# FURTHER TRIGONOMETRY

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1 0 YER The Improving Mthemtics Eduction in Schools (TIMES) Project FURTHER TRIGONOMETRY MESUREMENT ND GEOMETRY Module 24 guide for techers - Yer 0 June 20

2 Further Trigonometry (Mesurement nd Geometry: Module 24) For techers of Primry nd Secondry Mthemtics 50 over design, Lyout design nd Typesetting by lire Ho The Improving Mthemtics Eduction in Schools (TIMES) Project ws funded by the ustrlin Government Deprtment of Eduction, Employment nd Workplce Reltions. The views expressed here re those of the uthor nd do not necessrily represent the views of the ustrlin Government Deprtment of Eduction, Employment nd Workplce Reltions. The University of Melbourne on behlf of the Interntionl entre of Excellence for Eduction in Mthemtics (IE EM), the eduction division of the ustrlin Mthemticl Sciences Institute (MSI), 200 (except where otherwise indicted). This work is licensed under the retive ommons ttribution- Nonommercil-NoDerivs 3.0 Unported License

3 The Improving Mthemtics Eduction in Schools (TIMES) Project FURTHER TRIGONOMETRY MESUREMENT ND GEOMETRY Module 5 guide for techers - Yer 0 June 20 Peter rown Michel Evns Dvid Hunt Jnine McIntosh ill Pender Jcqui Rmgge 0 YER

4 {4} guide for techers FURTHER TRIGONOMETRY SSUMED KNOWLEDGE Fmilirity with the content of the module Introductory Trigonometry. Fmilirity with bsic coordinte geometry. Fcility with simple lgebr, formuls nd equtions. Fmilirity with surds. MOTIVTION In the module, Introductory Trigonometry, we showed tht if we know the ngles nd one side in right-ngled tringle we cn find the other sides using the trigonometric rtios sine, cosine nd tngent. Similrly, knowing ny two of the sides in right-ngled tringle enbles us to find ll the ngles. Not ll tringles contin right-ngle. We cn relte sides nd ngles in n rbitrry tringle using two bsic formuls known s the sine rule nd the cosine rule. rmed with these we cn solve greter rnge of problems in two dimensions nd extend these ides to three-dimensionl problems s well. This is n essentil tool for surveyors nd civil engineers. It soon becomes pprent tht in some cses we need to be ble to define the trigonometric rtio of n obtuse ngle. This llows us to del with broder rnge of problems nd pplictions. It will lso provide the model for extending the definition of the trigonometric rtios to ny ngle. This ide will be picked up in the module, The Trigonometric Functions.

5 The Improving Mthemtics Eduction in Schools (TIMES) Project {5} ONTENT In the module, Introductory Trigonometry, we defined the three stndrd trigonometric rtios sine, cosine nd tngent of n ngle, clled the reference ngle, in right-ngled tringle. hypotenuse opposite djcent These re defined by: sin = opposite hypotenuse, cos = djcent opposite hypotenuse, tn = djcent, where 0 < < 90. Students should lern these rtios thoroughly. One simple mnemonic tht might ssist them is SOH H TO, consisting of the first letter of ech rtio nd the first letter of the sides mking up tht rtio. In right-ngled tringle, the other two ngles re complements of ech other. s the digrm below shows, the side opposite one of these ngles is djcent to the other. 90 Thus, it cn be seen tht, sin = cos (90 ) nd cos = sin (90 ) if 0 < < 90 The cosine (co-sine) is so nmed since the cosine of n ngle is the sine of its complement. These rtios cn be used to find sides nd ngles in right-ngled tringles.

