IAS 2.4. Year 12 Mathematics. Contents. Trigonometric Relationships. ulake Ltd. Robert Lakeland & Carl Nugent

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1 Yea 12 Mathematics IS 2.4 Tigonometic Relationships Robet Lakeland & al Nugent ontents chievement Standad icula Measue c Length and the ea of a Secto ea of a Tiangle and Segment Sine Rule osine Rule pplications using the Sine and osine Rules Pactical Tigonomety Involving eaings Pactice Intenal ssessment Pactice Intenal ssessment Pactice Intenal ssessment nswes Ode Fom m 6.0 m Innovative Publishe of Mathematics Texts

2 2 IS 2.4 Tigonometic Relationships NE 2 Intenal chievement Standad 2.4 Tigonometic Relationships This achievement standad involves applying tigonometic elationships in solving poblems. chievement chievement with Meit chievement with Excellence pply tigonometic elationships in solving poblems. pply tigonometic elationships, using elational thinking, in solving poblems. pply tigonometic elationships, using extended abstact thinking, in solving poblems. This achievement standad is deived fom Level 7 of The New Zealand uiculum, and is elated to the achievement objective: apply tigonometic elationships, including the sine and cosine ules, in two and thee dimensions. pply tigonometic elationships in solving poblems involves: selecting and using methods demonstating knowledge of tigonometic concepts and tems communicating using appopiate epesentations. Relational thinking involves one o moe of: selecting and caying out a logical sequence of steps connecting diffeent concepts o epesentations demonstating undestanding of concepts foming and using a model; and also elating findings to a context, o communicating thinking using appopiate mathematical statements. Extended abstact thinking involves one o moe of: devising a stategy to investigate o solve a poblem identifying elevant concepts in context developing a chain of logical easoning, o poof foming a genealisation; and also using coect mathematical statements, o communicating mathematical insight. Poblems ae situations that povide oppotunities to apply knowledge o undestanding of mathematical concepts and methods. Situations will be set in eal-life o mathematical contexts. Methods include a selection fom those elated to: length of an ac of a cicle aea of a secto of a cicle sine ule cosine ule aea of a tiangle. Zincalum e 13.5 m 50 m x y m 44 60

3 IS 2.4 Tigonometic Relationships 3 icula Measue Radians The convention to divide a cicle into 360 appeas to have been used fom about 140. It pobably began fom dividing the 12 signs of the zodiac into smalle pats. cycle of the seasons of appoximately 360 days could be made to coespond to the 12 signs of the zodiac. s convenient as this is, it does not give a mathematical basis fo degees. Senio mathematics usually does not use degees but defines anothe measue fo angles called adians. Radians at fist appea clumsy (thee ae adians in a cicle) but they make the solving of many tigonometic poblems, and in paticula calculus, easie. adian is defined as that angle whose ac is the same length as the adius of the ac. onvesions Degees to Radians We use the exact convesion 2π ad. 360 o π ad. 180 to convet fom degees to adians and vice vesa. To convet θ degees to adians all we have to find is what faction of 180 it is and multiply by π. ix ngle in adians = 180 You can use the faction button on you calculato to simplify this faction quickly. Radians to Degees To convet θ adians to degees we divide by π and multiply by 180. n angle of one adian foms i x180 ngle in degees = ulake an ac length Ltd equal in length to θ the adius of the If you ae given an angle in adians which π secto. includes the symbol π (e.g. 6 ) it is easie to convet it to degees by just eplacing the π symbol with 180 (See the Example In θ = one adian on the next page). full cicle has an ac length of 2π and a adius of so the numbe of adians in a cicle is 2 o 2π. Thee ae 2π o appoximately adians in a cicle so appoximately 2π adians adians adian (1 dp) The symbol means equivalent to. It is used when two quantities ae the same but with diffeent units. Degees to adian convesion is also coveed in the Gaphical Models chievement Standad. If you have aleady done this thee is no need to epeat it. Example onvet the angles in degees to adians. a) 71.4 b) 135 leaving you answe in tems of π. ix a) θ ad. = 180 = 71.4 x 180 = 1.25 (2 dp) ix b) = 180 = 135 x 180 and simplify the faction. 3 = 4

