Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd pter 5 Worked Solutions to te Problems Hints. Strt by writing formul for te re of tringle. Note tt tere re two possible ngles. Now knowing te ngle nd two sides, find te tird side - wi rule pplies ere?. Drw digrm mke sure tt you ve it rigt. Join nd M, nd join E nd M. Now tere re lot of rigt ngles nd rigt-ngled tringles. Introdue one more by joining M to te side on te opposite side of te squre. Now everyting n be done using trigonometri rtios. 3. Drw mp of te pilot s trip. Hve te diretion of nort pointing towrds te top of te pge. Drw in te intended route - te one e sould ve tken. Drw in te new pt e needs to tke. You will ve tringle. 4. gin drw digrm sowing were te two bots re wen tey re on teir voyges. lulte nd mrk te position of bot t 6 pm in te fternoon. Drw in te vetor from one bot to te oter. lulte te lengt of te vetor nd ek if it is long enoug. 5.. Put equl to n ngle in wi is te rigt ngle in rigt ngled tringle b. Drw in te perpendiulr from, it s lengt. Relte, nd. 6. Use te digrm in question 5 but wit te perpendiulr oming down from. Ten pply te definition of os nd os nd you ll find lengts wi togeter mke te side. Now in ft if or is obtuse (between 90 or 80 ) ten tere s subtrtion involved, beuse ltoug te plus sign remins, eiter os or os is negtive. You n see tis by drwing tringle wit or obtuse. 7. Join te entre of te irle to e of te verties of te exgon. You ve 6 tringles. re tey identil? Do tey ve property for wi we ve speil nme? Find teir re, nd ten use little lgebr finises it off. 8. Drw "side-on" digrm wit te metres vertil segment representing te position of te piture nd noter point metre below orresponding to te level of te eye of te person. Ten drw lines from te person's eye position to te top nd bottom of te piture segment. Tese lines mke ngles wit te orizontl line from te piture to te eye, te ngle between tem ving to be 0. Using te definition of te tngent of n ngle you n express te orizontl distne in two wys e involving n unknown ngle. So n eqution for te ngle my be found. Find n pproximte solution of tis eqution, nd te distne from te piture n be lulted from tis.
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd 9.. Te sum of te two smll tringles res equls te re of te lrge one. Use tis nd pply some simple lgebr. b. Te rtios in prt () re osines. 0. You ve n equilterl tringle nd segments wit ngle 60 t te entre. Express te given re in terms of tese elements nd do some lgebr to get te required formul.. You need to remember te geometri properties of irles, tngents, rdii nd prllel lines. Te lengt required is te sum of four piees. Tere is q missing from te formul for belt lengt, it sould be: belt lengt π( + b) + θ( b) + osθ.. Te digrm for tis problem is relly te sme s tt used in problem 8 nd te sme formule my be used. Te mnipultions nd te objetive re different but te bsis of te problem is te sme.
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd pter 5 Worked Solutions to te Problems. Tink: Tere re two tringles tt meet te stted onditions. We will solve tem seprtely. For e I n find using n re formul. I ten know two sides nd te inluded ngle, so I n ten find side using te osine rule. m 85 m re bsin re sin b 85 8 0.8586 sin 0.8586 0 59.6 or 0.84 0 m 85 m 8 m 8 m Now we solve bot tringles. b + bos 0 + 8 8 os59.6 4.99 4.99 5.6 Te lengt of te 3 rd side of te tringle is 5.6 m. b + bos 0 + 8 8 os0.84 648.0 648.0 5.46 Te lengt of te 3 rd side of te tringle is 5.5 m.. Tink: Let te side of te squre be 6 units, so lengts M, DE nd E re ll wole numbers. Let EM x. Drw EF prllel to. I n find FEM nd ED using trig rtios. One I know tose I n find te required ngle. step. Find FEM. opp tnφ dj tnfem 6 FEM tn 6 F M 3 6 x E 4 D 3
