Tringles re nmed fter their verties - the ove tringle is lled tringle. The three ngles re ommonly referred to s ngles, nd. The length of the sides re

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1 Revision Topi 20: Sine nd osine Rules nd 3 Trigonometry rief rep of Grde nd mteril: Pythgors Theorem: This theorem, whih onnets the lengths of the sides in right-ngled tringles, sttes tht: where is the length of the hypotenuse (i.e. the side opposite the right-ngle) nd nd re the lengths of the other two sides. Note tht the hypotenuse is the longest side in right-ngled tringle. Trigonometry The following formule link the sides nd ngles in right-ngled tringles: O sin x H os x H O tn x O H x where H is the length of the hypotenuse; O is the length of the side opposite the ngle; is the length of the side djent to the ngle. These formule re often rememered using the ronym SOHHTO or y using mnemonis. Here is ommonly used mnemoni: Silly Old Hrry ouldn't nswer His Test On lger When finding ngles, rememer tht you need to use the SHIFT key. Further notes, exmples nd exmintion questions relting to Pythgors theorem nd trigonometry re ontined in seprte revision ooklets. Sometimes you need to lulte lengths nd ngles in tringles whih do not ontin ny rightngles. This is when the sine nd osine rules re useful. Lelling tringle To use the sine nd osine rules, you need to understnd the onvention for lelling sides nd ngles in ny tringle. onsider generl tringle: 1

2 Tringles re nmed fter their verties - the ove tringle is lled tringle. The three ngles re ommonly referred to s ngles, nd. The length of the sides re given lower se letters: Side is the side opposite ngle. It is sometimes referred to s side. Side is the side opposite ngle. It is equivlently lled side. Side is the side opposite ngle. It is lso known s side. tringle doesn t hve to e lelled using the letters, nd. For exmple, onsider the tringle PQR elow: R q p P r Q Sine Rule The sine rule onnets the length of sides nd ngles in ny tringle : It sttes tht:. sin sin sin n lterntive version of the formul is often used when finding the size of ngles: sin sin sin Exmple: Finding the length of side The digrm shows tringle. lulte the length of side m 72 o 2 55 o

3 Solution: To find the length of side using the sine rule, follow these steps: Step 1: Lel the tringle using the onventions outlined erlier. Step 2: Look to see whether ny dditionl informtion n e dded to the digrm (for exmple, n you dedue the length of ny other ngles?) Step 3: Sustitute informtion from the digrm into the sine rule formul. Step 4: elete the unneessry prt of the formul. Step 5: Rerrnge nd then work out the length of the required side. In our exmple, we egin y lelling the sides nd y working out the size of the 3 rd ngle (using the ft tht the sum of the ngles in ny tringle is m 55 o 72 o 53 o Sustituting into the formul, we get: sin sin sin 13.2 sin 55 sin 53 sin 72 s we wnt to lulte the length nd s the middle prt of the formul is ompletely known, we delete the first prt of the formul: 13.2 sin 53 sin 72 Rerrnging this formul (y multiplying y sin72) gives: 13.2 sin 72 sin 53 i.e. = 15.7 m (to 1 deiml ple). Exmple: Finding the length of n ngle The digrm shows tringle LMN. lulte the size of ngle LNM. L 17.5 m 6.9 m M 134 o N 3

4 Solution: Step 1: Lel the tringle using the onventions outlined erlier. Step 2: Sustitute informtion from the digrm into the sine rule formul Step 3: elete the unneessry prt of the formul. Step 4: Rerrnge nd then work out the size of the required ngle. sin sin sin. In our exmple, the lelled tringle looks like: L 6.9 m n M 17.5 m m 134 o l N The osine rule formul (djusted for our lettering) is: sin L sin M sin N l m n Sustituting into this gives: sin L sin134 sin N l We wnt to find ngle N nd we know the middle prt of the formul ompletely. We therefore delete the first prt of the formul, leving sin134 sin N If we multiply y 6.9, we get: 6.9 sin134 sin N So ngle N = SHIFT sin = 16.5 (to 1 d.p.) Exmintion style question In tringle, ngle = 65, ngle = 38, = 15 m. igrm NOT urtely drwn 65 o 38 o 15 m 4

