FURTHER TRIGONOMETRY

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "FURTHER TRIGONOMETRY"

Transcription

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.

Pythagoras theorem and trigonometry (2)

Pythagoras theorem and trigonometry (2) HPTR 10 Pythgors theorem nd trigonometry (2) 31 HPTR Liner equtions In hpter 19, Pythgors theorem nd trigonometry were used to find the lengths of sides nd the sizes of ngles in right-ngled tringles. These

More information

Section 5-4 Trigonometric Functions

Section 5-4 Trigonometric Functions 5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form

More information

Addition and subtraction of rational expressions

Addition and subtraction of rational expressions Lecture 5. Addition nd subtrction of rtionl expressions Two rtionl expressions in generl hve different denomintors, therefore if you wnt to dd or subtrct them you need to equte the denomintors first. The

More information

Sect 8.3 Triangles and Hexagons

Sect 8.3 Triangles and Hexagons 13 Objective 1: Sect 8.3 Tringles nd Hexgons Understnding nd Clssifying Different Types of Polygons. A Polygon is closed two-dimensionl geometric figure consisting of t lest three line segments for its

More information

STRAND I: Geometry and Trigonometry. UNIT I2 Trigonometric Problems: Text * * Contents. Section. I2.1 Mixed Problems Using Trigonometry

STRAND I: Geometry and Trigonometry. UNIT I2 Trigonometric Problems: Text * * Contents. Section. I2.1 Mixed Problems Using Trigonometry Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text STRND I: Geometry nd Trigonometry I Trigonometric Prolems Text ontents Section * * * I. Mixed Prolems Using Trigonometry I. Sine nd osine Rules

More information

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100

Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by

More information

Unit 6 Solving Oblique Triangles - Classwork

Unit 6 Solving Oblique Triangles - Classwork Unit 6 Solving Oblique Tringles - Clsswork A. The Lw of Sines ASA nd AAS In geometry, we lerned to prove congruence of tringles tht is when two tringles re exctly the sme. We used severl rules to prove

More information

Curve Sketching. 96 Chapter 5 Curve Sketching

Curve Sketching. 96 Chapter 5 Curve Sketching 96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of

More information

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES

LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of

More information

Two special Right-triangles 1. The

Two special Right-triangles 1. The Mth Right Tringle Trigonometry Hndout B (length of ) - c - (length of side ) (Length of side to ) Pythgoren s Theorem: for tringles with right ngle ( side + side = ) + = c Two specil Right-tringles. The

More information

Section 2.3. Motion Along a Curve. The Calculus of Functions of Several Variables

Section 2.3. Motion Along a Curve. The Calculus of Functions of Several Variables The Clculus of Functions of Severl Vribles Section 2.3 Motion Along Curve Velocity ccelertion Consider prticle moving in spce so tht its position t time t is given by x(t. We think of x(t s moving long

More information

Solutions to Section 1

Solutions to Section 1 Solutions to Section Exercise. Show tht nd. This follows from the fct tht mx{, } nd mx{, } Exercise. Show tht = { if 0 if < 0 Tht is, the bsolute vlue function is piecewise defined function. Grph this

More information

Quadrilaterals Here are some examples using quadrilaterals

Quadrilaterals Here are some examples using quadrilaterals Qudrilterls Here re some exmples using qudrilterls Exmple 30: igonls of rhomus rhomus hs sides length nd one digonl length, wht is the length of the other digonl? 4 - Exmple 31: igonls of prllelogrm Given

More information

Mathematics Higher Level

Mathematics Higher Level Mthemtics Higher Level Higher Mthemtics Exmintion Section : The Exmintion Mthemtics Higher Level. Structure of the exmintion pper The Higher Mthemtics Exmintion is divided into two ppers s detiled below:

More information

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.

Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions. Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd

More information

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.

