MATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 1 TRIGONOMETRIC RATIOS, TRIGONOMETRIC TECHNIQUES AND GRAPHICAL METHODS



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Transcription:

MATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 1 TRIGONOMETRIC RATIOS, TRIGONOMETRIC TECHNIQUES AND GRAPHICAL METHODS This is the ne f a series f basic tutrials in mathematics aimed at beginners r anyne wanting t refresh themselves n fundamentals. The tutrial cntains the fllwing. Units f angular measurement. The right angle triangle. Trignmetric Ratis Sine Csine Tangent Sinusidal functins Harmnic mtin Angular Velcity Angular frequency Frequency Peridic time D.J.Dunn www.freestudy.c.uk 1

1. ANGLES REVOLUTION A pint n a wheel that rtates ne revlutin traces ut a circle. One revlutin is the angle f rtatin. This is a bit crude fr use in calculatins and we need smaller parts f the revlutin. DEGREES Traditinally we divide ne revlutin int 360 parts and call this a degree with symbl. 1 revlutin = 360 The picture shws a circle divided int 360 parts. They are s clse they can hardly be seen individually. Even s, a single degree is nt accurate enugh fr many applicatins s we divide a degree up int smaller parts called minutes. 1 = 60 minutes r 60' A minute can be divided up int even smaller bits called secnds and 1 minute = 60 secnds r 60". In mdern times we use decimals t express angles accurately s yu are unlikely t use minutes and secnds. GRADS In France, they divide the circle up int 400 parts and this is called a Grad. 1 revlutin = 400 Grad. This makes a quarter f a circle 100 Grads whereas in degrees it wuld be 90. RADIAN In Engineering and Science, we use anther measurement f angle defined as the angle created by placing a line f length 1 radius arund the edge f the circle as shwn. In mathematical wrds it is the angle subtended by an arc f length ne radius. This angle is called the RADIAN. The circumference f a circle is 2πR. It fllws that the number f radians that make a cmplete circle is 2π R r 2π. R There are 2π radians in ne revlutin s 360 =2π radian 1 radian = 360/2π = 57.296 called the Radian. This is In the fllwing wrk we will be using degrees and radian s it is very imprtant that yu make sure yur calculatr is set t the units that yu are ging t use. Yu might find a buttn labelled DRG n yur calculatr. Press this repeatedly until the display shws either D (fr degrees) r R (fr Radian) r if yu are French, G (fr Grad). On ther calculatrs yu might have t d this by using the mde buttn s read yur instructin bk. D.J.Dunn www.freestudy.c.uk 2

2. THE RIGHT ANGLE TRIANGLE A triangle with a 90 angle is called a right angle triangle. The angle θ (theta) is used here as the reference angle but many ther symbls are used. The internal angles f a triangle always add up t 180. It fllws that the ther angle is θ 90 There are many ways t designate the three sides f the triangle. The Lngest side f the triangle is called the HYPOTENUSE. The side ppsite the angle is called the OPPOSITE The ther side is called the ADJACENT There are varius ways t identify the sides f a triangle. One f the easiest t remember is t letter the crners say A, B and C and identify the side ppsite the crner by the lwer case letter a, b and c as shwn. We culd identify the sides by the letters at the crners s a = BC, b = AB and c = AC Any letters may be used and we will nt always use the nes shwn. Nte that the angle is ften defined by the letter and the symbl fr s in the diagram θ culd be designated A PYTHAGORAS LAW Withut prf it can be shwn that c 2 = a 2 + b 2 WORKED EXAMPLE N.1 A right angle triangle has an adjacent side 50 mm lng and an ppsite side 30 mm lng. Calculate the hyptenuse. Using the ntatin shwn n the diagram c 2 = a 2 + b 2 = 30 2 + 50 2 = 900 + 2500 = 3400 (Use the square buttn n yur calculatr (buttn usually shwn as x 2 ) c = 3400 = 58.31 mm (Use the square rt buttn usually shwn as ) WORKED EXAMPLE N.2 Calculate the length f the unknwn side fr the triangle shwn. c = 260 a = 120 mm c 2 = a 2 + b 2 Subtract frm bth sides c 2 - a 2 = b 2 Evaluate 260 2-120 2 = z 2 b 2 = 53200 b = 53200 = 230.7 mm D.J.Dunn www.freestudy.c.uk 3

