 4 significant figures: (a) sin 231.78 ° (b) cos 151 ° 16 0 (c) tan 3 8 (3) 6. Sketch the following curves labelling relevant points: (a) y D 4cos C 45 °  (b) y D 5sin2t 60 °  (6) 7. Solve the following equations in the range 0 ° to 360 ° (a) sin 1 0.4161 D x (b) cot 1 2.4198 D  (6) 8. The current in an alternating current cir- cuit at any time t seconds is given by: i D 120 sin100t C 0.274 amperes. Determine (a) the amplitude, periodic time, fre- quency and phase angle (with ref- erence to 120 sin 100t) (b) the value of current when t D 0 (c) the value of current when t D 6ms (d) the time when the current first reaches 80 A Sketch one cycle of the oscillation. (17) 9. Change the following Cartesian co- ordinates into polar co-ordinates, correct to 2 decimal places, in both degrees and in radians: (a) (2.3, 5.4) (b) (7.6,  9.2) (6) 10. Change the following polar co-ordinates into Cartesian co-ordinates, correct to 3 decimal places: (a) (6.5, 132 ° ) (b) (3, 3 rad) (4) [...]... sin 4t] 0.58 06 0.89 64 0.52 04 D −0. 160 D cos P cos Q 20° D 4 sin 39 .44 9° D 2. 542 Thus sin Q D sin 63 .69 ° D 0.89 64 and tan Q D tan 63 .69 ° D 2.0225 D 0.8 142 0 .44 32 1. 368 0 cos x Since 4 sin x 20° D 5 cos x then 3.7588 sin x 1. 368 0 cos x D 5 cos x Rearranging gives: Thus cos P D cos 54. 51° D 0.58 06 and tan P D tan 54. 51° D 1 .40 25 Since cos Q D 0 .44 32, Q D cos B cos x 0. 342 0 ] [Check: LHS D 4 sin 59 .44 9°... 25.0 cos 64 ] D 1225 C 62 5 767 .1 D 1083 p from which, e D 1083 D 32.91 mm 2 2 Applying the sine rule: 32.91 25.0 D ° sin 64 sin F 25.0 sin 64 from which, sin F D D 0 .68 28 32.91 202 ENGINEERING MATHEMATICS Thus F D sin 1 0 .68 28 D 43 ° 40 or 1 36 560 6 0 F D 1 36 56 is not possible in this case since 1 36 560 C 64 is greater than 180° Thus only F = 43 ° 4 is valid 6 D D 180° 64 D 35.0 25.0 sin 64 D 393.2... in Fig 24. 15 [BF D 3.9 m, EB D 4. 0 m] 4m 2.5 m E F 50° A 5m B Figure 24. 15 C 48 ° A 30.0 m 44 ° B 4m D 2.5 m 50° 5m C Figure 24. 16 Similarly, from triangle BCD, BC D DC tan 44 ° 2 06 ENGINEERING MATHEMATICS Angle OAB D 180° For triangle ABC, using Pythagoras’ theorem: DC2 2 D 30.0 2 C 1 tan2 48 ° 1 tan2 44 ° DC2 1.072323 DC tan 48 ° 2 from which, D 30.02 0.810727 D 30.02 30.02 DC D D 344 0 .4 0. 261 5 96 p Hence,... D 4[ sin x cos 20° D 4[ sin x 0.9397 4 tan x 1 1 C tan x 1 tan x D −1 1 tan x Problem 4 If sin P D 0.8 142 and cos Q D 0 .44 32 evaluate, correct to 3 decimal places: (a) sin P Q , (b) cos P C Q and (c) tan P C Q , using the compound angle formulae D 3.7588 sin x 3.7588 sin x D 5 cos x C 1. 368 0 cos x D 6. 368 0 cos x 6. 368 0 sin x D D 1 .6 942 cos x 3.7588 and i.e tan x D 1 .6 942 , and x D tan 1 1 .6 942 D 59 .44 9°... is given by 4. 3 cos 3 6. 9 sin 3 [8.13 sin 3 C 2.5 84 ] 75.23° C 360 ° D 2 84. 77° A D 1 06. 35° – 148 .88° D Solve the following equations for values of  between 0° and 360 ° : Express this in the form R sin 3 š ˛ sin A C 148 .88° D A D 73 .65 ° – 148 .88° D 2.0 34 ] (a) 72.73° or 3 54. 