CHENG Chun Chor Litwin The Hong Kong Institute of Education

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1 PE-hing Mi terntionl onferene IV: novtion of Mthemtis Tehing nd Lerning through Lesson Study- onnetion etween ssessment nd Sujet Mtter HENG hun hor Litwin The Hong Kong stitute of Edution Report on using Sine Rule nd osine Rule to demonstrte the onnetion etween ssessment nd Sujet Mtter trodution, Hnd on ssessment It is diffiult to understnd mthemtil onept in one single lesson. Hene, it is good pproh to work on the exerise on the topi nd then through working the exmple to onsolidte the onepts. Suh working on the exerise llow us to pik up different piees of onepts nd onsolidte them through the prolems. It is the hnd on working nd ssessment tht help to uild the onepts. The pper would like to disuss the omponents of Sine Rule nd osine Rule, its sujet mtter nd the ssessment items relted, so tht ssessment items re used s tools to enhne the dvnement of the mthemtil onepts. Sujet Mtter Sine Rule, osine Rule Hnd on Exerise (simple lultion nd pplition) ssessment formtion of given tringle, pplition Explortion (mthemtis thinking nd extension, lterntive proof)

2 Sujet mtter in Sine Rule Sine Rule Speil se (right ngle tringle) Using right ngle, students n explore nd otin the following reltions. sin = =,sin = sin sin s sin =, so it is tre tht sin sin The next step is to estlish the formul = sin sin. sin = sin for ll tringles. the following proof, relted ssessment question is tthed. Sine Rule Proof Using the reltion of res of tringles, we hve Sin = Sin = Sin = = sin sin sin, This is the simplest wy to estlish the formul. Sine Rule Proof (sed on the logi of Proof ) Using the medin M of. M

3 re M = ()(M)sin re M = ()(M)sin s oth res re equl nd M = M, sin = sin sin = sin 3

4 Sine Rule Proof Using the height h of the tringle s referene, sin = h h,sin =, sin = sin sin = sin h Sine Rule Proof (sed on the logi of the Proof ) with ostue ngle. drw D D D, meeting the extension of t D, then sin, D sin ; sin( 80 ) sin, D sin, Hene sin sin, nd sin sin D Similrly, sin, sin Hene sin. sin sin Question:, ngle isetor D of meets t D. Prove tht D D β β α D s in digrm, using Sine Rule in D sin sin D sin sin(80 ) sin D nd D,

5 , D D onnetion nd Explortion = = sin sin sin =? Wht is the vlue of the Sttement: = = sin sin sin?, R sin sin sin = = sin sin sin = R, ut how this is done? Let O e the entre of irle insriing, onnet O nd extend O to meet the irle t D D = dimeter. D 90, D right ngle tringle D, D sin D D sin Rsin Rsin Similrly, Rsin, Rsin D O. sin sin sin R Disussion nd ssessment Only two of the three reltions in sine rule is independent. From. From sin sin sin,, we hve sin sin sin sin sin sin, there my exist positive numer nd ngle, so tht or suh result does not hold. sin 5

6 onnetion nd explortion The dedution of mthemtil onepts from given prolem The following result is given to student nd they re sked to relted them to Sine Rule. From Element, Theorem 0, ook 3 irle, the ngle sutends y the sme r t the entre is twie the ngle sutend t the irumferene. O Extension The ngle sutended t the irumferene re the sme (theorem of Element); Extension ngle sutended t the entre is right ngle. 3 3 O () Theroem () ngle sutended re right ngle Using the results, the lss n prove the Sine Rule. r O r r O r Sine Rule (ute ngle) Sine Rule (Otuse ngle) 6

7 When is ute, entre of the irle is lies inside the tringle. insried in the irle with rdius r,entre O. nd O. From O drws, the perpendiulr isetor, then sin = r sin ( ) / r, r sin sin sin When is otuse, the entre of the irle is outside the tringle. The r is more thn hlf of the irumferene, hene the ngle t O of O is 360 From O drws, the perpendiulr sin isetor, then ut Hene sin sin(80 = r. sin ( ) / r ) sin, osine Rule Proof of osine Rule Extension of Pythgors Theorem Using Pythgors Theorem ( Sin) + ( os + ) = () + () os = D Result, we hve os os os osine Rule Proof Through, onstrut D nd meet t D then D os, D os Hene D D D ( os ) os D os Similrly, os os 7

8 osine Rule Proof 3 os os, os os, os os, os os os os os os From the ove formul, os os os = os os Similrly, os, os. onnetion (Heron Formul) x [( ) ][ ( ) ] From the digrm, otin x re of tringle = = x [( ) x ][ ( ) ] = [( ) ][ ( ) ] Whih is the sme s p(p )(p )(p ), where p = ( ) From the following formt, numer of ssessments n e generted. 8