6 {6} guide for techers EXMPLE Find, correct to two deciml plces, the vlue of the pronumerl in ech tringle. b 8 cm x cm 2.2 cm 5 cm 28 Solution sin 5 = opposite hypotenuse b cos 28 = = x 8 x= 8 sin djcent hypotenuse = 2.2 = 2.2 cos 28 (correct to two deciml plces) 0.77 (correct to two deciml plces) EXMPLE lculte the vlue of, correct to one deciml plce. b 4m 7 cm 2 cm 8.2m SOLUTION cos = So, = cos b tn = (correct to one deciml plce) So, = tn (correct to one deciml plce)

7 The Improving Mthemtics Eduction in Schools (TIMES) Project {7} Note tht for 0 < x < the sttement sin - x = mens tht sin = x. This nottion is stndrd but it is essentil tht students do not confuse the inverse nottion with the usul mening of the index - used in lgebr. To help void this confusion, it is best to lwys red sin - x s inverse sine of x nd tn - x s inverse tngent of x. EXT VLUES The trigonometric rtios for the ngles 30, 45 nd 60 cn be expressed using surds nd occur very frequently in introductory trigonometry, in senior mthemtics nd in clculus. It is thus importnt for students to become fmilir with them. One wy to find them quickly is to sketch the following tringles nd then simply write down the rtios. right-ngled tringle contining 45 ngle will be isosceles, so we choose the two shorter sides to be unit in length nd use Pythgors theorem to find the hypotenuse. For the ngles 30 nd 60, we strt with n equilterl tringle of side 2 units in length nd drop perpendiculr s shown. Simple geometry nd Pythgors theorem gives the remining informtion s shown in the digrm The tble of vlues cn now be completed from these digrms. sin cos tn INDEX NOTTION There re severl devitions from the usul index nottion tht rise in trigonometry. Students my initilly find them confusing. We write, for exmple (tn ) 2 s tn 2, (sin ) 3 s sin 3 nd so on. This must not be confused with the inverse nottion discussed bove. We do not write, for exmple, sin -2 for (sin ) -2, since this would confuse the usul mening of indices with inverses.

8 {8} guide for techers EXERISE Simplify tn 2 30 cos b The top of tower hs n ngle of elevtion of 45 from point on the ground. From point 00m further on, the ngle of elevtion is 30, s shown in the digrm. Find the exct vlue, x, of the height of the tower. x m m THREE-DIMENSIONL PROLEMS We cn use our knowledge of trigonometry to solve problems in three dimensions. EXMPLE In the tringulr prism opposite, find: 3 cm the length F b the length F D c the ngle F, correct to one deciml plce. 4 cm SOLUTION F 5 cm E pplying Pythgors theorem to EF, F 2 = = 4 4 cm Hence F= 4 cm F 5 cm E b pplying Pythgors theorem to F, F 2 = ( 4) 2 = 50 3 cm So F= 5 2 cm F 4cm c To find the ngle F, drw F nd let F =. Now tn = 3 4 giving = tn 3 4 So F 25. (to one deciml plce) 5 2cm 3 cm F 4 cm

9 The Improving Mthemtics Eduction in Schools (TIMES) Project {9} EXERISE 2 Find EG in the cube shown below. D F G E 2 cm H THE SINE RULE In mny pplictions we encounter tringles tht re not right-ngled. We cn extend our knowledge of trigonometry to del with these tringles. This is done using two bsic formuls, the first of which is clled the sine rule. We will ssume, for the moment, tht we re deling with n cute-ngled tringle. s shown in the digrm, we drop perpendiculr P of length h from to. h b P c Then in P we hve sin = h b, so h = b sin. Similrly, in P we hve sin = h, so h = sin. Equting these two expressions for h we hve b sin = sin which we cn write s sin = b sin. The sme result holds for the side nd the ngle, so we cn write sin = b sin = c sin. This is known s the sine rule. In words it sys: ny side of tringle over the sine of the opposite ngle equls ny other side of the tringle over the sine of its opposite ngle.

10 {0} guide for techers We will soon see how to extend this result to obtuse-ngled tringles. EXMPLE In, = 9 cm. = 76 nd = 58. Find, correct to two deciml plces: b 9 cm SOLUTION pply the sine rule: sin 76 = 9 sin 58 nd so = 9 sin 76 sin cm (to two deciml plces) b To find, we need the ngle opposite it. Thus, by the sine rule: = = 46 sin 46 = 9 sin 58 = 9 sin 46 sin cm (to two deciml plces) ERINGS True berings were covered in the module, Introductory Trigonometry Yers. We cn now use the sine rule to solve simple surveying problems involving non-rightngled tringles.

