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HIGHER GRADES r(Lpar(Lj*(Fr8rJ:*~*H r"lf*WHEN YOU SEE THIS STAR *r"lf*PLEASE PRESS 'C' & ~@,"lf*TO MOVE FROM SCREEN TO SCREEN@j.Hr&"lf**tZ$rZ$p*C@D2j20Z$j*C@D.22Fr8V4r~lllrr~ll@lVT6"lf* Before studying r"lf* ALGEBRA T8"lf* it is essential to r"lf* understand several r"l f* aspects of 4:r~2lnll4v<"lf* + - ( ) r(`pr"l(`f*BASIC MATHS.rvX>&"lf**tZ$rZ$p*C@DBX0@Z$j*C@D>0(BF r8rJ:(DD"lf* DIRECTED NUMBERS rJ:DnF"lf* AS IS KNOWN 3 + 5 = 8 r"lf* SIMILARLY -3 + (-5) = -8 n8H"lf*AND -8 + (-12) = -20 8JL"l f* In short, if I lend you 10 and then J6N"lf* 6, I am now -16 6P"lf* NOW rTlxl2lr(\p2rV(\lxlrZR&"lf**tZ$rZ$p*C@DVZ2TZ$j*C@DR2VFr8X"lf*TO ADD TWO NUMBERS TOGETHER WHOSE SIGNSr"lf*DIFFERENT, SUBTRACT THE NUMERICALLYZ"lf*SMALLER FROM THE LARGER. NOW MAKE THEr"lf*SIGN OF THE RESULT THE SAME AS THATb\"l f*OF THE ORIGINAL LARGER NUMBER.r(Lpr(Lb<^"l f* e.g. (1) -12 +6 = -6<<`"l f* (2) 14 -5 = 9<Hb"lf* (3) -16 +10 -7 +3 +8 = -23 +21 = -2H^d&"lf**tZ$rZ$p*C@Dh^8fZ$j*C@Dd8(hFr8rJ:(Rj(\p@r(\l@lrRxl"lf* MULTIPLICATION AND DIVISION OF r"lf* DIRECTED NUMBERS FOLLOW xhn" lf* THE SIMPLE RULES r"lf* <---------- rJ:h^p"l f* (+) X (+) = (+) r"lf* (-) X (-) = (+) ^jr"lf* (+) X (-) = (-)r"lf* (-) X (+) = (-) rJjZt&" lf**tZ$rZ$p*C@DxZ8vZ$j*C@Dt8xFr8zz"lf*USING THESE RULES, THE REMOVAL OFr"lf*BRACKETS IS A SIMPLE PROCEDURE.zr|"lf*e.g. a) (+6) X (-3) = -18r"lf* b) (-9) X (-7) = +63rn~"l f* c) (+21) / (-7) = -3r"l f* d) (+16) / (+4) = +4n:"lf* ( / MEANS DIVISION ):`&"lf**tZ$rZ$p*C@D`2Z$j*C@D2Fr8bH rJ:r"lf*A L Gr"lf*E B R Ab`Tll<l<r(\pXrV(\llr`bTlll<r(\pxrV(\llrbTHr"lf*IS VERY DEPENDENT ON THIS CONCEPT OFTj"l f*DIRECTED NUMBERS SO....r"l f*HANDLE IT CORRECTLY !!j(p; r(prll(l(rrr~xlllhJ:r&"lf**tZ$rZ$p*C@Dh8Z$j*C@D84Fr8rr4<(pr(p2r(pFr( pZ<|(p,r(p2r(paxr( pZx|(l((l( r(p(vr(p(vr(p(vr( p( xrrrrVr~ldl@lrrllVFrllrJ:rH FF"lf* THERE IS NO MYSTERY ABOUT ALGEBRA