4.1 Right-angled Triangles Trigonometric Functions Trigonometric Identities Applications of Trigonometry to Triangles 53

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1 ontents 4 Trigonometry 4.1 Rigt-angled Triangles 4. Trigonometric Functions Trigonometric Identities pplications of Trigonometry to Triangles pplications of Trigonometry to Waves 65 Learning outcomes In tis Workbook you will learn about te basic building blocks of trigonometry. You will learn about te sine, cosine, tangent, cosecant, secant, cotangent functions and teir many important relationsips. You will learn about teir graps and teir periodic nature. You will learn ow to apply Pytagoras' teorem and te Sine and osine rules to find lengts and angles of triangles.

2 Rigt-angled Triangles 4.1 Introduction Rigt-angled triangles (tat is triangles were one of te angles is 90 ) are te easiest topic for introducing trigonometry. Since te sum of te tree angles in a triangle is 180 it follows tat in a rigt-angled triangle tere are no obtuse angles (i.e. angles greater tan 90 ). In tis Section we study many of te properties associated wit rigt-angled triangles. Prerequisites efore starting tis Section you sould... Learning Outcomes On completion you sould be able to... ave a basic knowledge of te geometry of triangles define trigonometric functions bot in rigt-angled triangles and more generally epress angles in degrees calculate all te angles and sides in any rigt-angled triangle given certain information HELM (008): Workbook 4: Trigonometry

3 1. Rigt-angled triangles Look at Figure 1 wic could, for eample, be a profile of a ill wit a constant gradient. 1 1 Figure 1 Te two rigt-angled triangles 1 1 and are similar (because te tree angles of triangle 1 1 are equal to te equivalent 3 angles of triangle ). From te basic properties of similar triangles corresponding sides ave te same ratio. Tus, for eample, = and 1 1 = (1) Te values of te two ratios (1) will clearly depend on te angle of inclination. Tese ratios are called te sine and cosine of te angle, tese being abbreviated to sin and cos. Key Point 1 sin = cos = Figure is te side adjacent to angle. is te side opposite to angle. is te ypotenuse of te triangle (te longest side). Task Referring again to Figure in Key Point 1, write down te ratios wic give sin and cos. Your solution nswer sin = cos =. Note tat sin = cos = cos(90 ) and cos = sin = sin(90 ) HELM (008): Section 4.1: Rigt-angled Triangles 3

4 tird result of importance from Figure 1 is = Tese ratios is referred to as te tangent of te angle at, written tan. () Key Point sin = cos = tan = Figure 3 = lengt of opposite side lengt of adjacent side For any rigt-angled triangle te values of sine, cosine and tangent are given in Key Point 3. Key Point 3 sin = cos = Figure 4 We can write, terefore, for any rigt-angled triangle containing an angle θ (not te rigt-angle) sin θ = lengt of side opposite angle θ lengt of ypotenuse = Opp Hyp cos θ = lengt of side adjacent to angle θ lengt of ypotenuse = dj Hyp tan θ = lengt of side opposite angle θ lengt of side adjacent to angle θ = Opp dj Tese are sometimes memorised as SOH, H and T O respectively. Tese tree ratios are called trigonometric ratios. 4 HELM (008): Workbook 4: Trigonometry

5 Task Write tan θ in terms of sin θ and cos θ. Your solution nswer tan θ = Opp dj = Opp dj.hyp Hyp = Opp Hyp.Hyp dj = Opp / dj Hyp Hyp i.e. tan θ = sin θ cos θ Key Point 4 Pytagoras Teorem a + b = c Figure 5 c a b Eample 1 Use te isosceles triangle in Figure 6 to obtain te sine, cosine and tangent of Figure 6 Solution y Pytagoras teorem () = + = so = Hence sin 45 = = = 1 cos 45 = = 1 tan 45 = = = 1 HELM (008): Section 4.1: Rigt-angled Triangles 5