Example 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers. 2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this

More information

Geometry 7-1 Geometric Mean and the Pythagorean Theorem

Geometry 7-1 Geometric Mean and the Pythagorean Theorem Geometry 7-1 Geometric Men nd the Pythgoren Theorem. Geometric Men 1. Def: The geometric men etween two positive numers nd is the positive numer x where: = x. x Ex 1: Find the geometric men etween the

More information

Mathematics in Art and Architecture GEK1518K

Mathematics in Art and Architecture GEK1518K Mthemtics in Art nd Architecture GEK1518K Helmer Aslksen Deprtment of Mthemtics Ntionl University of Singpore slksen@mth.nus.edu.sg www.mth.nus.edu.sg/slksen/ The Golden Rtio The Golden Rtio Suppose we

More information

8. Hyperbolic triangles

8. Hyperbolic triangles 8. Hyperoli tringles Note: This yer, I m not doing this mteril, prt from Pythgors theorem, in the letures (nd, s suh, the reminder isn t exminle). I ve left the mteril s Leture 8 so tht (i) nyody interested

More information

Basic Math Review. Numbers. Important Properties. Absolute Value PROPERTIES OF ADDITION NATURAL NUMBERS {1, 2, 3, 4, 5, }

Basic Math Review. Numbers. Important Properties. Absolute Value PROPERTIES OF ADDITION NATURAL NUMBERS {1, 2, 3, 4, 5, } ƒ Bsic Mth Review Numers NATURAL NUMBERS {1,, 3, 4, 5, } WHOLE NUMBERS {0, 1,, 3, 4, } INTEGERS {, 3,, 1, 0, 1,, } The Numer Line 5 4 3 1 0 1 3 4 5 Negtive integers Positive integers RATIONAL NUMBERS All

More information

1 Numerical Solution to Quadratic Equations

1 Numerical Solution to Quadratic Equations cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll

More information

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY

PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive

More information

Essential Question What are the Law of Sines and the Law of Cosines?

Essential Question What are the Law of Sines and the Law of Cosines? 9.7 TEXS ESSENTIL KNOWLEDGE ND SKILLS G.6.D Lw of Sines nd Lw of osines Essentil Question Wht re the Lw of Sines nd the Lw of osines? Disovering the Lw of Sines Work with prtner.. opy nd omplete the tle

More information

11. Fourier series. sin mx cos nx dx = 0 for any m, n, sin 2 mx dx = π.

11. Fourier series. sin mx cos nx dx = 0 for any m, n, sin 2 mx dx = π. . Fourier series Summry of the bsic ides The following is quick summry of the introductory tretment of Fourier series in MATH. We consider function f with period π, tht is, stisfying f(x + π) = f(x) for

More information

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS

RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is

More information

Lesson 10. Parametric Curves

Lesson 10. Parametric Curves Return to List of Lessons Lesson 10. Prmetric Curves (A) Prmetric Curves If curve fils the Verticl Line Test, it cn t be expressed by function. In this cse you will encounter problem if you try to find

More information

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )

Polynomial Functions. Polynomial functions in one variable can be written in expanded form as ( ) Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +

More information

Quadratic Equations. Math 99 N1 Chapter 8

Quadratic Equations. Math 99 N1 Chapter 8 Qudrtic Equtions Mth 99 N1 Chpter 8 1 Introduction A qudrtic eqution is n eqution where the unknown ppers rised to the second power t most. In other words, it looks for the vlues of x such tht second degree

More information

Graphs on Logarithmic and Semilogarithmic Paper

Graphs on Logarithmic and Semilogarithmic Paper 0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl

More information

Vectors 2. 1. Recap of vectors

Vectors 2. 1. Recap of vectors Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms

More information

Math 135 Circles and Completing the Square Examples

Math 135 Circles and Completing the Square Examples Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for

More information

Geometry Notes SIMILAR TRIANGLES

Geometry Notes SIMILAR TRIANGLES Similr Tringles Pge 1 of 6 SIMILAR TRIANGLES Objectives: After completing this section, you shoul be ble to o the following: Clculte the lengths of sies of similr tringles. Solve wor problems involving

More information

Section 7-4 Translation of Axes

Section 7-4 Translation of Axes 62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-4 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the

More information

Sequences and Series

Sequences and Series Centre for Eduction in Mthemtics nd Computing Euclid eworkshop # 5 Sequences nd Series c 014 UNIVERSITY OF WATERLOO While the vst mjority of Euclid questions in this topic re use formule for rithmetic

More information

Factoring Polynomials

Factoring Polynomials Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles

More information

Let us recall some facts you have learnt in previous grades under the topic Area.

