The angle sum of a triangle is 180, as one angle is 90 the other two angles must add to 90. α + β = 90. Angles can be labelled;

Transcription

1 Numercy Introuction to Trigonometry Pytgors Teorem n bsic Trigonometry use rigt ngle tringle structures. (Avnce Trigonometry uses non-rigt ngle tringles) Te ngle sum of tringle is 180, s one ngle is 90 te oter two ngles must to 90. α For tis tringle, it is possible to write α + β = 90 β Conventions for nming tringles involve using cpitl letters for vertices (corners) n lower cse letters for sies. You my ve lrey notice tt letters from te Greek Alpbet re use for nming ngles. A You will notice tt sie is osite ngle A, sie b is osite ngle B etc. c b Angles cn be lbelle; () using Greek letters θ = 3 B C (b) using te letter t te verte A = 3 (c) using tree letters BAC = 3 Centre for Tecing n Lerning Acemic Prctice Acemic Skills Digitl Resources Pge ctl@scu.eu.u [lst eite on 7 September 017]

2 Te tree letter meto cn remove mbiguity in more comple situtions. A A is mbiguous. BAC, CAD n BAD re epresse clerly. Note: BAD = BAC + CAD B C D Pge

3 Numercy Moule contents Introuction Pytgors Teorem Te Trigonometric Rtios Fining Sies Te Trigonometric Rtios Fining Angles Answers to ctivity questions Outcomes To use Pytgors Teorem to solve rigt ngle tringle problems. To solve rigt ngle tringles, ie: missing sies n missing ngles. To use Pytgors Teorem n trigonometry to solve problems. Ceck your skills Tis moule covers te following concepts, if you cn successfully nswer tese questions, you o not nee to o tis moule. Ceck your nswers from te nswer section t te en of te moule. 1. () Use Pytgors Teorem to clculte. (b) A rectngle s lengt of 6.7m n wit of 4.83m. Fin te lengt of te igonl (to.p.). 30 cm 10cm. () Clculte te vlue of in tis tringle. (b) Clculte te vlue of in tis tringle. (c) Clculte te vlue of θ in tis tringle m km 1.m 0.85m θ 3. A kite on te en of 30m string is flying t n ngle of elevtion of 7. Wt is te eigt of te kite irectly bove te groun? Centre for Tecing n Lerning Acemic Prctice Acemic Skills Digitl Resources Pge ctl@scu.eu.u [lst eite on 7 September 017]

4 Numercy Topic 1: Pytgors Teorem Pytgors Teorem sttes tt in rigt ngle tringle: Te squre of te ypotenuse is equl to te sum of te squres of te oter two sies Digrmmticlly: Hypotenuse c () Oter sies b Te Hypotenuse is te longest sie in te tringle n is lso osite te rigt ngle. c = + b Te squre of te Hypotenuse = Sum of te squres of te oter sies Alterntively, becuse te Hypotenuse is unique sie: = + b Tis mens tt Pytgors Teorem cn be use to fin te lengt of missing sie in rigt ngle tringle. Centre for Tecing n Lerning Acemic Prctice Acemic Skills Digitl Resources Pge ctl@scu.eu.u [lst eite on 7 September 017]

5 Emples Hypotenuse unknown (i) 7 9 = + b rule = substitute = = 130 = 130 squre root 11.4 (ii) In rigt tringle: sie = 5 cm, b = 10 cm n c is te ypotenuse. Determine te lengt of sie c. = + b rule = substitute = = 15 = 15 squre root 11. Oter sie unknown (i) 7 9 (ii) 5 metre ler = + b rule = substitute 841 = 79 + = rerrnge = 11 squre root = Or = + b rule b = rerrnge = s 9 7 ubstutite = = 11 squre root = How fr up te wll oes te ler rec? = + b rule = + b 5 1. substitute 5 = b = b rerrnge 3.56 squre root 3.56 b 4.85 = b b = 1. metres Te ler reces pproimtely 4.85m up te wll. Mie wore questions (i) A se s gble roof s rwn below. Clculte te lengt of seets of roofing iron require in its construction. l 1.6m 4.8 m Let te lengt of te seets be l. Pge

6 .4 m = + b l l l l = = = = l =.88 Te lengt of te seets of roofing iron is.88m (ii) A guy (support) wire is ttce 3. m up pole n t point.1 m from te pole. Te groun n te pole re perpeniculr (t rigt ngles). Wt is te lengt of te guy wire? Pole Guy wire Let te lengt of te guy wire be l. = + b l l = = l = l = 3.83 Te lengt of te guy wire is 3.83m Groun (iii) On softbll imon, te istnce between bses is 60 feet or 18.9m. How fr must te ctcer (t ome bse) trow te bll to te plyer on secon bse? Secon Bse 18.9m 18.9m First Bse Te ngle t first bse is rigt ngle. Let te istnce from Home plte to secon bse be. = + b = = = = 5.87 Te istnce from ome plte to secon bse is 5.87m Home Plte Pytgoren Triple or Tri Sometimes, te tree lengts of rigt ngle tringle re ll wole numbers. Wen tis occurs, tey re clle Pytgoren Triple or Tri. Te most commonly known of tese is (3,4,5) representing te tringle below: Pge 3