5 Work out the length of. Exmintion style question: In tringle, ngle = 115, = 5 m nd = 9 m. lulte the size of ngle. 115 o 5 m igrm NOT urtely drwn 9 m osine Rule The osine rule lso onnets the length of sides nd ngles in ny tringle : It sttes tht: 2 os Equivlently, we lso hve these formule: 2 os 2 os You need to e fmilir with the struture of these formule. In prtiulr note tht the letter tht ppers s the sujet of the formul lso ppers s the ngle. Note tht the osine rule n e onsidered s n extension of Pythgors theorem. Exmple: Finding the length of side The digrm shows tringle. lulte the length of side m 62 o 14.7 m Solution: To find the length of side using the osine rule, follow these steps: Step 1: Lel the tringle using the onventions outlined erlier. Step 2: Write down the pproprite version of the osine rule formul nd sustitute informtion from the digrm into it. Step 3: Work out the length of the required side. 5

6 Our lelled digrm here is: 17.9 m 62 o 14.7 m s we wish to find the length of, we need the formul with s the sujet: 2 os Sustitute in: os 62 Typing the right hnd side into lultor (in one long string nd pressing = only t the end) gives: Squre rooting gives = 17.0 m Exmple: Finding the length of n ngle The digrm shows tringle. lulte the length of ngle m 10.7 m Solution: 13.2 m To find the length of side using the osine rule, follow these steps: Step 1: Lel the tringle using the onventions outlined erlier. Step 2: Write down the pproprite version of the osine rule formul nd sustitute informtion from the digrm into it. Step 3: Rerrnge nd work out the length of the required ngle. The lelled digrm here looks like: 11.4 m d 10.7 m 13.2 m We wnt to find ngle. We therefore need to write down version of the osine rule formul tht ontins ngle. The sujet of the pproprite formul would therefore e d 2 : d 2 os 6

7 Sustituting into this formul gives: os = os (work out the squres of the terms) = os (dd together the first two numers on the right) = os (sutrt from oth sides) Therefore os i.e. = SHIFT os = 55.8 Worked Exmintion Question In tringle, = 9 m, = 15 m nd ngle = 110. igrm NOT urtely drwn. 9 m 110 o 15 m lulte the perimeter of the tringle. Give your nswer orret to the nerest m. Solution: In order to lulte the perimeter, we need to work out the length of the third side. Lelling the tringle: Using the osine rule: 9 m 15 m 110 o i.e. 2 os os = 20 m (to nerest m) So perimeter = = 44 m (to nerest m) Pst exmintion question (Edexel): 7

8 In tringle, = 7m, = 12 m nd ngle = 125. lulte the length of. 7 m 125 o 12 m Pst exmintion question (SEG): The digrm shows tringle. = 8.6 m, = 3.1 m nd = 9.7 m. Not to sle 8.6 m 3.1 m 9.7 m lulte ngle. When do you use the sine rule nd when do you use the osine rule? In n exmintion, you will need to deide whether to use the sine rule or the osine rule. It is helpful to rememer tht you will need to use the sine rule unless 1) you re told ll three sides, in whih you n use the osine rule to find ny ngle; 2) you re given 2 sides nd the ngle in etween, in whih se you n find the finl side using the osine rule. Worked exmple: 13 m 7 m 100 o 30 o 12 m lulte ) the length ; ) ngle. Solution onsider first tringle. In this tringle we know two sides nd the inluded ngle (i.e. the ngle in etween). We n therefore use the osine rule to find the third side,. 8