Let us recall some facts you have learnt in previous grades under the topic Area. 6 Are By studying this lesson you will be ble to find the res of sectors of circles, solve problems relted to the res of compound plne figures contining sectors of circles. Ares of plne figures Let us

More information

Net Change and Displacement

Net Change and Displacement mth 11, pplictions motion: velocity nd net chnge 1 Net Chnge nd Displcement We hve seen tht the definite integrl f (x) dx mesures the net re under the curve y f (x) on the intervl [, b] Any prt of the

More information

Name: Lab Partner: Section:

Name: Lab Partner: Section: Chpter 4 Newton s 2 nd Lw Nme: Lb Prtner: Section: 4.1 Purpose In this experiment, Newton s 2 nd lw will be investigted. 4.2 Introduction How does n object chnge its motion when force is pplied? A force

More information

Operations with Polynomials

Operations with Polynomials 38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply

More information

4.11 Inner Product Spaces

4.11 Inner Product Spaces 314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces

More information

1 PRECALCULUS READINESS DIAGNOSTIC TEST PRACTICE

1 PRECALCULUS READINESS DIAGNOSTIC TEST PRACTICE PRECALCULUS READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the smples, work the problems, then check your nswers t the end of ech topic. If you don t get the nswer given, check your work nd look

More information

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324

A.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324 A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................

More information

6.2 Volumes of Revolution: The Disk Method

6.2 Volumes of Revolution: The Disk Method mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of

More information

EQUATIONS OF LINES AND PLANES

EQUATIONS OF LINES AND PLANES EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint

More information

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.

5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one. 5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued

More information

Lines and angles. Name. Use a ruler and pencil to draw: a 2 parallel lines. c 2 perpendicular lines. b 2 intersecting lines. Complete the following:

Lines and angles. Name. Use a ruler and pencil to draw: a 2 parallel lines. c 2 perpendicular lines. b 2 intersecting lines. Complete the following: Lines nd s 1 Use ruler nd pencil to drw: 2 prllel lines 2 intersecting lines c 2 perpendiculr lines 2 Complete the following: drw in the digonls on this shpe mrk the interior s on this shpe c mrk equl

More information

SPECIAL PRODUCTS AND FACTORIZATION

SPECIAL PRODUCTS AND FACTORIZATION MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come

More information

The Math Learning Center PO Box 12929, Salem, Oregon 97309 0929 Math Learning Center

The Math Learning Center PO Box 12929, Salem, Oregon 97309 0929  Math Learning Center Resource Overview Quntile Mesure: Skill or Concept: 1010Q Determine perimeter using concrete models, nonstndrd units, nd stndrd units. (QT M 146) Use models to develop formuls for finding res of tringles,

More information

10.5 Graphing Quadratic Functions

10.5 Graphing Quadratic Functions 0.5 Grphing Qudrtic Functions Now tht we cn solve qudrtic equtions, we wnt to lern how to grph the function ssocited with the qudrtic eqution. We cll this the qudrtic function. Grphs of Qudrtic Functions

More information

CONIC SECTIONS. Chapter 11

CONIC SECTIONS. Chapter 11 CONIC SECTIONS Chpter 11 11.1 Overview 11.1.1 Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig. 11.1). Fig. 11.1 Suppose we

More information

NCERT INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS. Trigonometric Ratios of the angle A in a triangle ABC right angled at B are defined as:

NCERT INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS. Trigonometric Ratios of the angle A in a triangle ABC right angled at B are defined as: INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS (A) Min Concepts nd Results Trigonometric Rtios of the ngle A in tringle ABC right ngled t B re defined s: side opposite to A BC sine of A = sin A = hypotenuse

More information

r 2 F ds W = r 1 qe ds = q

r 2 F ds W = r 1 qe ds = q Chpter 4 The Electric Potentil 4.1 The Importnt Stuff 4.1.1 Electricl Potentil Energy A chrge q moving in constnt electric field E experiences force F = qe from tht field. Also, s we know from our study

More information

The Quadratic Formula and the Discriminant

The Quadratic Formula and the Discriminant 9-9 The Qudrtic Formul nd the Discriminnt Objectives Solve qudrtic equtions by using the Qudrtic Formul. Determine the number of solutions of qudrtic eqution by using the discriminnt. Vocbulry discriminnt

More information

Section 4.3. By the Mean Value Theorem, for every i = 1, 2, 3,..., n, there exists a point c i in the interval [x i 1, x i ] such that

Section 4.3. By the Mean Value Theorem, for every i = 1, 2, 3,..., n, there exists a point c i in the interval [x i 1, x i ] such that Difference Equtions to Differentil Equtions Section 4.3 The Fundmentl Theorem of Clculus We re now redy to mke the long-promised connection between differentition nd integrtion, between res nd tngent lines.