7 3 4 5 = + b 5 = = = 5 Tree oter Tris re (5,1,13), (7,4,5) n (8,15,17). Oter Tris cn be bse on multiples of te bse Tris; te Tris (6,8,10), (9,1,15), (1,16,0)... re bse te bse Tri (3,4,5). If it is importnt to ecie if tringle is rigt ngle tringle, ten Pytgors Teorem cn be use to ecie tis = 19.8 = b = = = b Tis is not rigt ngle tringle. = 10.4 = b = = = = + b Tis is rigt ngle tringle. Vieo Pytgors Teorem Pge 4

8 Activity 1. Use Pytgors Teorem to fin te missing lengt. () (b) 6 18 cm 1.5 cm 0 (c) Fin te ypotenuse wen =6.1 n b=3.4 () Fin te missing sie given =3.5 n =40 (e) (f) 6 m 6 m 8 m 10 mm Fin te eigt of te tringle. Also clculte te re. (g) A orienteering prticipnt runs 650m nort n ten turns n runs 1.4 km est. How fr from te strting point is te runner? () **A Rel Cllenge** A squre s igonl of 0 cm, wt is te sie lengt?. Do te following tringles contin rigt ngle? () 10, 4, 6 (b) 7, 8, 10 (c), 4.8, 5. () 1.4, 4.8, 5 (e) 6, 6, 8 (f) 5, 6, 7 Pge 5

9 Numercy Topic : Te Trigonometric Rtios Fining Sies Lbelling sies To use te Trigonometric Rtios, commonly clle te Trig Rtios, it is importnt to lern ow to lbel te rigt ngle tringle. Te ypotenuse of te tringle is still te longest sie n locte osite te rigt ngle. Te two oter sies re nme: te osite n te jcent. Te everyy lnguge menings of tese terms elp in using tese lbels. Te osite sie is locte osite te specifie ngle. Te jcent is locte jcent (net to) te specifie ngle. Wen nming tringle, lwys nme te Hypotenuse first. Te reson for oing tis is tt bot te Hypotenuse n te Ajcent re jcent (net to) te specifie ngle. By nming te ypotenuse first tere cn only be one sie tt is jcent to te specifie ngle. Emple: 18 Ajcent Hypotenuse Opposite Ajcent 0 Hypotenuse Opposite Ajcent 55 Hypotenuse Opposite If tere re two ngles, te osite n jcent re still reltive to te specifie ngle. Te igrm below sows te ie. Centre for Tecing n Lerning Acemic Prctice Acemic Skills Digitl Resources Pge ctl@scu.eu.u [lst eite on 0 November 017]

10 Tis is te specifie ngle α. Te Trigonometric Rtios Wit rigt ngle tringles ving tree sies, it is possible to ve 6 rtios. Tey re: osite jcent osite ypotenuse jcent ypotenuse,,,, n. jcent osite ypotenuse osite ypotenuse jcent Tere re bsiclly 3 unique rtios. Te 3 trig rtios commonly use re given te nmes Sine, Cosine n Tngent. Tese nmes re bbrevite to Sin, Cos n Tn. If te specifie ngle is θ, ten te rtios re written s: osite Sin θ = ypotenuse jcent Cos θ = ypotenuse osite Tn θ = jcent A cllenge for you is to evise wy to remember tese rtios. Te tble below sow ow te rtios re pplie to rigt ngle tringles. c For te specifie ngle θ: osite Sin θ = = ypotenuse c jcent b Cos θ = = ypotenuse c b θ osite Tn θ = = jcent b Pge

11 1 H 5 For te specifie ngle J: 5 Sin J = 13 For te specifie ngle I: 1 Sin I = I 1 Cos J = 13 5 Cos I = 13 J 5 Tn J = 1 1 Tn I = 5 14 m For te specifie ngle of 43 Sin Cos 43 y y Tn 43 y To elp unerstn trigonometry, te emple below gives n iniction of ow it works. Consier rigt ngle tringle wit n ngle of 45, tis mens tt te missing ngle is lso 45. Tis mens tt te osite n jcent sie must be equl, so te tringle is ctully n isosceles rigt ngle tringle. Te tringle is rwn below. ypotenuse 45 Ajcent 90-45=45 Opposite Sme lengt osite Te Trig rtio ssocite wit osite n jcent is =. Wit te osite n te jcent jcent being te sme lengt, te vlue of tis rtio woul be 1. So fr: In rigt ngle tringle wit specifie ngle of 45, te vlue of te rtio osite is 1. Tis mens Tn 45 is lwys 1. jcent Pge 3