9 Question Formt:, D is point on D Question:, point D on stisfy 5, 7, D 3, D 5, 5 7 then D =? 3 D 5 5, os nd in D, D D D os D 9 Question:, point D on stisfy 6, 3, D 5, find D. D 30 nd Let D x, re D D 6 x 3 x 6 3 sin 30 sin 5 3x 3x 6 x 3 9 sin 75 9

10 Question:, 8,, D is point on nd D 7, D 6. Find D 5 D 5, solving from D, nd otin from. D, os 85, 8 8os Question:, 7, 8, D is point on, nd D 3, D : D : 3 Find k D 3k Let D k, D 3k. y ommon ngle, the tringle D nd oth hs one ngle nd three sides, Using osine rule, nd k re solved. D,, k os os k 9 5k 3 k k k 7 3 k 7 5k 7 8 5k 8 k 9 k 5k 3 k 0

11 5k 0

12 ssessment, there re three types of questions. Two sides nd two ngles, find the third side. Q Length ngles nswer D D D 3 7 D : D : 3 D 60 D 30 5 D 5 30 = 50 7 D 5 D = 3 30 D = D 60 = 6 Three sides nd n ngle given, find nother length. Q Length ngles nswer D 6 7 D 3 D 5 8 D 7 0 D D D D 30 D = D = 37 D 60 = 5 D = 6 0 tn D = D 3 D 6 D D D 3 os D = Four lengths re given, find the fifth length Q Length nswers 7 D 3 D 7 D 8 D = 7 7 D 3 D 5 D = D D 9 D = ssessment Whih Rule to use, Sine Rule or osine Rule? The importnt prt of mthemtis is thinking, usully students lern theorem nd

13 then they pply the theorem t some given situtions. For exmple, the following two exerises require students to use osine Rule nd Sine Rule. Question:, 5, 8, 7. Find =. From the onditions given, student need to relte tht three sides given stisfies the requirement of the osine rule. nd os Question:, 05, 5, 8.? From the onditions given, two ngles n only relted to the using of Sine Rule, Hene sin 8 sin sin 30 9 sin 5 the following, students need to think of how to use the two theorems. Disussion y If sin sin,we hve =, then sin = sin sin sin sin os sin = os the rtio of sides nd is os. 3

14 Question:, 8, 5, Find the vlue of os. y, we hve 80 3, os os 80 3 os 3, the question is to find os. The rest is to use Sine Rule. 8 5 sin sin sin sin 5sin 8sin 0sin os 8sin os (ssin 0 ) 5 Hene os os80 3 os 3 os 3 5 3os Question: the trpzeum D, // D, = 0, = 5, D = 5, D= 6. Find. D 6 D x 5 θ 5 θ 0

15 From ommon side = x s //D, let D Using osine Rule, x 5 6 x 0 5 x 5 x 0 x x 75 x 97 x 97 x 5 6 os nd x 5 os x 0 5 x 0 Explortion Question: Prove tht the sum of produt of distnes nd the sine of the ngle from point inside tringle to the three sides is onstnt. Tht is, h sin h sin h sin is onstnt. Let P ny point inside, nd denote =, =, =, The distne from P to,, re h h h. onnet P, P nd P S P S h P S h P S h To llow the ommon prt of,, 中, using sine rule (R is the dimeter of the irle insrie ): R sin sin sin then Rsin h Rsin h Rsin h h sin h sin h sin R 5

16 6 Question: The sides of re,,, nd, find the lrgest ngle of. ) ( ) ( From >, nd, so >. Uisng osine Rule, os, = 0 Only Two of the three reltions in the osine Rule re independent, from two of the three, the third reltion ould e dedued. y os, nd os Tht is, os, nd os 80 Then, ) os( )] ( os[80 os os os sin sin os os os os ) ( ) ( ) ( os y os, there my or my not exist ngle, suh tht os

17 ssessment (Sine Rule), sin sin sin, prove tht is right ngle tringle., if os = os, wht kind of tringle is? 3, prove tht sin sin sin ssessment, osine Rule For, prove tht (os os os) Using osine Rule to prove The sum of squres of the sides of prlellgrm equls the sum of the squre of the digonls. Mixed ssessment, =, prove tht = os., =, prove tht sin 3 sin 3 在 中,sin(os+os) = sin + sin, prove tht is right ngle tringle, if = =, prove tht is n equilterl os os os 5 tringle. ( ),, find. 6, sin = sinos, show tht is issolsoles tringle. 7 The sum of two sides of tringle is 0, the inluded ngle is 60, find the minimum perimeter of this tringle. Referenes Hong Kong Exmintion nd ssessment uthority (008), Exminer Report. Hong 7

18 Kong SR. Edution Deprtment (999), Seondry Mthemtis urriulum, Hong Kong SR. 8

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