11 The Improving Mthemtics Eduction in Schools (TIMES) Project {} EXMPLE From the points nd, 800 metres prt, on stright North-South rod, the bering of house is 25 T nd 050 T respectively. Find how fr ech point is from the house, correct to the nerest metre. SOLUTION We drw digrm to represent the informtion. We cn find the ngles in H. N H= = nd H= = 75 pply the sine rule to H: H sin 55 = 800 sin sin 55 nd so H = sin m 50 H m (to two deciml plces) Thus is pproximtely 678 metres from the house. Similrly, H sin 50 = 800 sin 75 nd so 800 sin 50 H= sin m (to two deciml plces) Thus is pproximtely 634 metres from the house. EXERISE 3 From point t P, west of building O, the ngle of elevtion of the top of the building O is 28. From point Q 0m further west from P the ngle of elevtion is 20. Drw digrm nd then use the sine rule to find the distnce P nd hence the exct height of the building. Finlly, evlute the height O to the nerest centimetre. FINDING NGLES The sine rule cn be used to find ngles s well s sides in tringle. One of the known sides, however, must be opposite one of the known ngles.

12 {2} guide for techers EXMPLE ssuming tht ll the ngles re cute. G Find the ngle in the tringle FGH, to the nerest degree. 2 cm 8 cm SOLUTION F 75 H pply the sine rule to FGH, 8 2 sin = sin 75 To mke the lgebr esier, tke the reciprocl of both sides. sin sin 75 8 = 2 8 sin 75 Hence sin = 2 = Hence = 40 (to the nerest degree) s seen in the exmple bove, it is esier when finding ngles, to write the sine rule s sin = sin before substituting in the given informtion. b DELING WITH OTUSE NGLES oth the sine rule nd the cosine rule re used to find ngles nd sides in tringles. Wht hppens when one of the ngles is obtuse? To del with this we need to extend the definition of the bsic trigonometric rtios from cute to obtuse ngles. We use coordinte geometry to motivte the extended definitions s follows. We drw the unit circle centre the origin in the rtesin plne nd mrk the point on the circle in the first qudrnt. In the digrm shown, since OQ = cos, we cn see tht the x-coordinte of P is cos. Similrly, the y-coordinte of P is sin. Hence the coordintes of P re (cos, sin ). We cn now turn this ide round nd sy tht if is the ngle between OP nd the positive x-xis, then: y O P (cos, sin ) x Q the cosine of is defined to be the x-coordinte of the point P on the unit circle nd the sine of is defined to be the y-coordinte of the point P on the unit circle. This definition cn be pplied to ll ngles, both positive nd negtive, but in this module we will restrict the ngle to be between 0 nd 80.

13 The Improving Mthemtics Eduction in Schools (TIMES) Project {3} ONSISTENY OF THE DEFINITIONS In the module, Introduction to Trigonometry, we defined sin = opposite hypotenuse nd cos = djcent hypotenuse, where 0 < < 90. In the previous section we defined cos = OQ nd sin = PQ. We must show tht the two definitions gree. The digrm below shows the right-ngled tringle O, nd the tringle OPQ both contining the ngle. Tringle OPQ hs its vertex P on the unit circle. These tringles re similr nd so the rtio O = PQ = PQ, which is the y-coordinte of the point P. OP Similrly O O = OQ =, which is the x-coordinte of P. OP y P O Q x So we hve shown the two definitions gree. NGLES IN THE SEOND QUDRNT s n exmple, let us tke to be 30, so hs coordintes (cos 30, sin 30 ). Now move the point P round the circle to P so tht OP mkes n ngle of 50 with the positive x-xis. Note tht 30 nd 50 re supplementry ngles. The coordintes of P re (cos 50, sin 50 ). y P(cos 50, sin 50 ) P(cos 30, sin 30 ) Q 30 O Q x ut we cn see tht the tringles OPQ nd OP Q re congruent, so the y-coordintes of P nd P re the sme. Thus, sin 50 = sin 30.