12 Becuse te vlue of te rtio is obtine witout ctully knowing ny mesurements, noter key ie is: Key Ie For ny size rigt ngle tringle wit specifie ngle of 45, osite te rtio Tn 45 is lwys 1. If te specifie ngle is 45, te lengt jcent of te osite n te jcent must be te sme. Tere re oter rtios tt cn be clculte n wtever vlue tey equl, te vlue will be relte te 45 ngle. Tis ie cn be pplie to rigt ngle tringles wit oter sizes of specifie ngles. Using your clcultor Te net step is lerning ow to use te Trig functions on your scientific clcultor. At tis stge te trig rtio ssocite wit given ngle will be foun. Scientific clcultors llow te user to epress te ngle in tree ifferent wys. In tis moule, ngles re being epresse in egrees. Re te user guie for your clcultor to mke sure it is lwys set on egrees. On te Srp EL 531, tere is key lbelle DRG. Pressing tis key llows te user to cnge between te ifferent ngle types. Keep pressing until DEG is foun on te upper prt of te screen. Note: on tis clcultor it is esy to cnge tis setting ccientlly. Every time te user intens to use Trig functions, tis setting soul be cecke. On te Csio f-8au, press te SHIFT key ten te MODE SETUP key. Press 3 for Degrees. A smll D soul pper on te top of te isply s sown on te left. Alrey, it is known from erlier in tis topic tt Tn 45 = 1. Tis cn be confirme by pressing l45=, te isply on te clcultor soul re 1. (If not, ceck te egrees setting n try gin.) If you ve n oler style scientific clcultor, you my ve to o 45l=. Pge 4

13 Use your clcultor to fin te following: sin 30 (your clcultor soul isply 0.5) Tis mens tt in rigt ngle tringle wit specifie ngle of 30, te osite sie is 0.5 te lengt of te ypotenuse. cos 7 (your clcultor soul isply ) Tis mens tt in rigt ngle tringle wit specifie ngle of 7, te jcent sie is (ppro) 0.31 te lengt of te ypotenuse. tn 19 (your clcultor soul isply ) Tis mens tt in rigt ngle tringle wit specifie ngle of 19, te osite sie is (ppro) 0.34 te lengt of te jcent. Fining missing sies In tis rigt ngle tringle, clculte te lengt of te sie mrke. 37 Step 1: Determine wic rtio to use. In tis tringle, one ngle n one sie re given (37, 1m). Bse on te 37 (specifie) ngle, te 1m sie is te jcent n te sie is te osite. Te trig rtio to be use in tis question is Tn becuse it contins te sie lengts jcent n osite. Step : Write out te rtio n substitute in vlues. = j Tn37 1 Step 3: Do te necessry rerrnging. 1 Tn37 1 multiply bot sies by Tn37 Step 4: Clculte te nswer using your clcultor. 1[l37= Te missing sie lengt () is ppro 9.04m. 1 m Pge 5

14 Step 5: Try to ceck te resonbleness Te 1m sie is osite n ngle of (90-37) 53, becuse is osite te 37 ngle it soul be smller tn 1m, so 9.04 coul be correct. (smll sies re osite smll ngles, lrge sies re osite lrge ngles) A force of 1500N cts t 36 below te rigt orizontl s sown in te igrm below. Clculte te size of te verticl component of te force (v) 36 Force = 1500N Verticl Component (V) Step 1: Determine wic rtio to use. In tis tringle, one ngle n one sie re given (36, 1500N). Bse on te 36 (specifie) ngle, te 1500N sie is te ypotenuse n te v sie is te osite. Te trig rtio to be use in tis question is Sin becuse it contins te sie lengts ypotenuse n osite. Step : Write out te rtio n substitute in vlues. Sinθ = yp Sin Step 3: Do te necessry rerrnging. v 1500 Sin multiply bot sies by Sin36 v Step 4: Clculte te nswer using your clcultor. 1500[j36=881.7 Te verticl component of force (v) is ppro 881.7N. Step 5: Try to ceck te resonbleness Te ypotenuse of tis tringle is 1500N. Te ypotenuse is lso te longest sie. Obtining v s N is consistent wit tis informtion. Te net problem is sligtly ifferent in meto. Step 3: rerrnging is ifferent becuse te vrible is in te enomintor. Pge 6

15 455 m 19 Step 1: Determine wic rtio to use. In tis tringle, one ngle n one sie re given (19, 455m). Bse on te 19 ngle, te 455m sie is te jcent n te sie is te ypotenuse. Te trig rtio to be use in tis question is Cos becuse te question contins te two sie lengts jcent n te ypotenuse. Step : Write out te rtio n substitute in vlues. j Cosθ = yp 455 Cos19 Step 3: Do te necessry rerrnging. 455 Cos19 multiply bot sies by Cos19 Cos = ivie bot sies by Cos19 Cos = Cos19 Step 4: Clculte te nswer using your clcultor. 455Pk19= Te missing sie lengt is ppro 481.m Step 5: Try to ceck te resonbleness Te missing sie is te ypotenuse wic is te longest sie on te tringle. Te nswer of 481.m is longer tn te oter given sie (455m) so te nswer coul be correct. Some problem solving questions refer to te Angle of Elevtion or te Angle of Depression Te ngle of elevtion is te ngle forme by line of sigt bove te orizontl Line of sigt Angle of elevtion Horizontl Pge 7