6 Engineering Eample 1 Noise reduction by sound barriers Introduction udible sound as muc longer wavelengts tan ligt. onsequently, sound travelling in te atmospere is able to bend around obstacles even wen tese obstacles cause sarp sadows for ligt. Tis is te result of te wave penomenon known as diffraction. It can be observed also wit water waves at te ends of breakwaters. Te etent to wic waves bend around obstacles depends upon te wavelengt and te source-receiver geometry. So te efficacy of purpose built noise barriers, suc as to be found alongside motorways in urban and suburban areas, depends on te frequencies in te sound and te locations of te source and receiver (nearest noise-affected person or dwelling) relative to te barrier. Specifically, te barrier performance depends on te difference in te lengts of te ypotetical ray pats passing from source to receiver eiter directly or via te top of te barrier (see Figure 7). T R receiver source S s s H W U barrier r V r Problem in words Figure 7 Find te difference in te pat lengts from source to receiver eiter directly or via te top of te barrier in terms of (i) te source and receiver eigts, (ii) te orizontal distances from source and receiver to te barrier and (iii) te eigt of te barrier. alculate te pat lengt difference for a 1 m ig source, 3 m from a 3 m ig barrier wen te receiver is 30 m on te oter side of te barrier and at a eigt of 1 m. Matematical statement of te problem Find ST + T R SR in terms of s, r, s, r and H. alculate tis quantity for s = 1, s = 3, H = 3, r = 30 and r = 1. 6 HELM (008): Workbook 4: Trigonometry

7 Matematical analysis Note te labels V, U, W on points tat are useful for te analysis. Note tat te lengt of RV = r s and tat te orizontal separation between S and R is r + s. In te rigt-angled triangle SRV, Pytagoras teorem gives So (SR) = (r + s) + (r s) SR = (r + s) + (r s) (3) Note tat te lengt of T U = H s and te lengt of T W = H r. In te rigt-angled triangle ST U, (ST ) = s + (H s) In te rigt-angled triangle TWR, So (T R) = r + (H r) ST + T R = s + (H s) + r + (H r) (4) So using (3) and (4) ST + T R SR = s + (H s) + r + (H r) (r + s) + (r s). For s = 1, s = 3, H = 3, r = 30 and r = 1, ST + T R SR = 3 + (3 1) (3 1) (30 + 3) + (1 1) = = 0.67 So te pat lengt difference is 0.67 m. Interpretation Note tat, for equal source and receiver eigts, te furter eiter receiver or source is from te barrier, te smaller te pat lengt difference. Moreover if source and receiver are at te same eigt as te barrier, te pat lengt difference is zero. In fact diffraction by te barrier still gives some sound reduction for tis case. Te smaller te pat lengt difference, te more accurately it as to be calculated as part of predicting te barriers noise reduction. HELM (008): Section 4.1: Rigt-angled Triangles 7

8 Engineering Eample Horizon distance Problem in words Looking from a eigt of m above sea level, ow far away is te orizon? State any assumptions made. Matematical statement of te problem ssume tat te Eart is a spere. Find te lengt D of te tangent to te Eart s spere from te observation point O. D O R R Figure 8: Te Eart s spere and te tangent from te observation point O Matematical analysis Using Pytagoras teorem in te triangle sown in Figure 8, (R + ) = D + R Hence R + R + = D + R (R + ) = D D = (R + ) If R = m, ten te variation of D wit is sown in Figure Horizon D (m) Heigt (m) Figure 9 8 HELM (008): Workbook 4: Trigonometry

9 t an observation eigt of m, te formula predicts tat te orizon is just over 5 km away. In fact te variation of optical refractive inde wit eigt in te atmospere means tat te orizon is approimately 9% greater tan tis. Task Using te triangle in Figure 10 wic can be regarded as one alf of te equilateral triangle D, calculate sin, cos, tan for te angles 30 and D Figure 10 Your solution nswer y Pytagoras teorem: () = () () = 4 = 3 so 4 Hence sin 60 = = 3 3 = sin 30 = = = 1 cos 30 = = 3 3 = tan 60 = = 3 tan 30 = = cos 60 = = = 1 3 Values of sin θ, cos θ and tan θ can of course be obtained by calculator. Wen entering te angle in degrees ( e.g. 30 ) te calculator must be in degree mode. (Typically tis is ensured by pressing te DRG button until DEG is sown on te display). Te keystrokes for sin 30 are usually simply sin 30 or, on some calculators, 30 sin peraps followed by =. Task (a) Use your calculator to ceck te values of sin 45, cos 30 and tan 60 obtained in te previous Task. (b) lso obtain sin 3., cos 86.8, tan ( denotes a minute = 1 60 ) HELM (008): Section 4.1: Rigt-angled Triangles 9

