MATHEMATICS AND STATISTICS 1.6

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1 MTHMTIS N STTISTIS 1.6 pply geometri resoning in solving prolems ternlly ssessed 4 redits S inding unknown ngles When finding the size of unknown ngles in figure, t lest two steps of resoning will e required. The following rules for finding unknown ngles re lredy fmilir. Rule igrm mple djent ngles on stright line dd to 180 ( s on line) ngles t point dd to 360 ( s t point) Vertilly opposite ngles re equl (vert opp s) + = = 360 = 50 = 40 ( s on line) = 140 ( s t point) 80 = 80 (vert opp s) n revited form of eh rule is eptle when giving geometri resons. mple ind the vlue of nd y in the digrm elow. Give geometril resons for your nswers. y 20 Solution = 180 ( s on line) = 180 [simplifying] 2 = 110 [sutrting 70 ] = 55 y = 20 (vert opp s) = y = 35 erise : inding unknown ngles 1. ind the vlue of in the digrm elow. Give geometril reson for your nswer ind the vlue of in the digrm elow. Give geometril reson for your nswer ns. p. 49 S Pulitions (NZ) Ltd, reephone , ISN

2 2 hievement Stndrd (Mthemtis nd Sttistis 1.6) S Light rys re refleted from mirror s shown in the figure. 70 ind the size of ngle. Give resons nd show ll working.. ind the vlue of. ind the vlue of y (give reson for your nswer) 4. Tom uts ke into two piees so he gets twie s muh s his little sister. He mkes sketh of the figure to work out the ngle t the entre of his slie of ke ind the vlue of in the digrm elow. Give geometril reson for your nswer. orm n eqution nd solve it to find this ngle. Give resons. 5. The figure shows lptop with its lid open. ngles nd prllel lines trnsversl is line whih uts through given pir of lines to form vrious pirs of ngles. trnsversl 3 2 orresponding ngles lternte ngles ind the ngle through whih the lid hs een opened. orm n eqution nd solve it to find this ngle. Give resons. 6.. Give geometri reson for the reltionship + 2 = y o-interior ngles When the trnsversl uts pir of prllel lines, the following rules result (mthing rrows show prllel sides). Rule igrm mple orresponding ngles on prllel lines re equl. (orr s // lines) = 50 = 50 (orr s // lines) S Pulitions (NZ) Ltd, reephone , ISN

3 pply geometri resoning in solving prolems 3 Rule igrm mple lternte ngles on prllel lines re equl 110 (lt s // lines) = = 110 (lt s // lines) o-interior ngles on prllel lines dd to 180 (o-int s // lines) + = = 30 (o-int s // lines) erise : ngles nd prllel lines 1. lulte the size of the mrked ngles.. 48 =. 36 S ns. p. 49 In reverse, if trnsversl uts pir of lines nd the resulting pir of orresponding ngles re equl, then the lines re prllel. Similr results hold true for equl lternte ngles nd supplementry o-interior ngles.. = mple 50 ind the unknown ngles mrked on the digrms d. = d = d e e e = z f. 28 f 36 Solutions: 1. = 72 (vert opp s) = 108 (o-int s // lines) 2. = 120 ( s t point) g. f = 126 y = 120 (lt s // lines) 3. = 54 (orr s // lines) z = 83 (o-int s // lines) g = 39 g S Pulitions (NZ) Ltd, reephone , ISN

4 4 hievement Stndrd (Mthemtis nd Sttistis 1.6) S h. h = 112 h d. G ind the sizes of the ngles mrked in the digrms. Give resons for eh step In eh digrm (not drwn to sle) trnsversl uts pir of lines. re the lines prllel? Justify your nswer d S Pulitions (NZ) Ltd, reephone , ISN