4 6 IS 2.4 Tigonometic Relationships Meit nswe the following. 41. ca has a weak spot on one of its tyes. 42. solid cylinde (adius 652 mm) has a wedge of It will eventually fail and you ae to exploe how 51.0 cut out of it so it sot of looks simila to a many times the weak spot is in contact with the Pac-man. oad. The cylinde is m in depth. The ca is tavelling at an aveage speed of m 90 km/h and each wheel is 37 cm in adius. You will need to conside the following: the speed of the ca in m/s 51 the distance tavelled by the ca in one evolution of the wheel the numbe of evolutions of the wheel in m one second. The ca is being diven fom Napie to Wellington which will take 3.75 hous. How many times do you expect the weak spot The side suface aea (ignoing both ends) is to to be in contact with the oad? be coveed in expensive gold foil. You will need to Make sue you explain each step of you woking when solving this poblem. find the geen angle in adians find the ac length of the geen cicula secto the total distance aound the geen cicula secto the aea of the gold foil. The foil costs $75 a squae mete. What is the estimated cost of the foil equied? Make sue you explain each step of you woking when solving this poblem.

5 IS 2.4 Tigonometic Relationships 11 Excellence Solve these moe complex poblems explaining what you ae doing at each step. 56. conical hat is made by cutting a secto fom a cicle of cad and joining the edges. The hat has a slant height of 32 cm and the esulting hat has a diamete of 22 cm. Find the angle of the secto cut fom the cicle and the aea of the esulting hat. Slant height 32 cm 57. t a sideshow at an & P Show thee is a modified dat boad with fou sections shaded geen. The sideshow offes a $100 pize if one dat (costing $2) lands in any of the fou geen sections. You want to find the pobability that by chance a dat could land in a winning aea. The dat boad has a adius of 312 mm and an aea of mm 2 (4 sf). close up of one of the sectos gives you the 32 cm following dimensions. The inne geen section (of 223 mm Note: Not which two ae geen) goes 212 mm dawn to scale. fom 132 mm to 143 mm 22 cm in while the oute section (of diamete which two ae geen) goes fom 212 mm to 223 mm. 143 mm You will need to conside: 132 mm the angula measue of each wedge the aea of the two inne and oute geen sections the atio of the winning aea to the total boad aea in deciding whethe this game offes a good andom chance of winning.

6 IS 2.4 Tigonometic Relationships 17 Sine Rule Deiving the Sine Rule The Sine Rule Fomula The angles of a non-ight-angled tiangle ae a b c = = labelled with uppecase lettes and each angle s sin sin sin opposite side with the coesponding lowecase whee a, b and c ae the lengths of the thee sides lette. and, and ae the angles opposite each of the coesponding sides. c a The sine ule can be expessed in two foms. The fom below is useful when we ae equied to calculate an angle. b sin sin sin a = = To deive the sine ule an altitude is tempoaily b c constucted to one of the sides. Only two of the thee pats of these fomulae ae eve used to solve a single poblem. It is easie if we always epesent ou c a h unknown with the lette a o if the unknown is an angle, the lette. Fo example, to X b calculate x in 34 mm x this tiangle we stat by 41 Fom tiangle X we find that labelling 34 side x h sin = a as a. h = a sin We then label its and fom tiangle X we get opposite angle and the known h sin = side b and its c 34 mm x a angle. b h = c sin 41 Theefoe 34 a sin = c sin o a c = sin sin simila appoach is used to extend the elation to side b and sin. Giving b c = sin sin It does not matte how we label the othe sides as long as we have sides opposite thei coesponding angle. You can use you gaphics calculato fo sine and cosine ule poblems by enteing the poblem in the solve and getting the calculato to solve fo the unknown.

7 IS 2.4 Tigonometic Relationships 27 Excellence Solve this tigonometic poblem explaining what you ae doing at each step lint and Jenny enjoy long distance cycling. On a paticula weekend they both head off at ight angles to each othe. They agee to stop and ing each othe afte about 75 minutes. lint cycles at 4.1 m/s and afte 32.3 minutes must tun 81.0 ight and then staight fo 43.5 minutes. Jenny (with Molly on the back) cycles staight at 3.9 m/s fo 21.6 minutes then tuns 42.3 left and continues fo 54.8 minutes. 2 E L J2 Not to scale D K 1 J1 S a) How fa ae lint and Jenny fom the stat point when they stop? b) How fa ae they fom each othe when they stop?