14 {4} guide for techers lso, the x-coordintes of P nd P hve the sme mgnitude but opposite sign, so cos 50 = cos 30. From this typicl exmple, we see tht if is ny obtuse ngle, then its supplement, 80 is cute, nd the sine of is given by sin = sin (80 ), where 90 < < 80. Similrly, if is ny obtuse ngle then the cosine of is given by cos = cos (80 ), where 90 < < 80. In words this sys: the sine of n obtuse ngle equls the sine of its supplement, the cosine of n obtuse ngle equls minus the cosine of its supplement. EXMPLE Find the exct vlue of: sin 50 b cos 50 c sin 20 d cos 20 SOLUTION sin 50 = sin (80 50) b cos 50 = cos (80 50) = sin 30 = cos 30 = 2 = 3 2 c sin 20 = sin (80 20) d cos 20 = cos (80 20) = sin 60 = cos 60 = 3 2 = 2 Note: You cn verify these results using your clcultor. The sine rule is lso vlid for obtuse-ngled tringles. EXERISE 4 Reprove the sine rule ngle is obtus. sin = b for tringle in which sin h b M c

15 The Improving Mthemtics Eduction in Schools (TIMES) Project {5} THE NGLES 0, 90, 80 We cn use the extended definition of the trigonometric functions to find the sine nd cosine of the ngles 0, 90, 80. EXERISE 5 Drw digrm showing the point on the unit circle t ech of the bove ngles. Use the coordintes of to complete the entries in the tble below sin cos THE TNGENT OF N OTUSE NGLE For in the rnge 0 < < 90 or 90 < < 80 we define the tngent of n ngle by tn = sin cos, for cos 0. In the cse when cos = 0, the tngent rtio is undefined. This will hppen, when = 90. If is in the rnge 0 < < 90, this definition grees with the usul definition of tn = opposite djcent Hence, if is n obtuse ngle, then tn = sin cos = sin (80 ) cos (80 ) (from the definition) (since is obtuse) = tn (80 ) (from the definition). Hence the tngent of n obtuse ngle is the negtive of the tngent of its supplement. Note tht tn 0 = 0 nd tn 80 = 0 since the sine of these ngles is 0 nd tht tn 90 is undefined since cos 90 = 0. EXERISE 6 Find the exct vlues of tn 50 nd tn 20.

16 {6} guide for techers THE MIGUOUS SE In our work on congruence, it ws emphsized tht when pplying the SS congruence test, the ngle in question hd to be the ngle included between the two sides. Thus, the following digrm shows two non-congruent tringles nd with two pirs of mtching sides shring common (non-included) ngle Suppose we re told tht tringle PQR hs PQ = 9, PQR = 45, nd PR = 7. Then the ngle opposite PQ is not uniquely determined. There re two non-congruent tringles tht stisfy the given dt. P Q 45 R R pplying the sine rule to tringle we hve sin 9 = sin 45 7 nd so sin = 9 sin Thus 65, ssuming tht is cute. ut the supplementry ngle = 5. The ngle PR Q lso stisfies the given dt. This sitution is sometimes referred to s the mbiguous cse. Since the ngle sum of tringle is 80, in some circumstnces only one of the two ngles clculted is geometriclly vlid. EXERISE 7 Find the vlue of in the following digrm, explining why the nswer is unique. 3 5