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd step. Find ED. opp tnφ dj 6 tned 4 6 ED tn 4 step 3. Find x. Sine DEF is rigt ngle, 6 x 90 (tn + tn ) 6 4 0 0 4. 4 3'40'' 3. Tink: In order to find te bering from to we need to first find. I know two sides nd te inluded ngle so I n use te osine rule to find. I n ten use te sine rule to find. Ten I n find te bering. step. Find. 400 + 50 400 50 os0 0 59.8 N D 400 km 50 km 0 0 step. Find. sin sin sin sin 400sin0 59.8 0 sin 0.4346 0 0 5.8 or 54. It is obvious from te digrm tt is 54. 0. step 3. Find te bering. is on bering of 00. Sine D 54., te bering of D is 54. 4
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd 4. Tink: t 6 p.m., te first sip s trvelled 48 km wile te seond sip s trvelled 36 km. From te digrm, I need to find. I know two sides nd te inluded ngle so I n use te osine rule. N 48 km + b bos 478 478 0 48 + 36 48 36 os0 69. Sine te rnge of rdio ommunition is 75 km, te sips will be ble to ommunite t 6 p.m. 5 0 0 0 45 0 36 km 5. Tink: ompring te two formuls, tey bot ve ommon ftor of ½b. Terefore I need to sow tt sin. I strt by drwing digrms.. In te digrm longside,. ut lso opp sin yp sin sin b ut sin Te two re formuls re equivlent for ny rigt-ngled tringle. b. In te digrm longside, D is drwn perpendiulr to. From tringle D, opp sin yp sin Te two re formuls re equivlent for ny tringle. D b 5
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd 6. We sll prove insted b os + os (sme ting, just different lbels for te verties nd sides). In te digrm longside te bse onsists of two prts, D nd D. D is te bse of te rigt ngled tringle wit nd ypotenuse. D is te bse of te rigt-ngled tringle wit nd ypotenuse. So nd so nd nd so D os D os D os D os D b Now D + D b, so we ve os+ os b s required. 7. Te irle n be tougt of s six ongruent setors. One of te urved blue res equls te re of setor minus te re of te tringle formed by te two rdii nd te ord. Te required re required is six times tis. Now te re of te irle is: re πr π(7) 49 π sq metres. We ten lulte te re of e tringle: re bsin 7 sin(60 ) 7 metres 7 metres 60 0 O 3 49 4 Te exgon onsists of six tringles, so te re of te exgon equls 3 re 6 49 4 3 3 49 Finlly, lulte te re of te irle minus te re of te exgon: 3 3 3 3 re 49π 49 49( π ) 6.63 sq.metres 6
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd 8. P metres Q metre L E Tere re two solutions, toug only seems fesible for person wit norml eyesigt. Let ngle LEQ equl θ If te ngle of viewing is 0 π /9, ten PEL θ +π /9. Sine QL metre, pplying te definition of te tngent of n ngle to tringle QLE, we ve QL/LE tn(θ), so LE QL/tn(θ) /tn(θ). Now te sme rgument pplied to tringle PLE gives LE PL/tn(θ + π/9) 3/tn(θ + π/9). Tus te ngle θ (i.e. QEL) stisfies te eqution 3 tnθ tn( θ + π /9) We nnot solve tis eqution nlytilly, so we will solve it grpilly. Te grp of 3 y tnθ tn( θ + π /9) ppers below. Te grp uts te x xis t x 0. nd. For x 0., LE is pproximtely 4.9 metres, wile for x, LE is pproximtely.64 metres. Te ltter is surely too lose, so we onlude tt te person sould stnd bout 5 metres from te piture. 7
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd 9.. Now d sinθ so d sinθ Similrly e sinφ so e sinφ d θ φ N b e Now we will use re sin b to find te re of te tree tringles. re sin( ) θ + φ re N d sinθ re N e sinφ We see tt re ren + ren so sin( θ + φ) sinθ + sinφ {substitute} sin( θ + φ) sinθ + sinφ {multiply troug by } sin( θ + φ) sinθ sinφ + sin( θ + φ) sinθ + sinφ b. Now from te digrm osφ nd osθ. Substituting sin( θ + φ) osφsinθ + osθ sinφ {divide troug by } Re-writing sin( θ + φ) sinθ osφ + sinφosθ 8