9 7 m 30 o 12 m Using the osine rule: 2 os os = 6.89 m Therefore = 6.9 m (to 1 d.p.) () To find ngle, we now onsider tringle : We re not in either of the situtions where the osine rule n e used, so here we will e using the sine rule: sin sin sin d 13 m 6.89 m 100 o d Sustituting into this formul gives: sin 6.89 sin sin Rerrnging gives: 6.89 sin100 sin So = 31.5 Exmintion Question (Edexel June 2001) 8 m 80 o 10 m igrm NOT urtely drwn. 9

10 ) lulte the length of. Give your nswer in entimetres orret to 3 signifint figures. ) lulte the size of ngle. Give your nswer orret to 3 signifint figures. Exmintion Question (Edexel Novemer 1998) In the qudrilterl, = 6 m, = 7 m, = 12 m, ngle = 120, ngle = 70. lulte the size of ngle. Give your nswer orret to 3 signifint figures. 6 m 120 o 7 m 70 o 12 m re of tringle The re of tringle n e found using this lterntive formul: re of tringle = 1 sin 2 lterntive versions re: re = 1 sin 2 or re = 1 sin 2 10

11 This n equivlently e thought of s re = ½ produt of two sides sine of the inluded ngle. Exmple: Find the re of the tringle: 8.7 m 112 o 9.8 m We n use the ove formul to find the re of this tringle s we hve two sides nd the inluded ngle (i.e. the ngle in etween the given sides): re = sin m 2 (to 3 s.f.) 2 Worked exmintion question 150 o 60 m igrm NOT urtely drwn. ngle = 150. = 60 m. The re of tringle is 450 m 2. lulte the perimeter of tringle. Give your nswer orret to 3 signifint figures. We n use the formul for the re of tringle to find the length of : 150 o 60 m re = 1 sin 2 1 So sin sin s sin150 = 0.5 So = 30 m. 11

12 To find the perimeter, we lso need the length. We n use the osine rule: So 2 os os = 87.3 m Therefore the perimeter is = = 177 m 2 to 3 SF. Exmintion question (June 2004) igrm NOT urtely drwn. 3.2 m 8.4 m = 3.2 m. = 8.4 m. The re of tringle is 10 m 2. lulte the perimeter of tringle. Give your nswer orret to 3 SF. Exmintion style question Q 9 m 67 o P 60 o R In tringle PQR, PQ = 9 m, ngle PQR = 67 nd ngle QPR = 60. lulte the re of tringle PQR. Prolem style questions Questions re often set involving erings or ngles of elevtion. If digrm hs not een drwn in the question, you will need to egin y skething digrm to illustrte the sitution. There will usully e severl steps required in order to get to the solution. Worked exmintion question (Edexel June 2006) The digrm shows vertil tower on horizontl ground. is stright line. The ngle of elevtion of from is 28. The ngle of elevtion of from is

13 = 25 m. igrm NOT urtely drwn. 28 o 54 o 25 m lulte the height of the tower. Give your nswer to 3 signifint figures. Solution: Step 1: Use tringle to find the length. Step 2: Use tringle to find the height. Step 1: From the originl digrm, we n dedue tht ngle = 126 nd ngle = o 25 m d 126 o Using the sine rule: 25 sin 28 sin126 sin 26 25sin 28 So 26.77m sin 26 Step 2: m 54 o So the tower is 21.7 m tll. Using trigonometry for right-ngled tringles: sin sin m Exmintion question (NE) heliopter leves heliport H nd its mesuring instruments show tht it flies 3.2 km on ering of 128 to hekpoint. It then flies 4.7 km on ering of 066 to its se. H 128 o 3.2 km o 4.7 km