17 The Improving Mthemtics Eduction in Schools (TIMES) Project {7} THE OSINE RULE We know from the SS congruence test, tht tringle is completely determined if we re given two sides nd the included ngle. However, if we know two sides nd the included ngle in tringle, the sine rule does not help us determine the remining side. The second importnt formul for generl tringles is the cosine rule. Suppose is tringle nd tht the ngles nd re cute. Drop perpendiculr from to nd mrk the lengths s shown in the digrm. c h b x D x b In D, Pythgors theorem gives c 2 = h 2 + (b x) 2. lso in D, nother ppliction of Pythgors theorem gives h 2 = 2 x 2. Substituting this expression for into the first eqution nd expnding, c 2 = 2 x 2 + (b x) 2 = 2 x 2 + b 2 2bx + x 2 = 2 + b 2 2bx. Finlly, from D, we hve x = cos nd so c 2 = 2 + b 2 2bcos This lst formul is known s the cosine rule. y relbeling the sides nd ngle, we cn lso write 2 = b 2 + c 2 2bc cos, nd b 2 = 2 + c 2 2c cos. Notice tht if = 90 then, since cos = 0, we obtin Pythgors theorem, nd so we cn regrd the cosine rule s Pythgors theorem with correction term. The cosine rule is lso true when is obtuse, but note tht in this cse the finl term in the formul will produce positive number, becuse the cosine of n obtuse ngle is negtive. Some cre must be tken in this instnce.

18 {8} guide for techers EXMPLE Find the vlue of x to one deciml plce. SOLUTION pplying the cosine rule: x 2 = cos cm = so x= 2.3 (to one deciml plce) 8 cm x cm EXERISE 8 Prove tht the cosine rule lso holds in the cse when is obtuse. FINDING NGLES We know from the SSS congruence test tht if the three sides of tringle re known then the three ngles re uniquely determined. gin, the sine rule is of no help in finding them since it requires the knowledge of (t lest) one ngle, but we cn use the cosine rule insted. We cn substitute the three side lengths, b nd c into the formul c 2 = 2 + b 2 2b cos where is the ngle opposite the side c, nd then re-rrnge to find cos nd hence. lterntively, we cn re-rrnge the formul to obtin cos = 2 + b 2 c 2 2b nd then substitute. Students my cre to rerrnge the cosine rule or lern further formul. Using this form of the cosine rule often reduces rithmeticl errors. Recll tht in ny tringle, if > b then >.

19 The Improving Mthemtics Eduction in Schools (TIMES) Project {9} EXMPLE tringle hs side lengths 6 cm, 8 cm nd cm. Find the smllest ngle in the tringle. SOLUTION The smllest ngle in the tringle is opposite the smllest side. pplying the cosine rule: 6 2 = cos 8 6 cos = = nd so 32.2 (correct to one deciml plce) EXTENSION THE LONGEST SIDE ND THE LRGEST NGLE OF TRINGLE In the module ongruence, we proved n importnt reltionship between the reltive sizes of the ngles of tringle nd the reltive lengths of its sides: The ngle of tringle opposite longer side is lrger thn the ngle opposite shorter side. For sclene tringles, this cn be restted in terms of inequlities of ll three sides s follows: If is tringle in which > b > c, then > >. This result cn be proved in n interesting wy using either the sine rule or the cosine rule. THE LONGEST SIDE ND THE SINE RULE The following exercise uses the fct tht sin increses from 0 to s increses from 0 to 90. EXERISE 9 Let be tringle in which > b > c. Wht cn you conclude bout the reltive sizes of sin, sin nd sin using the sine rule? b If no ngle is obtuse, wht cn you conclude bout the reltive sizes of, nd? c If tringle PQR hs n obtuse ngle P = 80, where is cute, use the identity sin (80 ) = sin to explin why sin P is lrger thn sin Q nd sin R. d Hence prove tht if the tringle hs n obtuse ngle, then > >.

20 {20} guide for techers THE LONGEST SIDE ND THE OSINE RULE This exercise uses the fct tht cos decreses from to s increses from 0 to 80. EXERISE 0 Let be tringle in which > b > c. Write down cos nd cos in terms of, b nd c, nd express ech in terms of their common denomintor 2bc. b Show tht cos cos = (2 b b 2 ) (c 2 bc) 2 ( 3 b 3 ) 2bc c Hence explin why cos > cos. = ( b)(b c2 ) + ( 3 b 3 ) 2bc. d Similrly explin why cos > cos, nd hence show tht > >. EXERISE b Use Pythgors theorem to show tht the hypotenuse is the longest side of rightngled tringle Why is the result of prt specil cse of the theorem bove? RE OF TRINGLE We sw in the module, Introductory Trigonometry tht if we tke ny tringle with two given sides nd bout given (cute) ngle, then the re of the tringle is given by re = 2 b sin. EXERISE 2 Derive this formul in the cse when is obtuse. EXERISE 3 tringle hs two sides of length 5 cm nd 4 cm contining n ngle. Its re is 5 cm 2. Find the two possible (exct) vlues of nd drw the two tringles tht stisfy the given informtion. EXERISE 4 Write down two expressions for the re of tringle nd derive the sine rule from them.