10 Your solution (a) (b) nswer (a) , , to 4 d.p. (b) sin 3. = cos 86.8 = to 4 d.p., tan 8 15 = tan 8.5 = to 4 d.p. Inverse trigonometric functions (a first look) onsider, by way of eample, a rigt-angled triangle wit sides 3, 4 and 5, see Figure Figure 11 Suppose we wis to find te angles at and. learly sin = 3 5, cos = 4 5, tan = 3 so we 4 need to solve one of te above tree equations to find. Using sin = 3 ( ) 3 5 we write = sin 1 (read as is te inverse sine of ) Te value of can be obtained by calculator using te sin 1 button (often a second function to te sin function and accessed ( ) using a SHIFT or INV or SEOND FUNTION key). 3 Tus to obtain sin 1 we migt use te following keystrokes: 5 INV SIN 0.6 = or 3 5 INV SIN = We find sin = (to 4 significant figures). Key Point 5 Inverse Trigonometric Functions sin θ = implies θ = sin 1 cos θ = y implies θ = cos 1 y tan θ = z implies θ = tan 1 z (Te alternative notations arcsin, arccos, arctan are sometimes used for tese inverse functions.) 10 HELM (008): Workbook 4: Trigonometry

11 Task eck te values of te angles at and in Figure 11 above using te cos 1 functions on your calculator. Give your answers in degrees to d.p. Your solution nswer = cos = = cos = Task eck te values of te angles at and in Figure 11 above using te tan 1 functions on your calculator. Give your answers in degrees to d.p. Your solution nswer = tan = = tan = You sould note carefully tat sin 1 1 does not mean sin. 1 Indeed te function as a special name te cosecant of, written cosec. So sin cosec 1 (te cosecant function). sin Similarly sec 1 cos cot 1 tan (te secant function) (te cotangent function). HELM (008): Section 4.1: Rigt-angled Triangles 11

12 Task Use your calculator to obtain to 3 d.p. cosec 38.5, sec.6, cot 88.3 (Use te sin, cos or tan buttons unless your calculator as specific buttons.) Your solution nswer cosec = sin 38.5 = sec.6 = cot = tan 88.3 = cos.6 = Solving rigt-angled triangles Solving rigt-angled triangles means obtaining te values of all te angles and all te sides of a given rigt-angled triangle using te trigonometric functions (and, if necessary, te inverse trigonometric functions) and peraps Pytagoras teorem. Tere are tree cases to be considered: ase 1 Given te ypotenuse and an angle We use sin or cos as appropriate: y θ (a) ssuming and θ in Figure 1 are given ten Figure 1 cos θ = wic gives = cos θ from wic can be calculated. lso sin θ = y so y = sin θ wic enables us to calculate y. learly te tird angle of tis triangle (at ) is 90 θ. 1 HELM (008): Workbook 4: Trigonometry

13 ase Given a side oter tan te ypotenuse and an angle. We use tan: (a) If and θ are known ten, in Figure 1, tan θ = y so y = tan θ wic enables us to calculate y. (b) If y and θ are known ten tan θ = y gives = y tan θ from wic can be calculated. Ten te ypotenuse can be calculated using Pytagoras teorem: = + y ase 3 Given two of te sides We use tan 1 or sin 1 or cos 1 : (a) y tan θ = y so θ = tan 1 ( y ) θ (b) Figure 13 y sin θ = y so θ = sin 1 ( y ) θ Figure 14 (c) cos θ = so θ = cos 1 ( ) θ Figure 15 Note: since two sides are given we can use Pytagoras teorem to obtain te lengt of te tird side at te outset. HELM (008): Section 4.1: Rigt-angled Triangles 13

14 Engineering Eample 3 Vintage car brake pedal mecanism Introduction Figure 16 sows te structure and some dimensions of a vintage car brake pedal arrangement as far as te brake cable. Te moment of a force about a point is te product of te force and te perpendicular distance from te point to te line of action of te force. Te pedal is pivoted about te point. Te moments about must be equal as te pedal is stationary. Problem in words If te driver supplies a force of 900 N, to act at point, calculate te force (F ) in te cable. Matematical statement of problem Te perpendicular distance from te line of action of te force provided by te driver to te pivot point is denoted by 1 and te perpendicular distance from te line of action of force in te cable to te pivot point is denoted by. Use trigonometry to relate 1 and to te given dimensions. alculate clockwise and anticlockwise moments about te pivot and set tem equal. cable F mm N mm Figure 16: Structure and dimensions of vintage car brake pedal arrangement Matematical nalysis Te distance 1 is found by considering te rigt-angled triangle sown in Figure 17 and using te definition of cosine. 10 mm 40 cos(40 ) = ence 1 = 161 mm. Figure HELM (008): Workbook 4: Trigonometry