F$J:rH$`&"lf**tZ$rZ$p*C@D`6Z$j*C@D6(Fr8r(B"lf*SIMPLIFYING 5x + (-6x) - (+2x)B<"lf*WE OBTAIN 5x - 6x -2x<B"lf* = 5x - 8x = -3xBD"lf*SIMILARLY (-4x) X 7y = -28xyDB"l f*WHILE (-28x) / 4 = -7xBB"l f*AND (+28x) / 4 = +7xBVH rJ: r"lf*NEXT WE MUST CONSIDER....VzHrJ:r&" lf**tZ$rZ$p*C@Dz8Z$j*C@D8(Fr8r(F~ll l rTlllFp(\pr(`pFvdd(\|fz(frV(\l(`l rp8Hr"lf* INDICES 8lHr&"lf**tZ$rZ$p*C@Dl8Z$j*C@D8Fr8D"lf*THE GENERAL RULES ARErHD0"lf*a X a = a 00"lf* 3 5 80."lf*a / a = a.0"lf* 9 6 30F"lf*When MULTIPLYING indices ADD them.FH"lf*When DIVIDING indices SUBTRACT them.HR"l f*a = 1r"l f*a = ---RL"l f* or"l f* -4 1L@"l f*ar"l f* 4@6"lf*AND (a ) = a68"lf* 3 4 128`&"lf**tZ$rZ$p*C@D`8Z$j*C@D8Fr8|(l<l(r<l((lFr(lFl<rl<(l|N(l2lrrl< lZNl>(lPr(lL lZrlZlZrlZ(lZ^8ll r4l.Hl.r0l6Hl6^Nr~Hl$lnlBr" lf* ? NTllxl_rV<lZlr(Lp r(Lrr(\p,rl(\ldr`T2l(lNlr(\p<xrV(\l(lr`N(\p<rV(\l(lrrNtH rJ: r"lf*NOW TRY THEr"lf*FOLLOWING EXAMPLEStzHrJ:r&" lf**tZ$rZ$p*C@Dz2Z$j*C@D2BFr8rJ:rHrBH"lf*TYPE THE VALUE OF THE INDICES ONLY !!H. "lf* 3 5 . &"lf*a x a tA$r(pdA$fr(ndA$f@DF(pz@"lf*CORRECTFF(jz@"lf*INDICE = 8F."lf* 4 -5.V&"lf*a / a tB$r(pdB$fV8(ndB$f@D8D(pz @"lf*CORRECTDT (jz @"lf*INDICE = (4)-(-5) = 9T(""l f* x(p$&"l f*If 3 = 81, what is the value of x tC$r(pdC$fp8&(ndC$f@D$8F&(pz@"lf*CORRECTF^((jz@"lf*ANSWER = 4 (3 X 3 X 3 X 3 = 81)^B*"lf*Remember that 4a PLUS 24a = 28aB`,&"lf**tZ$rZ$p*C@D0`8.Z$j*C@D,80Fr8B2~lFllZrTlAll_B`4(\pxr(`prV(\l(`lr`>6Tll"lrVll>F8rllrH rJ: F^<"lf* E Q U A T I O N r"lf*S O L V I N G ^v>HrJ:r&"lf**tZ$rZ$p*C@DBv4@Z$j*C@D>4BFr8`DHr"lf* W H E N r(p r(`TF(`p 4rTl(`l(`rrTHHr"lf* LETTERS OR NUMBERS ARE MOVED r"l f* FROM ONE SIDE OF AN EQUATION J" l f* TO THE OTHER rJ:r"lf* THE SIGN MUST BE CHANGED rJ:BL"lf*e.g. If x-12 = 0, then x = 12 B`N&"lf**tZ$rZ$p*C@DR`0PZ$j*C@DN0*RFr8rJ:*PT"lf*CONSIDER THE FOLLOWING EXAMPLESrJ:PjV"lf*3x - 7 = 2 => 3x = 9, therefore x = 3r(Lpr(LjJX"lf* 3 - x = 6 + 2x => -x - 2x = 6 - 3 JJZ"l f*SO -3x = 3, therefore -x = 1 or x = -1J*\(Lpr(L*6^"lf*6(x + 1) - 10 = -46J`"lf*So 6x + 6 = -4 + 10 => 6x = -6 -4 + 10J@b"lf*Therefore 6x = 0, thus x = 0@Zd&" lf**tZ$rZ$p*C@DhZ2fZ$j*C@Dd2LhFr8r(p r(r(LDj~lFllZrTlAll_Dl(`pr(\pxr(`p(`vr(\p(\xrV(\l(`lrrFnH rJ:r"lf* NEXT SOME FZp"lf* EQUATIONS FOR YOU r"lf* TO SOLVE ZvrHrJ:r&" lf**tZ$rZ$p*C@Dvv0tZ$j*C@Dr02vFr8rr2Hx"lf*TYPE SOLUTION FOR a IN THE FOLLOWINGHhz&"lf*4(a - 4)- 2(a + 7) = 0 a = tA$r(pdA$fh6|(ndA$f@Dz6@~(pz@"!lf*O.K@@(jz@"!lf*ANS=15@h&"lf*22 - 5x - (x + 10) = 0 a = tB$r(pdB$fh<(ndB$f@D<D(pz@"!lf*O.K.D>(jz@"!lf*ANS=2>L"l f*NOW, WHAT HAPPENS WHEN WE USE FRACTIONS L(pr(r(pr(Dp <rll(Dlll(rrr2Hr" lf* ? 2fHr&"lf**tZ$rZ$p*C@Df0Z$j*C@D0(Fr8r(p"lf*In simple equations,r"lf*CROSS MULTIPLICATION may be used.pl"lf*This may be compared tor"lf* a BISHOP'S MOVE in Chess.lz<lP2ldr<lPFldr2ld6lnr6ln2lsz^2lsFlsrFlsBlnrBlnFld^F2luFlurr<ldlFT#lFlUlxr(`pFr(\pUsrV(\l(`lr(`p(`vrH"l f*WHEN A Cr"l f*AHJ"l f* --- = ---r"lf*DJ0"l f* B D0L"lf*THENr"lf*A X D = C X BLX&"lf**tZ$rZ$p*C@DX0Z$j*C@D0Fr88"lf*CONSIDER THE FOLLOWING8*"lf* X 12*,"lf*--- = ---,*"lf* 3 4*b"lf*BY CROSS-MULTIPLICATIONr"lf*4X = 36 AND X=9b4"l f*SOLVE: 3X 12044"l f* -- = ---46"l f* 30 46X&"l f* VALUE OF X = tA$r(pdA$fX8(ndA$f@D8B(pz,@"lf*CORRECTBF(jz,@" lf*//ANS = 300//FX&"lf**tZ$rZ$p*C@DX2Z$j*C@D2Fr8\(Dp<r(phr(p r(\TlSl(Dl<l(lr(Dp(DxrrTVr(\pF~r(\l(\lrVP(\p@x~r(\l(\lrP\6 r, ll*MORE COMPLICATEDr,l<l*EQUATIONS\V&"lf**tZ$rZ$p*C@DV0Z$j*C@D0Fr84"lf*SOLVE THE EQUATION44" lf* 5x - 2x x + 442" lf*---- ---- = -----2f" lf* 6 7 2r"lf*AS THE COMMON DENOMINATOR IS 42fD"l f*BOTH SIDES OF THE EQUATION MUST BEDD"l f*MULTIPLIED BY THIS VALUE TO GIVE:D6" l f*35x - 12x = 21x + 8464" lf*=> 35x-12x-21x = 844:" lf*=> 2x = 84 => x=42:Z&"lf**tZ$rZ$p*C@DZ0Z$j*C@D0Fr84"lf*x + 1 x - 144"lf*----- - 2 = -----4d"lf* 7 4r"lf*HERE THE C.D. IS 28 - HENCEd6" l f*4x + 4 - 56 = 7x - 768"l f*=> 4x - 7x = -7 -4 +5686"l f*=> -3x = +4562"lf*=> X = -152V&"lf**tZ$rZ$p*C@D V2Z$j*C@D2 Fr8J H r" lf*SOLVE THE EQUATIONrHJ." lf*3x - 1 4.." lf*------ = ---.." l f*3x + 4 5.X&" l f*Value of x = tA$r(pdA$fX6(ndA$f@D6H(pz@" lf* C O R R E C THR(jz@"lf*The answer is > 6 and < 8RV&"lf**tZ$rZ$p*C@D"V0 Z$j*C@D04"Fr8rr4R$(pxr(pPvzdd(|<ffRz&(l(l(l(rr"lf* TRANSFORMATION OF r"lf* FORMULAE zV(&"lf**tZ$rZ$p*C@D,V0*Z$j*C@D(0&,Fr8r&,." lf* PRT,,0" lf*If I = ---,f2" lf* 100r"lf*Make R the subject of the formula.fZ4"lf*By cross-multiplicationr"l f*PRT = 100IZ.6" l f* => R = 100I..8" l f* ----.,:" lf* PT,H<(\pr"(\lf*~~~>rHV>&" lf**tZ$rZ$p*C@DBV0@Z$j*C@D>0BFr8TDJ: r"lf*CONSIDER THE FOLLOWINGrJ:TJF"lf*If ax + b = cx + d, make x the subject. J4H"lf*=> ax - cx = d - b46J"lf*=> x(a - c) = d - b64L"l f* d - b44N"l f*=> x = -----44P"l f* a - c4~R"lf*WHEN MOVING LETTERS/NUMBERS FROM ONEr"lf*SIDE OF AN EQUATION TO THE OTHER,~>T"lf*REMEMBER TO CHANGE THE SIGN.>VV&" lf**tZ$rZ$p*C@DZV0XZ$j*C@DV0^ZFr8rr~ll@lrTll@l^D\(`pxrVl(`l rDp^r~#l2lldrH r"lf* F A C T O R I S A T I O N p `Hr Xb&"lf**tZ$rZ$p*C@DfX0dZ$j*C@Db0fFr8hh"lf*FIRST OF ALL, CONSIDER THE REMOVALr" lf*OF BRACKETS.hDj"lf*(x + 5)(x + 4) = x + 5x + 4x + 20D6l"lf* 2 6>n"l f* = x + 9x + 20>4p"l f* 24Dr"lf*(x - 6)(x - 2) = x - 6x - 2x + 12D4t"l f* 24>v"lf* = x - 8x + 12>4x"lf* 24Vz&"lf**tZ$rZ$p*C@D~V2|Z$j*C@Dz2:~Fr8rr l:Xl(l rTl l#l-r(\pXN(`p vd(\| fzrV(\l(`lrNH"l f* FACTORISING IS THE EXACT OPPOSITE OF HF"lf* THIS PROCEDURE AND IS THE BASIS OF F6"lf* QUADRATIC EQUATIONS 6X&" lf**tZ$rZ$p*C@DX6Z$j*C@D6Fr8,"lf*S O L V E: ,0"lf*x - 6x + 8 = 00r"lf* 2rr(`prl(`@l(`rr2"lf* Brackets first 2z"lf*( )( )r(Lp'r(Lr"lf* Now the x values zP"lf*(x )(x )r(Lp'r(LP4"lf*Next the multiples 4`"l f* of 8 which when r"l f* MULTIPLIED together `v"lf* give 8 & when r"lf*ADDED give - 6 r(Lp'r(LvP"lf* i.e. 4 and 2 r(Lp'r(LPP"lf*(x 4)(x 2)r(Lp'r(LPN"l f*Now the signsr(Lp'r(LNP"lf*(x - 4)(x - 2)r(Lp'r(LP."l f*SO x - 4 = 0.."l f*OR