More information

Number Systems & Working With Numbers

Number Systems & Working With Numbers Presenting the Mths Lectures! Your best bet for Qunt... MATHS LECTURE # 0 Number Systems & Working With Numbers System of numbers.3 0.6 π With the help of tree digrm, numbers cn be clssified s follows

More information

Scalar Line Integrals

Scalar Line Integrals Mth 3B Discussion Session Week 5 Notes April 6 nd 8, 06 This week we re going to define new type of integrl. For the first time, we ll be integrting long something other thn Eucliden spce R n, nd we ll

More information

Exponentiation: Theorems, Proofs, Problems Pre/Calculus 11, Veritas Prep.

Exponentiation: Theorems, Proofs, Problems Pre/Calculus 11, Veritas Prep. Exponentition: Theorems, Proofs, Problems Pre/Clculus, Verits Prep. Our Exponentition Theorems Theorem A: n+m = n m Theorem B: ( n ) m = nm Theorem C: (b) n = n b n ( ) n n Theorem D: = b b n Theorem E:

More information

Volumes of solids of revolution

Volumes of solids of revolution Volumes of solids of revolution We sometimes need to clculte the volume of solid which cn be obtined by rotting curve bout the x-xis. There is strightforwrd technique which enbles this to be done, using

More information

Square Roots Teacher Notes

Square Roots Teacher Notes Henri Picciotto Squre Roots Techer Notes This unit is intended to help students develop n understnding of squre roots from visul / geometric point of view, nd lso to develop their numer sense round this

More information

ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI

ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI ON THE FRAME-STEWART ALGORITHM FOR THE TOWER OF HANOI MICHAEL RAND 1. Introduction The Tower of Hnoi puzzle ws creted over century go by the number theorist Edourd Lucs [, 4], nd it nd its vrints hve chllenged

More information

A new algorithm for generating Pythagorean triples

A new algorithm for generating Pythagorean triples A new lgorithm for generting Pythgoren triples RH Dye 1 nd RWD Nicklls 2 The Mthemticl Gzette (1998); 82 (Mrch, No. 493), p. 86 91 (JSTOR rchive) http://www.nicklls.org/dick/ppers/mths/pythgtriples1998.pdf

More information

PHY 140A: Solid State Physics. Solution to Homework #2

PHY 140A: Solid State Physics. Solution to Homework #2 PHY 140A: Solid Stte Physics Solution to Homework # TA: Xun Ji 1 October 14, 006 1 Emil: jixun@physics.ucl.edu Problem #1 Prove tht the reciprocl lttice for the reciprocl lttice is the originl lttice.

More information

11. PYTHAGORAS THEOREM

11. PYTHAGORAS THEOREM 11. PYTHAGORAS THEOREM 11-1 Along the Nile 2 11-2 Proofs of Pythgors theorem 3 11-3 Finding sides nd ngles 5 11-4 Semiirles 7 11-5 Surds 8 11-6 Chlking hndll ourt 9 11-7 Pythgors prolems 10 11-8 Designing

More information

Lesson 12.1 Trigonometric Ratios

Lesson 12.1 Trigonometric Ratios Lesson 12.1 rigonometric Rtios Nme eriod Dte In Eercises 1 6, give ech nswer s frction in terms of p, q, nd r. 1. sin 2. cos 3. tn 4. sin Q 5. cos Q 6. tn Q p In Eercises 7 12, give ech nswer s deciml

More information

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes

9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is so-clled becuse when the sclr product of two vectors

More information

Binary Representation of Numbers Autar Kaw

Binary Representation of Numbers Autar Kaw Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy

More information

fraction arithmetic. For example, consider this problem the 1995 TIMSS Trends in International Mathematics and Science Study:

fraction arithmetic. For example, consider this problem the 1995 TIMSS Trends in International Mathematics and Science Study: In recent yers, mthemtics eductors hve begun to relize tht understnding frctions nd frctionl rithmetic is the gtewy to dvnced high school mthemtics. 1 Yet, US students continue to do poorly when rnked