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd 0. Look t te digrm below, on te left. Te spe mrked by te points, nd onsists of entrl equilterl tringle nd tree ongruent segments. We need to find ll of tese res. Te lengts,, re ll equl to te rdii of te irle, r. Sine is n equilterl tringle ll its ngles re 60. Tus te re of te tringle (ll te re V) is 0 3 3 V r sin60 r r 4 Te re of e segment equls te re of te setor minus te re of te tringle. Sine te setor is one-sixt of irle of rdius r, we n find n expression for re G: π G πr r 6 6 so te re of te segment, S, is given by S G V π 3 r r 6 4 π 3 ( ) r 6 4 Te re we re to evlute equls te re of te blk tringle plus 3 times te re of setor, i.e. V + 3S Tus 3 π 3 V + 3S r + 3( r ( )) 4 6 4 3 π 3 r ( + 3( )) 4 6 4 3 3 π r ( 3 + 3 ) 4 4 6 3 π r ( + ) 4 9
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd 3 π 3 r ( + 3( )) 4 6 4 π 3 r ( ) Te required re is given by: re π 3 r ( ). Tere is n error in te question in te book. Te belt lengt in tis problem is given by ( + b)p + ( - b)q +osq. Note te extr ftor θ. In te digrm bove, te lengt of O, te lengt of OP, nd te lengt of P b. euse te belt is wrpped round e pulley nd te lengts nd D re surely strigt, e of nd D re tngents to te irles wi represent te pulleys. euse is tngent to e irle, te rdii O nd P re perpendiulr to. Tis mens tt O nd P re prllel. To find te lengt of te belt we need to find te lengts of te two strigt line segments, nd D, nd two irulr rs, speifilly, te longer r from to, nd te sorter r from to D. Te key to finding tese lengts is to drw line from P to O prllel to s sown in te digrm. Ten, sine te tngents nd rdii re perpendiulr, te qudrilterl PF is retngle. pplying Pytgors teorem to tringle OFP we n find te lengt of FP nd terefore. Wen tis is done we will know te lengts of ll te 0
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd strigt line segments in te digrm, so we just need to find te lengts of te irulr rs. To find te lengts of te irulr rs we need to know te ngles subtended by te rs nd D t te entre of te irle, i.e. we need to know te ngles O nd PD. From te geometry, one we know one ngle in te digrm we n find ny of tem. If we let ngle FPO be q, sine OF - b b we ve sinθ. y symmetry ngle PE is lf te ngle PD. lso, beuse FP is rigt ngle, ngles FPO nd PE sum to 90. Tus ngle PE p/ - q nd ngle PD p - q. Finlly we need te reflex ngle P. Te obtuse ngle P equls te obtuse ngle PD (tese being orresponding ngles of trnsversl utting prllel lines). Tus te reflex ngle P p - (p - q) p + q. Now to omplete te problem, belt lengt ( ) + longer r + sorter r D. Now FP nd from tringle FPO, FP ( b) + or FP os θ, so FP osθ Sine te r of irle equls te produt of te ngle t te entre nd te rdius, we ve longer r ( π+ θ) nd sorter r D ( π- θ ) b tus finlly belt lengt ( π + θ) + ( π θ) b+ osθ nd tis n be rerrnged to give belt lengt π( + b) + θ( b) + osθ. Looking t te digrm te distne d is lbelled, but it doesn t pper in te nswer, so tis suggests tt te vlue of d will be prt of te solution. Now we n relte to te lengts nd d sine te ground nd te liff re te sides in rigt ngled tringle Similrly we n relte, d nd t+.
Mtemtis for Queenslnd, Yer Mtemtis, Grpis lultor ppro 00. Kiddy olger, Rex oggs, Rond Frger, Jon elwrd If te definition of te tngent of n ngle is pplied to, we ve tn d Similrly pplying te formul to we ve t+ tn d So d tn nd substituting for d in te seond eqution gives t tn + tn We now simplify te bove expression. First be bring te term tn into te numertor of te rigt side ( to divide by frtion invert te frtion nd multiply ). We ve ( t+ ) tn tn Now multiplying e side by gives tn t+ tn ( ) nd dividing bot sides by tn gives tn ( t ) tn + tn nd ten t tn Finlly ftoring out gives te quoted formul. tn t tn