x - 2 = 0.,"lf*i.e. x = 4,*"lf* or x = 2*Z&" lf**tZ$rZ$p*C@DZ0Z$j*C@D0Fr8NJ: r"lf*S I M I L A R L YrJ:N&"lf* 2&B"lf*x - 2x - 35 => (x - 7)(x + 5)B$"lf* 2$D"l f*x + 8x - 48 => (x + 12)(x - 4)D$"l f* 2$D"l f*8x - 10x - 25 => (2x - 5)(4x + 5)DH"lf*CHECK THE LATTER BY REMOVAL OF BRACKETSH6"lf*=> 8x +10x -20x -256("lf* 2(X&"lf**tZ$rZ$p*C@DX0Z$j*C@D0Fr8TJ:r"lf*THE SUM OF TWO SQUARES IS RECOGNISEDTH"lf*BY THE GENERAL EQUATION :rJ:H<"lf*A - B = (A - B)(A + B)<J"lf* 2 2r(Lppr(LJD"l f*e.g. 64x - 16 = (8x - 4)(8x + 4)D*"l f* 2*H"lf*IF THE SIGN IS +VE THE ONLY SOLUTION ISH>"lf* 64x + 16 = 8(8x + 2)>8"lf* 2 28V&" lf**tZ$rZ$p*C@DV0Z$j*C@D04Fr8rr4x(p<r(phr(pd(|fz<r(l(lrxXH r"lf*SIMULTANEOUSr"l f*EQUATIONSXbHr&" lf**tZ$rZ$p*C@Db0Z$j*C@D0Fr8F"lf*SIMULTANEOUS EQUATIONS MAY BE SOLVEDFv"lf*IN THE FOLLOWING MANNER:rr lr~l(l@lv0"lf*7x + 4y = 2 (1)00"lf*3x - y = 9 (2)0z"lf*EQUALISE THE +VEr"lf*SIGNS & SUBTRACT.r(Lppr(Lzx"l f*MULTIPLY (1) BY 3r"l f* AND (2) BY 7 r(Lppr(Lx2"l f*21x + 12y = 62| "l f*21x - 7y = 63r(Lppr(Lr"lf*IN SUBTRACTION THERE|V "lf*IS A HIDDEN -VE SIGNr(Lppr(LVX"lf*(CHANGE BOTTOM SIGNr"lf* AND ADD !) X*ll~ll*0"lf* 19y = -570P"lf* y = -3r(Lppr(LP\"lf*BY SUBSTITUTION (2)r"lf*3x+3 = 9 SO x = 2\bDr&""lf**tZ$rZ$p*C@Db0Z$j*C@D0(Fr8rD(bJ: r"lf*EXAMPLE TAKEN FROM A GRADE A/B PAPERrJ:b. "lf*5x + 3y = 27..""lf*2x + y = 10.T$&"lf*Value of x tA$r(pdA$fT6&(ndA$f@D$6F((pz@" lf*O.K.FD*(jz@" lf*NO..DT,&"lf*Value of y tB$r(pdB$fT6.(ndB$f@D,6>0(pz@" lf*O.K.>@2(jz@" lf*NO..@4&"l f*DO YOU WISH TO TRY THIS AGAIN (Y/N) tA$rA$p*Y@DN6A$j*Y@" lf*//x = 3: y = 4//NV8&"lf**tZ$rZ$p*C@D<V0:Z$j*C@D80<Fr8B>"lf*It should be remembered that theBB@"lf*answer need not necessarily be aB>B"lf*a whole number. Fractions and>FD"lf*decimal fractions are also possible.FFF"l f*CURRENTLY MORE CONSIDERATION IS GIVENFBH"l f*TO ALGEBRAIC QUESTIONS REQUIRINGB0J" lf*T H O U G H T.06L"lf*For example ========>6XN&"lf**tZ$rZ$p*C@DRX0PZ$j*C@DN0RFr8@T"lf*Entrance fees to an exhibition@@V"lf*totalled 6.8O for 3 adults and@@X"lf*2 children; for 2 adults and 5@<Z"lf*children the fee was 8.20.