More information

Lesson 18.2: Right Triangle Trigonometry

Lesson 18.2: Right Triangle Trigonometry Lesson 8.: Right Tringle Trigonometry lthough Trigonometry is used to solve mny prolems, historilly it ws first pplied to prolems tht involve right tringle. This n e extended to non-right tringles (hpter

More information

Uniform convergence and its consequences

Uniform convergence and its consequences Uniform convergence nd its consequences The following issue is centrl in mthemtics: On some domin D, we hve sequence of functions {f n }. This mens tht we relly hve n uncountble set of ordinry sequences,

More information

Arc Length. P i 1 P i (1) L = lim. i=1

Arc Length. P i 1 P i (1) L = lim. i=1 Arc Length Suppose tht curve C is defined by the eqution y = f(x), where f is continuous nd x b. We obtin polygonl pproximtion to C by dividing the intervl [, b] into n subintervls with endpoints x, x,...,x

More information

Algebra Review. How well do you remember your algebra?

Algebra Review. How well do you remember your algebra? Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then

More information

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.

addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix. APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The

More information

Lecture 5. Inner Product

Lecture 5. Inner Product Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right

More information

4.0 5-Minute Review: Rational Functions

4.0 5-Minute Review: Rational Functions mth 130 dy 4: working with limits 1 40 5-Minute Review: Rtionl Functions DEFINITION A rtionl function 1 is function of the form y = r(x) = p(x) q(x), 1 Here the term rtionl mens rtio s in the rtio of two

More information

Physics 43 Homework Set 9 Chapter 40 Key

Physics 43 Homework Set 9 Chapter 40 Key Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x

More information

Integration by Substitution

Integration by Substitution Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is

More information

Answer, Key Homework 8 David McIntyre 1

Answer, Key Homework 8 David McIntyre 1 Answer, Key Homework 8 Dvid McIntyre 1 This print-out should hve 17 questions, check tht it is complete. Multiple-choice questions my continue on the net column or pge: find ll choices before mking your

More information

Review guide for the final exam in Math 233

Review guide for the final exam in Math 233 Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered

More information

Anti-derivatives/Indefinite Integrals of Basic Functions

Anti-derivatives/Indefinite Integrals of Basic Functions Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: x n+ x n n + + C, dx = ln x + C, if n if n = In prticulr, this mens tht dx = ln x + C x nd x 0 dx = dx = dx = x + C Integrl of Constnt:

More information

5.2 The Definite Integral

5.2 The Definite Integral 5.2 THE DEFINITE INTEGRAL 5.2 The Definite Integrl In the previous section, we sw how to pproximte totl chnge given the rte of chnge. In this section we see how to mke the pproximtion more ccurte. Suppose

More information

9.1 PYTHAGOREAN THEOREM (right triangles)

9.1 PYTHAGOREAN THEOREM (right triangles) Simplifying Rdicls: ) 1 b) 60 c) 11 d) 3 e) 7 Solve: ) x 4 9 b) 16 80 c) 9 16 9.1 PYTHAGOREAN THEOREM (right tringles) c If tringle is right tringle then b, b re the legs * c is clled the hypotenuse (side

More information

Triangles, Altitudes, and Area Instructor: Natalya St. Clair

Triangles, Altitudes, and Area Instructor: Natalya St. Clair Tringle, nd ltitudes erkeley Mth ircles 015 Lecture Notes Tringles, ltitudes, nd re Instructor: Ntly St. lir *Note: This M session is inspired from vriety of sources, including wesomemth, reteem Mth Zoom,

More information

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding

Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde

More information

Chapter 6 Solving equations

Chapter 6 Solving equations Chpter 6 Solving equtions Defining n eqution 6.1 Up to now we hve looked minly t epressions. An epression is n incomplete sttement nd hs no equl sign. Now we wnt to look t equtions. An eqution hs n = sign

More information

Geometry and Measure. 12am 1am 2am 3am 4am 5am 6am 7am 8am 9am 10am 11am 12pm

Geometry and Measure. 12am 1am 2am 3am 4am 5am 6am 7am 8am 9am 10am 11am 12pm Reding Scles There re two things to do when reding scle. 1. Mke sure you know wht ech division on the scle represents. 2. Mke sure you red in the right direction. Mesure Length metres (m), kilometres (km),