16 Horizontl Te ngle of epression is te ngle forme by line of sigt below te orizontl Angle of epression Line of sigt Te net problem is simple problem solving question. Te etr skill ere is to re te informtion given n construct igrm. Te ngle of elevtion of plne t n ltitue of 4500m is 7 to te orizontl. In irect line, ow fr wy is te plne. 4500m 7 Step 1: Determine wic rtio to use. In tis tringle, one ngle n one sie re given (7, 4500m). Bse on te 7 ngle, te 4550m sie is te osite n te sie is te ypotenuse. Te trig rtio to be use in tis question is Sin. tep : Write out te rtio n substitute in vlues. Sinθ = yp 4500 Sin7 Step 3: Do te necessry rerrnging Sin7 multiply bot sies by Sin7 Sin = ivie bot sies by Sin7 Sin = Sin7 Step 4: Clculte te nswer using your clcultor. 4500Pj7= Te plne is 991m wy. Pge 8

17 Step 5: Try to ceck te resonbleness Te missing sie is te ypotenuse, wic is te longest sie on te tringle. Te nswer of 991m is longer tn te oter given sie (4500m) so te nswer coul be correct. Vieo Trigonometry Rtios Fining Sie Lengts Te sections below my or my not be relevnt to your stuies. Activity questions re locte t te en. Angles less tn 1 or contining prt ngle Angles wit prt tt is smller tn 1 egree cn be epresse eiter in egrees s eciml or in egrees, minutes n secons. Angle mesurement in egrees, minutes n secons is just like time (ours, minutes n secons). 60 secons equlling 1 minute (60 = 1 ) 60 minutes equlling one egree (60 = 1 ) Fining te tn of 3.7 is l3.7=, te isply soul re Fining te sin of 67 4 (67 egrees, 4 minutes) is: On Csio f-8: j674= Te key represents egrees( ), minutes( ) n secons( ) On Srp EL531: j67 Or Te isply soul re DM ' S4= Fining te cos of (37 egrees, minutes, 41 secons ) is: On Csio f-8: k3741 On Srp EL531: k37 Or DM ' S DM' S Te isply soul re Pge 9

18 Use your clcultor to fin te following: sin 14.5 (your clcultor soul isply ) cos (your clcultor soul isply ) tn 7.1 (your clcultor soul isply ) sin (your clcultor soul isply ) Pge 10

19 Compss n True Berings Tis problem involves compss irections. Tere re two metos for epressing irections. Te first meto is True Bering were Nort is 0 n te ngle increses in clockwise irection, giving Est s 90, Sout s 180 n West s 70. N 40 T = 303 T = 39 T Te secon meto is Compss Bering suc s S40 W. Te first compss irection stte is eiter N or S, followe by n ngulr mesurement in n E or W irection. N N 57 W N 40 E 31 S 59 W Vieo Berings A group of buswlkers wlk 5.4 km nort n ten turn. Tey wlk in n esterly irection until tey rec tower. Te bering of te tower from te originl point is N44 E. Clculte te istnce wlke in te esterly irection by te wlkers. Te igrm for te problem is rwn below. Pge 11

20 Tower 1. Determine wic rtio to use. In tis tringle, one ngle n one sie re given (44, 5.4km). Bse on te 44 ngle, te 5.4km sie is te jcent n te sie is te osite. Te trig rtio to be use in tis question is Tn. 5.4 km 44. Write out te rtio n substitute in vlues. = j Tn Do te necessry rerrnging. 5.4 Tn Tn44 4. Clculte te nswer using your clcultor. 5.4Ol44= Te tower is 5.km to te est. 5. Try to ceck te resonbleness Te missing sie soul be bout te sme lengt s te given sie. Te net emple is problem solving question tt contins multiple steps. Cre must be tken to re te question n ccurtely trnsfer tis informtion into igrm. A sip is 1km out to se from te bse of cliff. On top of te cliff is ligtouse. From te sip, te ngle of elevtion to te bse of te ligtouse is 16 n te ngle of elevtion to te top of te ligtouse is Clculte te eigt of te ligtouse. Te igrm is; L t c km Remember tt only rigt ngle tringles cn be use to solve tis question. Te strtegy to solve tis problem is: 1. Clculte te eigt of te cliff. Clculte te totl eigt (cliff + ligtouse) Pge 1

21 3. Clculte te eigt of te ligtouse by subtrcting te eigt of te cliff from te totl eigt. Clculting te eigt of te cliff: (1km = 1000m) c Tn Tn16 Clculting te totl eigt: Clculting te eigt of te ligtouse: 86.7 = t Tn Tn = t c = L t c L = L = 67.4m Te eigt of te ligtouse is 67.4 m. Tis is pproimtely equivlent to 14 level builing! c t Activity 1. For te following tringles, ientify te jcent, osite n ypotenuse for ec tringle. () (b) (c) 5 z y 1 b c Pge 13