SELF ASSESSMENT EXERCISE N.1 1. A right angle triangle has an adjacent side 200 mm lng and an ppsite side 60 mm lng. Calculate the hyptenuse t ne significant decimal pint. (208.8 mm) 2. A right angle triangle has a hyptenuse 300 mm lng and an ppsite side 100 mm lng. Calculate the adjacent side t ne significant decimal pint. (282.8 mm) 3. A right angle triangle has a hyptenuse 440 mm lng and an adjacent side 250 mm lng. Calculate the ppsite side t ne significant decimal pint. (362.1 mm) 4. What is the length f the hyptenuse fr the triangle shwn? (94.8 mm) 3. TRIGONOMETRIC RATIOS The ratis f the lengths f the sides f a triangle are always the same fr any given angle θ. These ratis are very imprtant because they allw us t calculate lts f things t d with triangles. SINE A1B1 A2B2 A3B3 If yu measured all the lengths n the diagram yu wuld find that = = OA1 OA2 OA3 This rati is called the sine f the angle. It is best t remember this as the rati f the ppsite side Oppsite t the hyptenuse and we write it as sin( θ ) = (nte we usually drp the e n sine) Hyptenuse Befre the use f calculatrs, the values f the sine f angles were placed in tables but all yu have t d is enter the angle int yur calculatr and press the buttn shwn as sin. Fr example if yu enter 60 int yur calculatr in degree mde and press sin yu get 0.8660 If yu enter 0.2 int yur calculatr in radian mde and press sin yu get 0.1987 D.J.Dunn www.freestudy.c.uk 4

COSINE OB1 OB2 OB3 Fr all the triangles shwn we wuld als find that = = OA1 OA2 OA3 This rati is called the csine f the angle and it is abbreviated t cs. It is best t remember it as Adjacent the rati f the adjacent side t the hyptenuse. cs( θ ) = Hyptenuse On yur calculatr the buttn is labelled cs. Fr example enter 60 int yur calculatr in degree mde and press the cs buttn. Yu shuld btain 0.5 If yu enter 0.2 int yur calculatr in radian mde and press cs yu get 0.9800 TANGENT A1B1 A2B2 A3B3 Fr all the triangles shwn we wuld als find that = = OB1 OB2 OB3 This rati is called the tangent f the angle and it is abbreviated t tan. It is best t remember it as Oppsite the rati f the ppsite side t the adjacent. tan( θ ) = Adjacent On yur calculatr the buttn is labelled tan. Fr example enter 60 int yur calculatr in degree mde and press the tan buttn and yu shuld btain 1.732. If yu enter 0.2 int yur calculatr in radian mde and press tan yu get 0.2027 SELF ASSESSMENT EXERCISE N. 2 Using ruler and prtractr draw a right angle triangle with an angle f 60 as shwn and the vertical side 100 mm lng. Measure the length f the adjacent side and the hyptenuse as accurately as yu can. Using the measurements calculate the sin, csine and tangent f 60 and check the answers with yur calculatr. WORKED EXAMPLE N. 3 Calculate the ppsite and adjacent sides f the triangle shwn. Oppsite Oppsite = sin 30 The hyptenuse is 50 mm = sin 30 hyptenuse 50 Enter 30 in the calculatr and press sin, the result is 0.5 Oppsite = 0.5 Oppsite = 50 x 0.5 = 25 mm 50 Adjacent Adjacent = cs 30 The hyptenuse is 50 mm = cs 30 hyptenuse 50 Enter 30 in the calculatr and press cs, the result is 0.8660 Adjacent = 0.8660 Oppsite = 50 x 0.866 = 43.4 mm 50 D.J.Dunn www.freestudy.c.uk 5

WORKED EXAMPLE N.4 Calculate the ppsite and hyptenuse sides f the triangle shwn. Adjacent 75 = cs 20 The adjacent is 75 mm hyptenuse hyptenuse Enter 20 in the calculatr and press cs, the result is 0.9394 75 75 = 0.9394 = hyptenuse = 79.81 mm hyptenuse 0.9394 Adjacent 75 = sin 20 The adjacent is 75 mm hyptenuse hyptenuse Enter 20 in the calculatr and press sin, the result is 0.3420 75 75 = 0.3420 = hyptenuse = 2.19 mm hyptenuse 0.3420 = cs 20 = sin 20 SELF ASSESSMENT EXERCISE N. 3 1. Calculate the unknwn side in each f the three triangles shwn. Answers (i) b = 3.83, a = 3.22 m (ii) c = 3.19 m, b = 1.09 m (iii) a = 7 m, c = 9.9 m 2. A tall mast has a guy rpe attached t the tp. The guy rpe makes an angle f 80 t the grund and it is fixed t the grund at a distance f 30 m frm the ft f the mast. What are the height f the mast and the length f the guy rpe? (Answers 170.1 m and 172.8 m) D.J.Dunn www.freestudy.c.uk 6