More information

Generalized Inverses: How to Invert a Non-Invertible Matrix

Generalized Inverses: How to Invert a Non-Invertible Matrix Generlized Inverses: How to Invert Non-Invertible Mtrix S. Swyer September 7, 2006 rev August 6, 2008. Introduction nd Definition. Let A be generl m n mtrix. Then nturl question is when we cn solve Ax

More information

QUANTITATIVE REASONING

QUANTITATIVE REASONING Guide For Exminees Inter-University Psychometric Entrnce Test QUNTITTIVE RESONING The Quntittive Resoning domin tests your bility to use numbers nd mthemticl concepts to solve mthemticl problems, s well

More information

Quadratic Equations - 1

Quadratic Equations - 1 Alger Module A60 Qudrtic Equtions - 1 Copyright This puliction The Northern Alert Institute of Technology 00. All Rights Reserved. LAST REVISED Novemer, 008 Qudrtic Equtions - 1 Sttement of Prerequisite

More information

CHAPTER 11 Numerical Differentiation and Integration

CHAPTER 11 Numerical Differentiation and Integration CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods

More information

Right Triangles and Trigonometry

Right Triangles and Trigonometry 9 Right Tringles nd Trigonometry 9.1 The Pythgoren Theorem 9. Specil Right Tringles 9.3 Similr Right Tringles 9.4 The Tngent Rtio 9.5 The Sine nd osine Rtios 9.6 Solving Right Tringles 9.7 Lw of Sines

More information

COMPONENTS: COMBINED LOADING

COMPONENTS: COMBINED LOADING LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of

More information

Introduction to Mathematical Reasoning, Saylor 111

Introduction to Mathematical Reasoning, Saylor 111 Frction versus rtionl number. Wht s the difference? It s not n esy question. In fct, the difference is somewht like the difference between set of words on one hnd nd sentence on the other. A symbol is

More information

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1

PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1 PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.

More information

The Parallelogram Law. Objective: To take students through the process of discovery, making a conjecture, further exploration, and finally proof.

The Parallelogram Law. Objective: To take students through the process of discovery, making a conjecture, further exploration, and finally proof. The Prllelogrm Lw Objective: To tke students through the process of discovery, mking conjecture, further explortion, nd finlly proof. I. Introduction: Use one of the following Geometer s Sketchpd demonstrtion

More information

State the size of angle x. Sometimes the fact that the angle sum of a triangle is 180 and other angle facts are needed. b y 127

State the size of angle x. Sometimes the fact that the angle sum of a triangle is 180 and other angle facts are needed. b y 127 ngles 2 CHTER 2.1 Tringles Drw tringle on pper nd lel its ngles, nd. Ter off its orners. Fit ngles, nd together. They mke stright line. This shows tht the ngles in this tringle dd up to 180 ut it is not

More information

Vectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a.

Vectors. The magnitude of a vector is its length, which can be determined by Pythagoras Theorem. The magnitude of a is written as a. Vectors mesurement which onl descries the mgnitude (i.e. size) of the oject is clled sclr quntit, e.g. Glsgow is 11 miles from irdrie. vector is quntit with mgnitude nd direction, e.g. Glsgow is 11 miles

More information

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity

Babylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University

More information

Content Objectives: After completing the activity, students will gain experience of informally proving Pythagoras Theorem

Content Objectives: After completing the activity, students will gain experience of informally proving Pythagoras Theorem Pythgors Theorem S Topic 1 Level: Key Stge 3 Dimension: Mesures, Shpe nd Spce Module: Lerning Geometry through Deductive Approch Unit: Pythgors Theorem Student ility: Averge Content Ojectives: After completing

More information

NUMBER SYSTEMS CHAPTER 1. (A) Main Concepts and Results

NUMBER SYSTEMS CHAPTER 1. (A) Main Concepts and Results CHAPTER NUMBER SYSTEMS Min Concepts nd Results Rtionl numbers Irrtionl numbers Locting irrtionl numbers on the number line Rel numbers nd their deciml expnsions Representing rel numbers on the number line

More information

2 If a branch is prime, no other factors

2 If a branch is prime, no other factors Chpter 2 Multiples, nd primes 59 Find the prime of 50 by drwing fctor tree. b Write 50 s product of its prime. 1 Find fctor pir of the given 50 number nd begin the fctor tree (50 = 5 10). 5 10 2 If brnch

More information