22 () (e) (f) 45.34mm 45 30mm 34mm k g. For te following tringles, fin te vlue of te missing sie(s). () (b) (c) m.5 5 f b 58 () (e) (f) 14.7m b e m 33 75cm 4 45' m Pge 14

23 (g) () (i) ' 345mm 6.5 km 57 4 cm 57 Fin n ten te re of te tringle. 3. Epress te following irections s true bering. () S15 E (b) NE (c) N45 W () 10 W of N (e) SSW (f) A person stning on top of 5 m cliff sees rower out to se. Te ngle of epression of te bot is How fr is te bot from te bse of te cliff? 5. A stuent wises to fin te eigt of builing. From istnce of 50m on perfectly level groun, te ngle of elevtion to te top is 4.6. Fin te eigt of te builing? 6. A plne is flying t ltitue of 5000m. Te pilot observes bot t n ngle of epression of 1, clculte te orizontl istnce wic plces te plne irectly bove te bot. 7. A wlker ecies to tke irect route to lnmrk. Tey wlk 1.7 km t bering of 78 T. How fr i tey wlk in norterly n esterly irection? 8. Fin te perimeter of tis trpezium. 8 cm cm 9. A kite is ttce to 45m line. On winy y, te kite flies t n ngle of elevtion of 8. Clculte te eigt of te kite bove te groun. Pge 15

24 10. A plne flying t n ltitue of m is flying wy from person. Te ngle of elevtion of te plne is 76 wen initilly observe. After 1 minute 15 secons, te plne is t n ngle of elevtion of 9. Ignoring te eigt of te person, wt is te spee of te plne in km/r? 76 9 Pge 16

25 Numercy Topic 3: Te Trigonometric Rtios Fining Angles In te previous section, trig rtio for given ngle ws use to fin missing sie. In tis topic, trig rtio from informtion given bout sies is use to fin te size of n ngle. Erlier in tis moule, it ws iscovere te Tn 45 = 1. So fr te moule s use tis fct to fin missing sie, eiter te osite or te jcent. In tis topic, te tinking is te osite wy roun. Te vlue of te Tn rtio is 1(obtine from te lengts of te osite n jcent), so te question is Te Tn of wt ngle is 1? osite In rigt ngle tringle, it is known tt te vlue of is equl to 0.5. It follows from tis tt ypotenuse te Sin of te unknown ngle is equl to 0.5 or te sin of wt ngle is equl to 0.5? Te emple below will elp in unerstning ow to fin te unknown ngle. A typicl emple is: 7 θ 10 Step 1: Determine wic rtio to use. In tis tringle, two sies re given (7,10). Bse on te θ ngle, te 7 sie is te osite n te 10 sie is te jcent. Te trig rtio to be use in tis question is Tn. Step : Write out te rtio n substitute in vlues. = j 7 = 10 = 0.7 Centre for Tecing n Lerning Acemic Prctice Acemic Skills Digitl Resources Pge ctl@scu.eu.u [lst eite on 0 November 017]

26 So fr it is known tt te Tn of θ is 0.7 or te Tn of wt ngle is 0.7? How is te ngle θ foun? Becuse tis is te osite process to fining te Tn of given ngle, te? key on your scientific clcultor is use. Te -1on tis key mens tt tis is te osite process to fining te Tn of given ngle. Te net line of working is θ = Tn Step 3: Clculte te nswer using your clcultor. On scientific clcultor, te < >? keys re use by first pressing eiter te q or te nf first. Te key use will epen upon te mke n moel of clcultor use. For emple; to get? on Csio f-8, ql keys re use. On Csio f-8, te ngle θ is foun by pressing ql0.7= On Srp EL-531, te ngle θ is foun by pressing nf l0.7= Te complete setting out is: = j 7 = 10 = 0.7 θ = Tn θ = Te ngle θ is pproimtely 35. Answers foun on your clcultor will nee to be roune s tey re usully epresse to mny eciml plces. Rouning to one eciml is usully sufficient. Anoter lterntive is to epress te nswer (ngle) in egrees, minutes n secons. To obtin te nswer in egrees, minutes n secons ( ); On Csio f-100, pressing gives If is presse gin it cnges bck to te eciml form. On Srp EL-531, pressing nf D M ' S gives If nf D M ' S is presse gin it cnges bck to te eciml form. Step 4: Try to ceck te resonbleness Becuse smller sies re osite smller ngles, te ngle osite te 7 sie soul be less tn 45. Te nswer coul be correct. Pge