63 ° (b) 11.15° or 311.98° D 6. 7 74 sin A C 148 .88° D 6. 5 Thus, 6 cos ωt (a) 3 cos A C 2 sin A D 2.8 (b) 12 cos A 4 sin A D 11... cm, area D 0.9 16 cm 6 k D 46 mm, l D 36 mm, L D 35°   K D 47 ° 80 , J D 97° 520 ,  j D 62 .2 mm, area D 820.2 mm2 OR   K D 132° 520 , J D 12° 80 ,  2 j D 13.19 mm, area D 1 74. 0 mm 36. 5 29 .6 sin 10° 270 D 97.98 mm2 Triangle PQR for case 2 is shown in Fig 24. 5 133°33′ 9.1 34 mm Q d D 32 .6 mm, e D 25 .4 mm, D D 1 04 220 4 Case 2 P D 133° 330 , Q D 36 , R D 10° 270 , p D 36. 5 mm and q D 29 .6 mm From the... the second, as shown in Fig 26. 7 Now try the following exercise Exercise 99 In Problems 1 to 4, change the functions into the form R sin ωt š ˛ 1 90° 5 sin ωt C 8 cos ωt [9 .43 4 sin ωt C 1.012 ] 2 3 4 sin ωt 4 a 0° q −5.8 Figure 26. 7 5.8 2 D 6. 7 74 and 6 5.8 sin A Á or 6. 5 , from which, 6. 7 74 6. 5 D sin 1 6. 7 74 D 73 .65 ° or 1 06. 35° 7 Á Solve the following equations for 0° < A < 360 ° : 8 The displacement x... 22.31 mm and AB D 17.92 mm Find also its area Triangle ABC is shown in Fig 24. 3 c = 17.9 2 mm A b = 22.31 mm 78°51′ Triangle PQR is shown in Fig 24. 4 P r q = 29 .6 mm 36 Q p = 36. 5 mm R Figure 24. 4 Applying the sine rule: 29 .6 36. 5 D sin 36 sin P 36. 5 sin 36 from which, sin P D D 0.7 248 29 .6 a C 1 0.7 248 D 46 ° 270 Figure 24. 3 Applying the sine rule: 17.92 22.31 D ° 510 sin 78 sin C 17.92 sin 78° 510... –31.12° D 148 .88° A C 148 .88° 0. 64 4 ] (a) 2 sin  C 4 cos  D 3 (b) 12 sin  9 cos  D 7 (a) 74. 43° or 338.70° (b) 64 . 68 ° or 189.05° 270° From Fig 26. 7, R D 3.52 C 3.5  D tan 1 D 31.12° 5.8 3 sin ωt [6. 708 sin ωt 360 ° 5 Hence, [5 sin ωt [8. 062 sin ωt C 2 .62 2 ] 3.5 Thus 3.5 cos A 3 cos ωt 7 sin ωt C 4 cos ωt R 180° Further problems on the conversion of a sin !t Y b cos !t into R sin.!t Y a/ 219 220 ENGINEERING. .. 29° , B D 68 ° , b D 27 mm C D 83° , a D 14. 1 mm, c D 28.9 mm, area D 189 mm2 2 B D 71° 260 , C D 56 320 , b D 8 .60 cm A D 52° 20 , c D 7. 568 cm, a D 7.152 cm, area D 25 .65 cm2 In Problems 3 and 4, use the sine rule to solve the triangles DEF and find their areas 3 d D 17 cm, f D 22 cm, F D 26 D D 19° 48 0 , E D 1 34 120 , e D 36. 0 cm, area D 1 34 cm2 f = 25.0 mm e 64 E d = 35.0 mm F Figure 24. 6 Applying . 5) [(5.83, 59. 04 ° or (5.83, 1.03 rad)] 2. (6. 18, 2.35) [ (6. 61, 20.82 ° or (6. 61, 0. 36 rad)] 3. (2, 4) [ (4. 47, 1 16. 57 ° or (4. 47, 2.03 rad)] 4. (5 .4, 3.7) [ (6. 55, 145 .58 ° or (6. 55, 2. 54 rad)] 5 MATHEMATICS Thus 6 F D sin 1 0 .68 28 D 43 ° 4 0 or 1 36 ° 56 0 F D 1 36 ° 56 0 is not possible in this case since 1 36 ° 56 0 C 64 ° is greater than 180 ° . Thus only F = 43 ° 4  is valid. 66 D D 180 °  64 ° . 75 ° ) [(1.2 94, 4. 830)] 2. (4. 4, 1.12 rad) [(1.917, 3. 960 )] 3. (7, 140 ° )[(5, 362 , 4. 500)] 4. (3 .6, 2.5 rad) [(2.8 84, 2.1 54) ] 5. (10.8, 210 ° )[(9.353, 5 .40 0)] 6. (4, 4 rad) [(2 .61 5, 3.207)] 7.