21 The Improving Mthemtics Eduction in Schools (TIMES) Project {2} LINKS FORWRD The sine nd cosine rules cn be used to solve rnge of prcticl problems in surveying nd nvigtion. THREE-DIMENSIONL PROLEMS EXMPLE The point M is directly cross the river from the bse of tree. From point 7 metres upstrem from M, the ngle of elevtion of the top of the tree is 7. From point Q, 5 metres downstrem from M the ngle of elevtion of the top of the tree is 9. ssuming tht PMQ is stright line nd tht the tree is on the edge of the river, we wish to find the width w metres of the river. SOLUTION In problems such s these, it is impertive to drw creful digrm. h b 7 w m 9 Q 5 m M 7 m P Let P =, Q = b, nd = h, nd then pplying Pythgors theorem to tringles MP nd MQ, we hve = 49 + w 2, b = 25 + w 2. From tringles P nd Q we hve h = sin 7, h = b sin 9. Equting these, substituting in the vlues of nd b nd squring we rrive t (49 + w 2 )sin 2 7 = (25 + w 2 )sin 2 9 We cn now mke w 2 the subject nd obtin w 2 = 49sin2 7 25sin 2 9, nd so sin 2 9 sin 2 7 w 8.66 m correct to 2 deciml plces. Note: Do not evlute until the lst step to retin full clcultor ccurcy. ivil engineers nlyzing forces nd stresses in buildings nd other structures often use vectors to represent the direction nd mgnitude of these forces. vector is n rrow which hs both direction nd mgnitude. The sine nd cosine rules re used in vector digrms to find resultnt forces nd stresses. This is n importnt ppliction.

22 {22} guide for techers TRIGONOMETRI IDENTITIES s well s hving prcticl uses, the sine nd cosine rules cn be used to derive theoreticl results, known s trigonometric identities tht hve importnt implictions nd pplictions in lter work. mong these re the double ngle results which we will describe below. Using the re formul, = 2 b sin, for tringle with two sides nd b, contining n ngle, we cn do the following: Fix cute ngles nd nd let ngle = +. From the point drw D of length y nd construct tringle s shown in the digrm, where is perpendiculr to D. From the digrm, we hve y = cos => y = cos () nd y = cos b => y = b cos b (2). b y b ompring res, D 2 b sin ( + ) = 2 y sin + 2 by sin. Substituting in the vlue of y from (2) into the first term nd tht from () into the second, we hve, fter some simplifiction, sin ( + ) = sin cos b + cos sin. In the discussion bove we ssumed tht, b were cute ngles. This identity holds for ll nd but to show this requires different pproch. Note tht sin ( + ) sin + sin. For exmple, sin( ) = sin 90 =, wheres ( + 3) sin 60 + sin 30 =. 2 EXERISE 5 Use the bove formul to show tht the exct vlue of sin 75 is Putting = = in the bove formul, we obtin the double ngle formul for sine, nmely sin 2 = 2sin cos. There is similr double ngle formul for cosine, cos 2 = cos 2 sin 2.