27 Te net emple is similr ecept it uses ifferent rtio. 4m 0m θ Step 1: Determine wic rtio to use. In tis tringle, two sies re given (4m, 0m). Bse on te θ ngle, te 4m sie is te jcent n te 10 sie is te ypotenuse. Te trig rtio to be use in tis question is Cos. Step : Write out te rtio n substitute in vlues. j Cosθ = yp Cosθ = Cosθ = θ = Cos 1 0. Step 3: Clculte te nswer using your clcultor. On Csio f-8, te ngle θ is foun by pressing qk0.= On Srp EL-531, te ngle θ is foun by pressing nf k0.= Step 4: Try to ceck te resonbleness In tis question it is r to ceck te resonbleness. Te sie osite te ngle clculte woul be muc lrger tn 4m (coul be confirme by using Pytgors Teorem) so te ngle woul be greter tn 45. Te net emple is problem solving question. A tree metre ler is plce ginst brick wll. Te bse of te ler is 900mm from te bse of te wll. Fin te ngle te ler mkes wit te wll. Before strting te process of solving tis question, igrm is require. Re te informtion crefully. In tis question, it woul esy to inicte te wrong ngle. Pge 3

28 Step 1: Determine wic rtio to use. Cnge 900mm to 0.9m In tis tringle, two sies re given (0.9m, 3m). Bse on te θ ngle, te 0.9m sie is te osite n te 3 sie is te ypotenuse. Te trig rtio to be use in tis question is Sin. Step : Write out te rtio n substitute in vlues. Sinθ = yp Sinθ = Sinθ = θ = Sin Step 3: Clculte te nswer using your clcultor. On Csio f-8, te ngle θ is foun by pressing qj0.3=17.46 On Srp EL-531, te ngle θ is foun by pressing nf j0.3=17.46 Te ngle between te ler n te wll is 17.5 Step 4: Try to ceck te resonbleness In tis question it is r to ceck te resonbleness. Tis ngle is osite smll sie so te ngle soul be smller tn 45. In te net question te instruction is to Complete tis Tringle. Tis mens tt ll missing ngles n sies must be clculte. Emple: Complete tis tringle. As two sies re given, te tir sie cn be foun using Pytgors Teorem. Pge 4

29 + b = c + 5 = = = = 144 = 144 = 1 Te net step is to fin one of te two missing ngles. Te ngle α is foun using Trig rtio. It is lwys goo to use te originl sies mesurements given just in cse te clculte sie is inccurte. As te given sies re te osite n te ypotenuse, te Sin rtio is use. Sinα = yp 5 Sinα = 13 Sinα = α = Sin Using te ngle properties of rigt ngle tringle; α =.6 (to 1.p.) α + β = 90 α = 90.6 α = 67.4 In tis rigt ngle tringle, te sie mesurements re 5, 1, 13 n te ngles re.6 n Te tble below summries ow to fin ll missing sies n ngles. Given Do Sies Use Pytgors' Teorem to clculte te tir sie. Use Trig rtio to fin one of te ngles. Ten use your knowlege of rigt ngle tringles to fin te oter ngle. 1 Sie n 1 ngle Use Trig Rtio to clculte one missing sie. Ten, eiter Trig or Pytgors cn be use to fin te oter missing sie. Ten use your knowlege of rigt ngle tringles to fin te oter ngle. Vieo Trigonometry Rtios Fining Angles Pge 5

30 Activity 1. Fin te size of te ngle in te questions below to.p. () Sin θ = (b) Cos θ = (c) Tn θ =.4604 () Cos β = 0.63 (e) Tn μ = (f) Sin σ = Fin te size of te ngle in te questions below in egrees, minutes n secons. () Sin θ = (b) Cos θ = (c) Tn θ =.4604 () Cos β = 0.63 (e) Tn μ = (f) Sin σ = For te following tringles, fin te size of te missing ngle. () (b) (c) () (e) (f) Pge 6

31 4. Complete te tringle below: 5 cm 3 cm 5. A sip sils 400 km nort n ten 800 km west. How fr is it wy from te port it strte from n t wt true bering? 6. It is recommene tt rmps for weelcirs ve rise to run rtio of 1:15. Wt ngle woul te rmp surfce mke to te orizontl if rmp follows tis specifiction? 7. Two flg poles re in prk on level surfce 30m prt. One pole is 10m tll n te oter is 19m tll. Wt is te ngle of elevtion of te top of te tll pole from te top of te smller pole? 8. A 10m ntenn is supporte by guie wires on four sies. Ec wire is ttce to te groun 6 metres from te bse of te ntenn. (i) Fin te lengt of ec wire (ii) Wt ngle oes te wire mke wit te groun? 9. Clculte te pitc of tis roof. 10. In ABC : A = 90, = 16.9, b = 6.5, clculte B. Pge 7

32 Numercy Answers to ctivity questions Ceck your skills 1. () Use Pytgors Teorem to clculte. = + b 30 cm 10cm = = = 1000 = 1000 = 31.6cm (b) A rectngle s lengt of 6.7m n wit of 4.83m. Fin te lengt of te igonl (to.p.).. () Clculte te vlue of in tis tringle. Digonl m 6.7m 4.83m Let te lengt of te igonl be. = + b = = = = = 8.8cm j Tn Tn40 = (to.p.) (b) Clculte te vlue of in tis tringle km Sinθ yp 7.3 Sin49 Sin = Sin49 = 9.67km Centre for Tecing n Lerning Acemic Prctice Acemic Skills Digitl Resources Pge ctl@scu.eu.u [lst eite on 7 September 017]