4. SINUSOIDAL FUNCTIONS Cnsider a pint P n a wheel at radius R. Let the pint start at the hrizntal psitin and then rtate the wheel anticlckwise by angle θ degrees. The vertical height f the pint is x = R sin θ as shwn. Let R = 1 s the vertical height is x = sin θ. If we nw plt x frm 0 t 90 we get the graph shwn and this is simply a plt f the Sine values btained frm tables r yur calculatr. Fr example at 30 x = sin (30 ) = 0.5 and s the pltted pint is fund. If we allw the wheel t rtate further than 90 we can see the pint must start t cme back dwn until after 180 ( ½ revlutin) the pint is back t the hrizntal psitin with x = 0. The values f x at angle θ must be the same as the value at 180 θ. Fr example if θ = 150 the sine is the same as sin 30. Useful t remember sin θ = sin (180 θ) If we allw the wheel t rtate further, the value f x will g negative as the pint ges belw the hrizntal and will eventually reach -1 at 270 and then it will cme back t the starting pint as shwn. We can see that that the sine f θ is the negative f the sine (360 θ) Useful t remember sin θ = -sin (360 θ) The graph has nw cmplete ne cycle and further rtatin prduces the same value s it fllws that sin 360 = sin 0 If the radius f the wheel was any ther value than 1 then x = R sin θ and the graph wuld g between R and R. The graph represents a typical scillatry cnditin where the value is directly prprtinal t the sine f the angle. This is als called SIMPLE HARMONIC MOTION. The fllwing are sme examples f things that bey this rule. D.J.Dunn www.freestudy.c.uk 7

SCOTCH YOKE This is a device as shwn. A wheel rtates anticlckwise. On the wheel is a pin at radius R. The pin slides in the slt but the yke can nly mve up and dwn in the guide. The displacement f the yke frm the hrizntal psitin is x = R sin θ. ANGULAR VELOCITY Fr thse wh have nt yet studied velcity we need t explain angular velcity here. Angular velcity is the angle turned by the wheel in ne secnd. We use the symbl ω (small mega). ω =angle/time = θ/t It is nrmal t use radians and nt degrees s the units f ω are radian/s. It fllws that θ = ω t. The displacement f the yke can be written as x = R sin (ω t) Fr any given speed we culd plt x against time and get the sinusidal graph shwn. If we rtated the wheel at twice the speed it wuld cmplete tw cycles in the same time. Each cmplete revlutin represents 1 cycle n the graph. There are many examples in Engineering and Science where smething has a value that varies sinusidally with time. They d nt have t have a physical wheel as in the Sctch Yke and ω is simple a cnstant equivalent t angular velcity. Fr example if a mass is hung n a spring as shwn it will rest at the rest psitin. If it is then pulled dwn and released, it will scillate up and dwn and the distance x frm the rest level at any time is x = A sin ωt where ω is a cnstant depending n the mass and spring stiffness. A is the amplitude. Anther example is hw the vltage varies in ur mains electrical system. The vltage at any mment in time is given by the equatin v = V sin(ωt) where V is the maximum vltage (amplitude) in the cycle. WORKED EXAMPLE N. 5 The mains vltage in British husehld varies with time that v = 339.5sin(100π t) where t is the time frm the start f a cycle. What is the vltage when t = 0.004 secnds? v = 339.5sin(100π x 0.003) = 339.5 sin(0.942) Put the calculatr in RADIAN mde and evaluate. v = 339.5 x 0.809 = 274.7 V D.J.Dunn www.freestudy.c.uk 8

ANGULAR FREQUENCY, FREQUENCY AND PERIODIC TIME ω is the angular velcity f the wheel but in any vibratin such as the mass n the spring, it is called the angular frequency as n physical wheel exists. The frequency f the wheel in revlutins/secnd is equivalent t the frequency f the vibratin. If the wheel rtates at 2 rev/s the time f ne revlutin is 1/2 secnds. If the wheel rtates at 5 rev/s the time f ne revlutin is 1/5 secnd. If it rtates at f rev/s the time f ne revlutin is 1/f. This frmula is imprtant and gives the peridic time. Peridic Time T = time needed t perfrm ne cycle. f is the frequency r number f cycles per secnd r Hertz (Hz). It fllws that T = 1/f and f = 1/T Each cycle f an scillatin is equivalent t ne rtatin f the wheel and 1 revlutin is an angle f 2π radians. When θ = 2π, t = T. It fllws that since θ = ωt then 2π = ωt Rearrange and ω = 2π/T. Substitute T = 1/f and ω =2πf COSINUSOIDAL FUNCTION Cnsider again the Sctch Yke but this time placed as shwn. The angle θ is measured frm the hrizntal psitin and the displacement x f the yke is nw x =R cs (θ) The graph f x against time shws the plt f the csine curve which is cyclic and identical t the sine curve except that it starts at +R and nt zer. If we plt the sine and csine f the angle tgether ver ne cycle we see that sin (θ) = cs (θ 90 ) D.J.Dunn www.freestudy.c.uk 9