5 pply geometri resoning in solving prolems The ross-setion of prt of liming frme is shown. is horizontl nd // = 120, = 32 nd G = Is the stright line through G (representing the ground) horizontl? Give resons for your nswer. G ind, with resons, the size of ngle. The figure shows the roof lines of ftory S ind, with resons, the size of ngle. The five-r gte t frmhouse is shown in the figure. 46 ind the size of the ngles,, nd. Give geometril resons for eh. 6.. The figure shows grden roller eing pulled up steep inline. The hndle is prllel to the inline. Some ngles re mrked on the figure. 126 o ind, with resons, the size of the ngle mrked on the digrm.. The figure shows prt of ike frmework. y 30 o 62 o 5.. Toy trvels from point to point on ering of 118. Toy wnts to work out the ering of point from point nd to do this, he needs to lulte ngle. N 118 N Prllel les re mrked with rrows. lulte, with resons, the size of the ngle mrked y on the digrm. S Pulitions (NZ) Ltd, reephone , ISN

6 6 hievement Stndrd (Mthemtis nd Sttistis 1.6) S Prove + + = 180 Polygons tringles polygon is 2- losed figure with stright sides, e.g. hegon is polygon with si sides. hegon regulr polygon hs ll sides the sme length nd ll interior ngles equl.. Prove + = 90 Tringles regulr hegon tringle is polygon with 3 sides nd 3 interior ngles. n equilterl tringle hs ll three sides the sme length. n isoseles tringle hs two sides of equl length. slene tringle hs no sides of equl length. In right-ngled tringle one ngle is of size 90. Mthing mrks on the sides of tringle re used to show equl lengths.. Prove + = 0 Tringle rules The following rules nd their revitions should e known. Rule igrm mple The ngle sum of tringle is 180 ( sum Δ) + + = = 50 ( sum ) The eterior ngle of tringle is equl to the sum of the two interior opposite ngles (et Δ) = = 70 (et S Pulitions (NZ) Ltd, reephone , ISN

7 pply geometri resoning in solving prolems 7 Rule igrm mple In n equilterl tringle ll three ngles re equl in size to 60 ( s in eq Δ) = = = 60 = 60 ( s in eq ). d. 26 S n isoseles tringle hs n is of symmetry so the two se ngles re equl (se s isos Δ) = 50 = 50 (se s isos ) Tringle rules re omined with previous rules when finding unknown ngles in geometry prolems. mple In the digrm shown // // G nd = = G = 100 y ind with resons: 1. ngle ngle y Solution G 1. = 100 (lt s // lines) = (se s isos Δ) = 40 ( sum Δ) [ = 180 ] 2. = 80 ( s on line) = 80 (se s isos Δ) y = 20 ( sum Δ) erise : Tringles 1. ind the size of the ngle in eh of the following digrms e. 48 f ind, with resons, the size of the mrked ngles ns. p S Pulitions (NZ) Ltd, reephone , ISN

8 8 hievement Stndrd (Mthemtis nd Sttistis 1.6) S d. d In the tringle = nd = Prove tht = 30 3 e. 52 e Prove + = 180 Give resons for eh step of working. f. Q f P R ind the size of ngle. 3. Prove tht is prllel to y ompleting the proof elow = ( ) is prllel to sine = (lt s equl) 4. plin why every tringle must hve t lest two ute ngles.. Prove tht + + = 360 in the digrm elow. S Pulitions (NZ) Ltd, reephone , ISN

9 NSWRS erise : inding unknown ngles (pge 1) 1.. = 50 (vert opp s nd s on line). = 54 ( s on line) 2.. = 47 (vert opp s). = 28 ( s on line) = 180 (ngles on line); = = 360 (ngles t point); = 120 Tom s ngle is = 180 (ngles on line); = 36 Lid opened through djent ngles on stright line ( s on line) 7. = 135 ( s t point) erise : ngles nd prllel lines (pge 3) d. 103 e. 65 f. 64 g. 15 h = 108 (o-int s // lines) = 146 ( s t point). = 26 (lt s // lines) = 26 (lt s // lines). = 80 (vert opp s) = 100 (o-int s // lines) = 100 (orr s // lines) d. = 55 (lt s // lines) = 55 (lt s // lines) = 125 ( s on line) 3.. No (orr s not =). Yes (lt s =). Yes (orr s =) d. No (o-int s not =) 4.. = 58 ( s on line) = 122 is not // to G (o-int s // lines) (lt s not = sine = 122 nd = 120 ) Ground not horizontl. = 46 (orr s // lines) = 46 (lt s // lines) = 134 (o-int s // lines or s on line) 5.. Mrk y the ngle t point with y = 62 (o-int ngles // lines) = 298 (ngles t point). = (lt ngles // lines (twie)) = Mrk the ngle o-interior with 126 = 54 (o-int ngles // lines) = 54 (vert opp s). y = 180 (o-int s // lines) y = = 180 (o-int s // lines) = = = 180 (lt s // lines). = 90 (lt s // lines). = 90 ( sum Δ) = 90 = + = 90 G rw prllel line through (vert opp s) S Pulitions (NZ) Ltd, reephone , ISN