33 (c) Clculte te vlue of θ in tis tringle. 1.m 0.85m θ Sinθ yp Sinθ Sinθ θ = Sin θ = or 45 5'48" 3. A kite on te en of 30m string is flying t n ngle of elevtion of 7. Wt is te eigt of te kite irectly bove te groun? Sinθ yp 7 30m Sin Sin7 = 8.53m Pytgors Teorem 1. Use Pytgors Teorem to fin te missing lengt. () 6 = + b = = = 1076 = 1076 = (b) 1.5 cm 18 cm = + b 18 = = = + = = 1.95cm (c) Fin te ypotenuse wen =6.1 n b=3.4 = + b = = = = = 6.98 Pge

34 () Fin te missing sie given =3.5 n =40 = + b 40 = = = = = 3.37cm (e) Let te lengt of te linking sie be l. 6 m = + b l = = + b = l m 8 m l l = = 100 l = 100 l = 10m = = 136 = 136 = 11.66m (f) 10 mm Fin te eigt of te tringle. Also clculte te re. (g) A orienteering prticipnt runs 650m nort n ten turns n runs 1.4 km est. How fr from te strting point is te runner? 650m = 0.65km Tis is n equilterl tringle. Hlf te tringle to obtin rigt ngle tringle. = + b 10 = = = = 75 = 8.66mm = + b = = = = = 1.54km () **A Rel Cllenge** A squre s igonl of 0 cm, wt is te sie lengt? 0cm = + b 0 = = = = = 14.14cm. Do te following tringles contin rigt ngle? () 10, 4, 6 (b) 7, 8, 10 (c), 4.8, 5. Pge 3

35 Remember te ypotenuse is te longest sie. + b = b = b = 676 = 6 = 676 As + b = te tringle must contin rigt ngle. () 1.4, 4.8, 5 + b = b = b = 113 = 10 = 100 As + b te tringle oes not contin rigt ngle. (e) 6, 6, 8 + b = b = b = 7.04 = 5. = 7.04 As + b = te tringle must contin rigt ngle. (f) 5, 6, 7 + b = b = b = 5 + b = b = b = 7 + b = b = b = 61 = 5 = 5 = 8 = 64 = 7 = 49 As + b = te tringle oes contin rigt ngle. As + b te tringle oes not contin rigt ngle. As + b te tringle oes not contin rigt ngle. Te Trigonometric Rtios Fining Sies 1. For te following tringles, ientify te jcent, osite n ypotenuse for ec tringle. () (b) (c) 5 z y 1 13 b c 60 jcent y - osite z - ypotenuse 5 jcent 1 osite 13 ypotenuse jcent b osite c ypotenuse Pge 4

36 () (e) (f) 45.34mm mm 34mm 450 k 30mm jcent 34mm osite 45.34mm ypotenuse g jcent osite 5 ypotenuse g 45 jcent 450 osite k ypotenuse Pge 5

37 . For te following tringles, fin te vlue of te missing sie(s). () (b) (c) m.5 5 f b 58 Sinθ yp Sin Sin58 = 1. j f Tn Tn63 f f = j b Tn Tn.5 b b= 6.63m () (e) (f) 14.7m b e m 33 75cm 4 45' m j Cosθ yp 14.7 Cos33 m m Cos m = Cos 33 m= 17.5m j Cosθ yp b Cos4 45' = Cos4 45' = b b = 658.4cm Sinθ yp 500 Sin68.5 e e Sin e = Sin68.5 e= 537.3m Using Pytgors = = = e = = = (g) () (i) Pge 6

38 ' 345mm j Cosθ yp 345 Cos9 Cos = Cos9 = 349.3mm 6.5 km Sinθ yp 6.5 Sin36 36' = Sin36 36' = = Sin36 36' = 104.8km 57 Hlf te spe is rigt ngle tringle. j Tn Tn57 = 18.48cm Are: 1 A = b A = A = 1.76cm 4 cm Epress te following irections s true bering. () S15 E = 165 T (b) NE = 45 T (c) N45 W = 315 T () 10 W of N = 350 T (e) SSW = 0.5 (f) 15 = 15 T 4. A person stning on top of 5 m cliff sees rower out to se. Te ngle of epression of te bot is How fr is te bot from te bse of te cliff? 5m 15.3 Angle of Depression by lternte ngles = 15.3 j 5 Tn15.3 Tn = Tn15.3 = 91.4m Pge 7