15 Te distance is found by considering te rigt-angled triangle sown in Figure mm 15 cos(15 ) = ence = 7 mm. Figure 18 Equating moments about : = F so F = 013 N. Interpretation Tis means tat te force eerted by te cable is 013 N in te direction of te cable. Tis force is more tan twice tat applied by te driver. In fact, watever te force applied at te pedal te force in te cable will be more tan twice tat force. Te pedal structure is an eample of a lever system tat offers a mecanical gain. Task Obtain all te angles and te remaining side for te triangle sown: c 4 5 Your solution nswer Tis is ase 3. To obtain te angle at we use tan = 4 5 so = tan 1 (0.8) = Ten te angle at is 180 ( ) = y Pytagoras teorem c = = HELM (008): Section 4.1: Rigt-angled Triangles 15

16 Task Obtain te remaining sides and angles for te triangle sown. b a Your solution nswer Tis is ase 1. Since = ten cos = a 15 so a = 15 cos = Te angle at is 180 ( ) = Finally sin = b b = 15 sin = (lternatively, of course, Pytagoras teorem could be used to calculate te lengt b.) Task Obtain te remaining sides and angles of te following triangle. 8 a c 34 0 Your solution nswer Tis is ase. Here tan = 8 a so a = 8 tan = 11.7 lso c = = and te angle at is 180 ( ) = HELM (008): Workbook 4: Trigonometry

17 Eercises 1. Obtain cosec θ, sec θ, cot θ, θ in te following rigt-angled triangle θ. Write down sin θ, cos θ, tan θ, cosec θ for eac of te following triangles: (a) 5 θ (b) y θ 3. If θ is an acute angle suc tat sin θ = /7 obtain, witout use of a calculator, cos θ and tan θ. 4. Use your calculator to obtain te acute angles θ satisfying (a) sin θ = 0.560, (b) tan θ =.4, (c) cos θ = Solve te rigt-angled triangle sown: α b c 10 β α = surveyor measures te angle of elevation between te top of a mountain and ground level at two different points. Te results are sown in te following figure. Use trigonometry to obtain te distance z (wic cannot be measured) and ten obtain te eigt of te mountain km z 7. s sown below two tracking stations S 1 and S sigt a weater balloon (W ) between tem at elevation angles α and β respectively. W S 1 α P β S c Sow tat te eigt of te balloon is given by = c cot α + cot β 8. veicle entered in a soap bo derby rolls down a ill as sown in te figure. Find te total distance (d 1 + d ) tat te soap bo travels. STRT 15 FINISH d HELM (008): Section 4.1: Rigt-angled Triangles d 1 00 metres 8 17

18 nswers 1. = =17, cosec θ = 1 sin θ = 17 8 sec θ = 1 cos θ = cot θ = 1 tan θ = 15 8 θ = sin (for eample)... θ = (a) sin θ = 5 cos θ = 1 5 tan θ = 1 1 cosec θ = 5 (b) sin θ = y + y cos θ = + y tan θ = y cosec θ = + y y 3. Referring to te following diagram θ 7 l Hence cos θ = tan θ = 3 5 = 5 15 l = 7 = 45 = (a) θ = sin = (b) θ = tan 1.4 = (c) θ = cos 1 0. = β = 90 α = 3.5, b = 6. tan 37 = z tan 41 = z 10 tan c = 10 sin from wic = (z + 0.5) tan 37 = z tan 41, so z tan 37 z tan 41 = 0.5 tan z = 0.5 tan 37 tan 37 tan km, so = z tan 41 = tan km 7. Since te required answer is in terms of cot α and cot β we proceed as follows: Using to denote te distance S 1 P cot α = 1 tan α = dding: cot α + cot β = + c = c... = c cot α + cot β cot β = 1 tan β = c as required. 8. From te smaller rigt-angled triangle d 1 = 00 = 46.0 m. Te base of tis triangle sin 8 ten as lengt l = 46 cos 8 = m From te larger rigt-angled triangle te straigt-line distance from STRT to FINISH is 00 sin 15 = 77.7 m. Ten, using Pytagoras teorem (d + l) = = m from wic d = m... d 1 + d = m 18 HELM (008): Workbook 4: Trigonometry