10 50 nswers NSWRS = = So = + + = 0 (lt s // lines) (lt s // lines) (djent ngles) (rerrnging) erise : Tringles (pge 7) d. 120 e. 22 f = 57 (vert opp s) = 49 (et Δ). = 52 ( sum Δ) = 52 = 52 (lt s // lines). = 45 (orr s // lines) = 90 ( s in isos Δ) d. = 20 (lt s // lines) d = 40 ( s in eq Δ) e. = 52 (lt s // lines) = 38 (sine = 90 ) = 110 ( sum Δ) e = 160 ( s t point) f. = 54 ( s in isos Δ) = 68 (se s isos Δ) f = 14 ( ) 3. = 64 (se s isos Δ) // ( = ) (lt s equl) 4. If there is only one ute ngle, then the other two ngles re otuse, so their sum is more thn = 180. ut the ngle sum of tringe is 180, so this is impossile. 5. = (se s isos Δ) = 2 (et Δ) = (vert opp s) = (se s isos Δ) = 180 ( sum Δ ) 6 = 180 = QPR = 180 ( s on line) PQR = = = 180 (vert opp s) (et Δ) (rerrnging) = + 50 (et Δ = sum int opp s) = 50 (sutrting from oth sides). Interior ngles re 180, 180, 180 ( s on line) = 180 ( sum Δ) 360 = 180 (rerrnging) + + = 360 (rerrnging). Interior ngles re 60, 360, 360 ( s t point) = 180 ( sum Δ) 780 = = 600 erise : Qudrilterls (pge 9) d = (symmetry) = 360 ( sum qud) = 124. = 38 (se s isos Δ) = 38 (lt s // lines) = 52 (se s isos Δ). Refle = 242 ( s t point) = 360 ( sum qud nd symmetry) = 37 d. = 112 (symmetry) d = 68 (o-int s // lines) 3. QPS = (vertilly opp s) PQR = 180 d SRQ = 180 ( s on line) ( s on line) QPS + PSR + SRQ + RQP = 360 ( sum qud) d = d = 0 + = + d 4. G = 180 (o-int s // lines) G = G = 180 G = 180 (o-int s // lines) (o-int s // lines) (o-int s // lines) S Pulitions (NZ) Ltd, reephone , ISN

11 INX djent (side) 35 djent ngles on stright line 1 lternte ngles 2 ngle of depression 44 ngle of elevtion 44 ngle sum of tringle 6 ngles t point 1 r 20 re 19 rrowhed 9 erings 41 hord 23 irle 20 irumferene 20 linometer 44 o-interior ngles 2 onyli points 26 orresponding ngles 2 osine 35 yli qudrilterl 26 equilterl tringle 6 eterior ngle 6 hegon 6 hypotenuse 29, 35 inverse trig funtions 37, 38 isoseles trpezium 9 isoseles tringle 6, 33 kite 9 prllel lines 2 prllelogrm 9 polygon 6, 11 Pythgors theorem 29 Pythgoren triple 31 qudrilterl 9 rdius (rdii) 23 retngle 9 regulr hegon 6 regulr polygon 6 rhomus 9 right-ngled tringle 6, 29 slene tringle 6 similr (polygons) 13 sine 35 SOH H TO 35 squre 9 sum of eterior ngles of polygon 11 sum of interior ngles of polygon 11 tngent (to irle) 23 tngent (trigonometry) 35 trnsversl 2 trpezium 9 tringle 6 trigonometri funtions 37 trigonometri rtios 35 vertilly opposite 1 volume 17 S Pulitions (NZ) Ltd, reephone , ISN

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