23 The Improving Mthemtics Eduction in Schools (TIMES) Project {23} oth formuls re extremely useful when clculus is pplied to the trigonometric functions. NGLES OF NY MGNITUDE ND THE TRIGONOMETRI FUNTIONS We sw in this module how to use the unit circle to give mening to the sine nd cosine of n obtuse ngle. This definition cn be extended to include ngles greter thn 80 nd lso to negtive ngles. Thus, for exmple, if is between 80 nd 270, then sin = sin ( 80 ) nd cos = cos ( 80 ). Once we cn find the vlues of sin nd cos for vlues of, we cn plot grphs of the functions y = sin, y = cos. y y = sin These ides will be developed in the module, The Trigonometric Functions. The grphs of the sine nd cosine functions re used to model wve motion nd electricl signls. They re n essentil prt of modern signl processing nd telecommunictions. This provides brethtking exmple of how simple ide involving geometry nd rtio ws bstrcted nd developed into remrkbly powerful tool tht hs chnged the world. HISTORY In the module, Introductory Trigonometry, we mentioned tht the Greeks hd version of trigonometry involving chords. This is shown in the digrm below. R 0 crd ( ) =

24 {24} guide for techers In the digrm, the chord of the ngle is the length of the chord tht subtends n ngle t the centre in the circle of rdius R. Ptolemy (85-85D), who lived nd worked in lexndri, wrote n extremely influentil book clled the Mthemticl Syntxis. It ws trnslted into rbic nd given the rbic title of lmgest. Ptolemy considered chords subtending n ngle on the circumference. Using modern nottion, if we tke dimeter of length unit s shown nd chord subtending n ngle t the circumference, then the lengths nd re respectively sin nd cos. cos sin 0 D Since ngles on the circumference subtended by the sme rc re equl, if is the ngle subtended by ny chord in this circle, then the length of tht chord is lwys sin. sin 0 cos Ptolemy lso showed tht if D is cyclic qudrilterl, then D E.D +.D =.D. D Tht is, the sum of the products of the opposite sides of cyclic qudrilterl is equl to the product of the digonls. This result is known s Ptolemy s theorem. pplying Ptolemy s theorem in the digrm below, where the circle hs dimeter, we obtin the result sin ( + b) = sin cos b + cos sin b, tht we derived bove.

25 The Improving Mthemtics Eduction in Schools (TIMES) Project {25} sin sin sin ( + ) cos cos D Using this, nd other formuls, Ptolemy ws ble to construct detiled tble of chords of ngles. Since chords re closely relted to the sine rtio, he essentilly hd tble of sines. THE SINE RULE IN THE IRLE Given tringle, we cn drw its circumcircle with dimeter D s shown. Let 2R be the dimeter of the circumcircle. R D R Then for cute, the sine of ngle D is given by sin D = sin = 2R. Re-rrnging gives sin = 2R. Thus the quntities sin, b sin nd c sin in the sine rule re ll equl to the dimeter of the circumcircle of the tringle. Regiomontnus ( ), who wrote the first modern Europen book on trigonometry, included the sine rule nd its derivtion in his work. THE REGULR PENTGRM ND REGULR PENTGON The regulr pentgrm nd the regulr pentgon hve lwys been source of fscintion nd is often used in strology. It is bsed on tringle whose properties re investigted in the following exercise. Regulr pentgon Regulr pentgrm

26 {26} guide for techers EXERISE x 4 D 4 2x 72 2x x 36 We begin with n isosceles tringle with ngles 72, 72, 36, s shown. This tringle occurs nturlly inside both the regulr pentgon nd the regulr pentgrm. For ese of clcultion we tke the two equl sides to be 4 units in length. We tke the point D on such tht D = 72. Finlly, we let = 2x. Show tht the informtion mrked on the digrm is correct. b Prove tht the tringles nd D re similr. c Deduce tht 4 2x 2x = 2x 4 nd solve this eqution to obtin x = 5. d y dropping perpendiculr from to show tht cos 72 = 5 4. e Use the identity cos 2 + sin 2 = to show tht sin 72 =

27 The Improving Mthemtics Eduction in Schools (TIMES) Project {27} NSWERS TO EXERISES EXERISE b 50( 3 + ) EXERISE 2 tn , correct to two deciml plces. EXERISE 3 P = m nd O =.54 m to the nerest cm. EXERISE 4 Let M = Therefore = 80 h = b sin (tringle M) nd h = sin (tringle ) h b Therefore sin = b sin = b sin (80 ) = b sin so sin = b sin M c hence the result. EXERISE 5 sin 0 = 0, sin 90 =, sin 80 = 0 cos 0 =, cos 90 = 0, cos 80 = EXERISE 6 3 3, 3 EXERISE 7 sin (correct to two deciml plces). The two sides nd the right ngle define unique tringle (RHS congruence).