TANGENT GRAPH If we plt the values f tan(θ) we get a ttally different graphs as shwn. The tangent f 0 is 0 and at 90 it is infinity. The value then jumps t - and repeats the pattern cyclically as shwn. We can deduce that : tan (θ) = tan (θ + 180 ) = tan((θ + 360 ) =... In ther wrds a given tangent value can represent many pssible angles each 180 apart. WORKED EXAMPLE N.6 1. The mvement f a mass scillating n a spring (in mm) is given by the frmulae: x = 60 sin (5t) where t is in secnds. (i) What is the amplitude? (ii) What is the angular frequency? (iii) What is the frequency f scillatin? (iv) Hw lng des it take t cmplete ne cycle? (v) What is the time taken fr the mass t travel frm zer t maximum mvement? It is nt necessary t plt the graph but it is included fr clarity. (i) The Amplitude is 60 mm (ii) The angular frequency is ω = 5 rad/s since θ = ωt = 5t (iii) The frequency is f = ω/2π = 5/2π = 0.7958 Hz (iv) The time f ne cycle is T = 1/f = 1.257 secnds (v) The time t travel frm 0 t 60 mm is ¼ f a cycle s t = 1.257/4 = 0.314 secnds D.J.Dunn www.freestudy.c.uk 10

SELF ASSESSMENT EXERCISE N. 4 1. Evaluate the fllwing t 3 significant places. (i) 12 sin(25 ) (ii) 27 cs(76 ) (iii) 160 tan (45 ) (iv) 7 sin(θ) where θ is 0.17 radian. (v) 16 cs(θ) where θ is 0.25 radian. (vi) 180 tan(θ) where θ is 0.8 radian. (Answers 5.071, 6.532, 160, 1.184, 15.503, 185.335) 2. Evaluate the fllwing expressin. 4sin(50 ) (i) + 2tan(70 ) 1+ cs(23 ) (ii) sin(60 )cs(30 tan30 (Answers -3.9 and 1.3) ) 3. The electric current flwing in a resistr is given by I = 20 sin (100 t) where I is in Amperes and t is in secnds. (i) What is the maximum current? (ii) What is the frequency f the current? (iii) What is the time taken t change frm zer t maximum current in each cycle? (Answers 20 A, 15.92 Hz and 0.0157 secnds) 4. A ladder is 10 m lng and rests against a wall at an angle f 70 t the grund. What is the height where the ladder tuches the wall and hw far is the bttm f the ladder frm the wall? (Answers 9.4 m and 3.42 m) D.J.Dunn www.freestudy.c.uk 11

5. INVERSE TRIGONOMETRY FUNCTIONS Suppse we knw the sides f a triangle but wish t find the angle. T d this we use the inverse functins usually written as sin -1, cs -1 and tan -1. Yu will usually find these are the same buttns n yur calculatr as sin, cs and tan but yu need t press the shift r functin buttn. WORKED EXAMPLE N.7 Calculate the A in degrees Using the sin value we have 55/75 = sin(a) Sin(A) = 0.7333 A = sin -1 (0.7333) = 47.167 WORKED EXAMPLE N.8 Calculate the A in degrees Using the tan value we have 100/50 = tan(a) tan(a) = 2 A = tan -1 (2) = 63.435 WORKED EXAMPLE N.9 Calculate the B and A in degrees and the The length f side b Using the cs value we have 120/150 =cs(b) cs(b) = 0.8 B = cs -1 (0.8) = 36.87 A = 180-36.87 = 143.13 Pythagras b 2 = c 2 a 2 = 150 2 120 2 = 8100 b = 8100 = 90 Check b = 150 sin B = 150 sin 36.87 = 90 D.J.Dunn www.freestudy.c.uk 12

SELF ASSESSMENT EXERCISE N. 5 1. A ladder is 15 m lng and rests against a wall with the bttm 2 m frm the wall. What is the angle f the ladder t the grund and the wall? (Answers 82.34 and 7.66 ) 2. A garden is rectangular being 22 m wide and 45 m lng. Hw lng is a line drawn frm ne crner t the ppsite crner and what angle des it make t the lng side? Answers 50.09 m and 26.05 ) 3. A tight guy rpe is stretched frm the tp f a twer 350 m high t the grund and the bttm f the rpe is 100 m frm the base. What is the angle f the rpe t the grund and hw lng is it? (354 m and 74 ) 4. The diagram shws an engineering drawing fr a cmpnent t be turned n a lathe (nt drawn t scale). Wrk ut the angles A and B. (Answers 33.7 and 68.2 ) D.J.Dunn www.freestudy.c.uk 13