8 40 IS 2.4 Tigonometic Relationships Pactice Intenal ssessment Task 2 Tigonometic Relationships 2.4 INTRODUTION The peimete of a quadilateal shaped fitness tack at a pimay school is maked out by the goundsman with fou pegs,, and D. plan of the tack is dawn below with all the bounday lengths and one angle maked. Each school level (junio, middle and senio) complete diffeent paths aound the pegs. The school would like to calculate the distances fo each section and to also calculate the aea enclosed by the pegs so they can ode the coect amount of lawn fetilise. 102 m 67.5 m m Diagam NOT to scale 183 m D The Yea 7 and 8 students (senios) un aound D (a distance of m), the middle school aound D and the junios un aound. Woking independently you ae equied to: find the distance the middle and junio school students un in one lap. calculate the aea enclosed by the fou pegs D. The staff would like to adjust peg so the senios complete a 500 m lap but they ae unable to change angle D as it is the limited by the bounday fence of the school. Find a new position fo peg and the instuctions that should be given to the goundsman. The quality of you easoning, using a ange of methods, and how well you link this context to you solutions will detemine you oveall gade. lealy communicate you method using appopiate mathematical statements and woking.

9 48 IS 2.4 Tigonometic Relationships nswes Page (3 dp) (3 dp) (3 dp) (3 dp) (3 dp) (3 dp) (3 dp) (3 dp) Page π 10 3π 4 26π π (1 dp) (1 dp) (1 dp) (1 dp) (1 dp) (1 dp) (1 dp) (1 dp) Page 5 cont Page The speed of the ca in m/s Speed = / 3600 = 25 m/s icumfeence of wheel = 2π x 0.37 = m Revolutions pe second Revs = 25 / = evs / s Time to Wellington Time = 3.75 x 3600 = s Revolutions to Wellington Revs = x = (4 sf) The weak spot will touch the oad about times (4 sf). 42. We need the angle in adians to calculate the ac length. ngle cut out ngle out = ngle = 2π = adians c length of geen c = x = m (5 sf) Peimete + wedge Dist. = x = m (5 sf) ea of gold ea = x = m 2 (5 sf) ost of foil ost = $ (2 dp) Page m (3 sf) adians (3 dp) 70.8 (1 dp) Page a = 6.3 m (2 sf) adians o 84.0 (3 sf) Page 9 cont c = 5.52 m (3 sf) Secto = 6.84 m 2 (3 sf) 48. c = 153 cm (3 sf) Secto = 1940 cm 2 (3 sf) 49. Peimete = 545 m (3 sf) 50. = 9.80 m (2 sf) mm 2 (3 sf) 52. c = 3.29 m (3 sf) ea = 2.39 m 2 (3 sf) Page a) 2.71 adians b) 0.61 m 2 (2 sf) 54. ngle = 4.5 adians o = ea = 6800 mm 2 (2 sf) 55. a) 1300 m 2 (2 sf) b) 152 m c) 105 m Page Let = adius hat, R = adius cad. θr = 2π 32θ = 2π11 11 θ = π (2.16) cut out = π (4.12) 16 ea = 1100 cm ngle of one secto. ngle = ad. ea in = 0.5π( ) = 475 mm 2 (0 dp) ea out = 0.5π( ) = 752 mm 2 (0 dp) Total winning aea = 2( ) = 2454 mm 2 Ratio win ulake to Ltd total = : 2454 = 125 : 1 Theefoe given odds of $100 : $2 o 50 : 1 ae not good paticulaly as you may miss the entie boad. Page ea = 32 m 2 (2 sf) 59. ea = 100 cm 2 (2 sf) 60. ea = 75.9 m 2 (1 dp) 61. ea = 427 cm 2 (3 sf) 62. ea = 26.8 m 2 (3 sf)