14 ) Show tht ngle H is 118. ) lulte the diret distne from the heliport H to the se. 3 trigonometry Grde / * questions often involve you finding distnes nd ngles in 3 dimensionl ojets. The key to these questions is to identify nd drw the relevnt 2 dimensionl tringle. Exmple 1: EFGH is uoid with dimensions 8m, 6m nd 5m (s shown in the digrm). X is the midpoint of side HG. X H G E F 5 m 8 m 6 m ) Find length. is length digonlly ross the se of the uoid. * We strt y skething the se nd we mrk on the length we wnt to find: 6 m 8 m * We identify relevnt right-ngled tringle (here oviously tringle ). * We n then use Pythgors theorem to find : So 2 = = = 100 = 10 m ) Find the length G * We egin y identifying relevnt right-ngled tringle. Here we use tringle G (we use this tringle euse is vertilly elow G). We mrk on the digrm ll the lengths we know: 14

15 We now use Pythgors theorem to find G: So G 2 = 2 + G 2 G 2 = G 2 = 125 G = 11.2 m ) Find ngle G. The letters mentioned in the nme of the ngle tell you whih tringle to drw (i.e. tringle G). This is the tringle drwn in prt (). We n use trigonometry to find ngle G. θ tnθ = 5 10 i.e θ = 26.6 d) Find the length X. * X is digonl length ross the uoid. Let Y e the point vertilly elow X. We drw tringle XY: * There is not yet enough informtion in the digrm to find length X. * We n work out length Y however if we drw out the se : 15

16 We n use Pythgors theorem to find Y: Y 2 = 2 + Y 2 Y 2 = Y 2 = 52 So Y = m (note tht we don t round too erly). Now we n find X from tringle XY: m So X 2 = Y 2 + XY 2 X 2 = X 2 = 77 Y = m = 8.8 m (to 1 P) e) Find ngle X We egin y drwing out tringle X (i.e. the tringle with the sme letters s the ngle we wnt). This is n isoseles tringle s X = X. α We egin y finding α: 4 sin 8.8 α = 27 16

17 So ngle X = 54 Exmple 2: E is squre-sed pyrmid. The length is 8m. Point E is vertilly ove point X, the entre point of squre. The height of the pyrmid, EX, is 7m. ) lulte length. * lies long the se of the pyrmid. We therefore egin y drwing out the se. ) lulte length E. Tringle is right-ngled tringle. We n therefore use Pythgors theorem to find : 2 = = 128 = 11.3 m * The point elow E is point X. We therefore drw out tringle EX. Note: X is hlf of, i.e m. We n find E using Pythgors theorem: E 2 = E 2 = 81 E = 9.00 m ) lulte ngle E. * This ngle is the sme s ngle EX. We n lulte this ngle using the digrm in (). Using trigonometry 7 tn( EX ) 5.65 EX 51.1 So ngle E = 51.1 d) lulte the re of fe E. * We egin y drwing out tringle E, n isoseles tringle (E = E). E 9 m 9 m 17

18 8 m * One wy to find the re of tringle E would e to find the height of this tringle (y splitting it into 2 right-ngled tringles) nd then using the formul ½ h. lterntively, we ould find ngle E (for exmple using the osine rule) nd then using the formul: re = ½ sin. If we use the first method, then we must egin y finding the height h of the tringle. Using Pythgors theorem: E h²+ 4²= 9² h²+ 16 = 81 h²= 65 h = 8.06 m h 9 m Therefore the re is m m Exmintion question (Edexel Novemer 2004) The digrm represents uoid EFGH. = 5 m, = 7 m, F = 3 m. ) lulte the length of G. Give your nswer orret to 3 signifint figures. ) lulte the size of the ngle etween G nd the fe. Give your nswer orret to 3 signifint figures. F 3 m E G 5 m H 7 m Exmintion question (Edexel Novemer 2005) V The digrm shows pyrmid. The se,, is horizontl squre of side 10 m. The vertex V is vertilly ove the midpoint, M, of the se. VM = 12 m. lulte the size of ngle VM. M 10 m 10 m Exmintion question (Edexel) F E The digrm represents prism. EF is retngle. is squre. 30 o m

19 EF nd F re perpendiulr to plne. = = 60 m. ngle E = 90 degrees. ngle E = 30 degrees. lulte the size of the ngle tht the line E mkes with the plne. 19