39 5. A stuent wises to fin te eigt of builing. From istnce of 50m on perfectly level groun, te ngle of elevtion to te top is 4.6. Fin te eigt of te builing? 4.6 j Tn Tn4.6 =.9m 50m 6. A plne is flying t ltitue of 5000m. Te pilot observes bot t n ngle of epression of 1, clculte te orizontl istnce wic plces te plne irectly bove te bot. 5000m 1 j 5000 Tn1 Tn = Tn1 = 353 m or 3.5km 7. A wlker ecies to tke irect route to lnmrk. Tey wlk 1.7 km t bering of 78 T. How fr i tey wlk in norterly n esterly irection? n N 78 e 1.7 km j Cosθ yp n Cos Cos78 n n = km or 353m Sinθ yp e Sin Sin78 e e = 1.663km Pge 8

40 8. Fin te perimeter of tis trpezium. 51 s 44-8 = 16cm Perimeter = s = = 44 cm 8 cm 117. cm or 1.17m j Tn Tn51 = 19.8cm j Cosθ yp 16 Cos51 s s Cos s = Cos51 s = 5.4cm 9. A kite is ttce to 45m line. On winy y, te kite flies t n ngle of elevtion of 8. Clculte te eigt of te kite bove te groun. Sinθ yp 8 45m Sin Sin8 = 1.13m 10. A plne flying t n ltitue of m is flying wy from person. Te ngle of elevtion of te plne is 76 wen initilly observe. After 1 minute 15 secons, te plne is t n ngle of elevtion of 9. Ignoring te eigt of te person, wt is te spee of te plne in km/r? Pge 9

41 10000m c i 76 9 f Initilly, te plne is i wy from te observer. After 1minute 15 secons, te plne is f wy from te observer. Te istnce covere in 1 minute 15 secons is f i. i j Tn76 Tn i = Tn76 = 493m i i f j Tn9 Tn f f = Tn9 = 18040m f f i = = m or km 1 minute 15 secons is 1.5 minutes. 1.5 minutes 60 = ours. istnce covere (km) spee( km / r) = time tken(r) km = r = 746 km / r (roune) Te Trigonometric Rtios Fining Angles 1. Fin te size of te ngle in te questions below to.p. () Sin θ = (b) Cos θ = (c) Tn θ =.4604 Pge 10

42 () Cos β = 0.63 β = θ = θ = 55 θ = (e) Tn μ = μ = 6.56 (f) Sin σ = σ = 71. Fin te size of te ngle in te questions below in egrees, minutes n secons. () Sin θ = θ = () Cos β = 0.63 β = (b) Cos θ = θ = (e) Tn μ = μ = (c) Tn θ =.4604 θ = (f) Sin σ = σ = For te following tringles, fin te size of te missing ngle. () (b) (c) 5 cm 3.7m cm 4.45m 800mm Sinα yp Sinα Sinα α or 61 38'33" j Cosθ yp Cosθ Cosθ θ = or 33 45'3" j 100mm θ 66.8 or 66 48'5" () (e) (f).4km 1800m 9. cm 14. cm 3000mm.4km = 400 m j θ 36.9 or 36 5'1" j Cosβ yp 9. Cosβ 14. Cosβ β = 49.6 or 49 37'3" Tis is n isosceles tringle. Hlf te bse is 1450mm. j 900mm θ 64. or 64 1'14" Pge 11

43 4. Complete te tringle below: 5 cm 3 cm Clculte using Pytgors. = + b = = = 34 = 34 = 5.83 Clculting α Tnα j Tnα Tnα α or 30 57'50" Clculting β β = 90 α β = β = or β = '50" β = 59 '10" 5. A sip sils 400 km nort n ten 800 km west. How fr is it wy from te port it strte from n t wt true bering? N 400 km 800 km Te sip is locte 894km wy t true bering of Fin using Pytgors. = + b = = = = = 894.4km Fining te true bering j θ 63.4 or 63 6'6" Pge 1

44 6. It is recommene tt rmps for weelcirs ve rise to run rtio of 1:15. Wt ngle woul te rmp surfce mke to te orizontl if rmp follows tis specifiction? Units re not require to nswer tis question. Te surfce of te rmp soul mke n ngle of bout 4 to te orizontl to be weelcir frienly j θ 3.8 or 3 48'51" 7. Two flg poles re in prk on level surfce 30m prt. One pole is 10m tll n te oter is 19m tll. Wt is te ngle of elevtion of te top of te tll pole from te top of te smller pole? j 10m 30m 19-10=9m 19m θ 16.7 or 16 41'57" 8. A 10m ntenn is supporte by guie wires on four sies. Ec wire is ttce to te groun 6 metres from te bse of te ntenn. 10m l 6m Fin te lengt of ec wire = + b = = = 136 = 136 = 11.66m Wt ngle oes te wire mke wit te groun? j θ or 59 '10" Pge 13

45 9. Clculte te pitc of tis roof. (Assuming symmetry).5m 8.m Tis is n isosceles tringle. Hlf te lrge tringle is rigt ngle tringle. j θ 31.4 or 31 '3" 10. In ABC : A = 90, = 16.9, b = 6.5, clculte B. Te first step is to rw te igrm to mtc te informtion given. B 16.9 Sinθ yp 6.5 Sinθ 16.9 Sinθ B = θ.6 or 37'1" A 6.5 C Pge 14