28 {28} guide for techers EXERISE 8 In tringle M, h 2 = 2 (c + M) 2 In tringle M, h 2 = b 2 M 2 Therefore, 2 (c + M) 2 = b 2 M 2 h b 2 = b 2 + c 2 + 2c M ut M = b cos (80 ) = b cos M c Hence 2 = b 2 + c 2 2bc cos EXERISE 9 sin > sin > sin (if > b nd sin = b sin then sin > sin b y the remrk t the beginning of this prgrph, > >. c The ngles P nd Q dd to, becuse the ngle sum of the tringle is 80. Hence P nd Q re smller thn, so sin P nd sin Q re less thn sin = sin P. d From prt c, the obtuse ngle is, which is therefore the lrgest of the three ngles. Hence sin > sin, where nd re cute, so > >. EXERISE 0 cos = b2 + c bc nd cos = c2 + 2 b 2 2c = b2 + c 2 3 2bc = bc2 + 2 b b 3 2bc b Grouping terms, cos cos = (2 b b 2 ) (c 2 bc 2 ) + ( 3 b 3 ) 2bc = b( b) c 2 ( b) + ( 3 b 3 ) 2bc = (b c2 ) ( b) + ( 3 b 3 ) 2bc c From > b > c > 0, it follows tht b > c 2 nd 3 > b 3. Hence cos > cos. d similr rgument proves tht cos > cos, so cos > cos > cos. y the remrk t the strt of this prgrph, > >. EXERISE Let be right-ngled t. Then 2 = Hence 2 is lrger thn both 2 nd 2, so is longer thn both nd.

29 The Improving Mthemtics Eduction in Schools (TIMES) Project {29} b In right-hnded tringle, the other two ngles re cute, becuse the ngle sum of the tringle is 80. Hence the right ngle is the lrgest ngle. EXERISE 2 We use the sme digrm s tht used for the obtuse ngle cse proof of the sine nd cosine rule. re = 2 c h = 2 cb sin = 2 cb sin h b The stted result is obtined by symmetry of rgument. M c EXERISE 3 = 30 or 50 EXERISE 4 For given tringle, 2 cb sin = 2 c sin. Therefore bsin = sin nd sin = b sin. EXERISE 5 sin 75 = sin ( ) = sin 45 cos 30 + sin 30 cos 45 = = EXERISE 6 D = 36 (ngle sum of tringle), D = 36 ( D = 36 nd D = 72 ) D = 2x (tringle D is isosceles), D = 2x (tringle D is isosceles), D = D = 4 2x. b Tringle is similr to tringle D () c D = hence 4 2x 2x = 2x 4 2 x x = x 2 d cos = x 4 e sin 2 = x2 6 = x 2 2x = 4 x 2 + 2x + = 5 (x + ) 2 = 5 x = + 5 s x > 0

30 The im of the Interntionl entre of Excellence for Eduction in Mthemtics (IE-EM) is to strengthen eduction in the mthemticl sciences t ll levelsfrom school to dvnced reserch nd contemporry pplictions in industry nd commerce. IE-EM is the eduction division of the ustrlin Mthemticl Sciences Institute, consortium of 27 university mthemtics deprtments, SIRO Mthemticl nd Informtion Sciences, the ustrlin ureu of Sttistics, the ustrlin Mthemticl Society nd the ustrlin Mthemtics Trust. The IE-EM modules re prt of The Improving Mthemtics Eduction in Schools (TIMES) Project. The modules re orgnised under the strnd titles of the ustrlin urriculum: Number nd lgebr Mesurement nd Geometry Sttistics nd Probbility The modules re written for techers. Ech module contins discussion of component of the mthemtics curriculum up to the end of Yer 0.

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