<j\"l f*WHAT IS THE COST FOR AN ADULT & A CHILD?r(Lp@r(Lj@^"l f*Set up a simultaneous equation:@0`" l f*3x + 2y = 6.8000b" lf*2x + 5y = 8.200Hd"lf*Equalise, subtract and find x/y valuesH8f"lf*S O L U T I O N ====>8Xj&" lf**tZ$rZ$p*C@DnX0lZ$j*C@Dj0nFr8:p" lf*3x + 2y = 6.80 (X 2)::r" lf*2x + 5y = 8.20 (x 3):Vt"lf*=> 6x + 4y = 13.60r(Lp@r(LVRv" lf*6x + 15y = 24.60r"l f*SubtractR0x" l f* -11y = -110:z"l f* So -y = -1 i.e. y = 1:>"l f*By substitution, x = 1.60>b"lf*ADULT TICKET = 1.60r"lf*CHILD'S TICKET = 1bV&" lf**tZ$rZ$p*C@DV0Z$j*C@D0Fr8&(DpP&2ldl(Dllr2~(pr(r(DpPxrldl(Dllrr~^J:rr"lf* DO NOT LOOK FOR HIDDEN DIFFICULTIES^@"lf* IN THE G.C.S.E. EXAMINATION. @H"lf* IN THE PIONEER PAPERS, THE QUESTIONS HH"lf* WERE SIGNIFICANTLY EASIER THAN IN THEHD"l f* ORIGINAL 'O' LEVEL EXAMINATIONS. D@"lf* THE ALGEBRAIC DATA GIVEN HERE @F"lf* IS ADEQUATE FOR GRADES A, B AND C. FdJ:r&"lf**tZ$rZ$p*C@Dd0Z$j*C@D0Fr8F"lf*THIS CONCLUDES PART 1 OF THE G.C.S.E.FX" lf*MATHEMATICS CURRICULUM.r(Lppr(LXD"lf*GEOMETRY AND TRIGONOMETRY ARE ALSOD`"lf* AVAILABLE IN THIS SERIES.r(Lppr(L`n"l f*THANK YOU FOR YOUR INTERESTr"lf*(E. MILLS, COPYRIGHT 1989)nV&"lf**tZ$rZ$p*C@DV0Z$j*C@D0Fr8B( p @ r( l/l( B`@l( @x( l@rl@x( l@x( `0( llr0:"l f* WOULD YOU LIKE A RE-RUN :t&"l f* OF THIS PROGRAM (Y/N) tA$rA$p*Y@Dt*A$p*N@*  JSET UP SCREEN$'lll$$'lll$"'r8r4"'                             ,PROG1.BSCPROG2.BSCPROG3.BSCPROG4.BSCPROG5.BSCPROG6.BSCPROG7.BSCPROG8.BSCPROG9.BSCPROG10.BSC<<<<<<<<<<A:\*.BSCab Hz/:NuH z0(2(@g4"P&1::g*g2g:ftgl`2g\`NgL`g<`.E\@f#`@Y@j0(Y@1@`,E0Y@j0(Y@XBhmBBhf1@LNu,X.6H*2( 6^VjTzVb&$N Hf($*  n| 60X2 `\V"8"&Z,NNN  $ L& " V ,>v" 2^nP ZV 4n  $ rT*0 0D>L  " >: RR8N".$B.d Dbn^jX2 8*  v2`ll|:D.D&|"(8BBBP&&LHJHNLHJH$., THJH*&.ZR PFHH@ (0\HxT0&TDL R$6.* $`&` ݠ 4MJy f2*O*m-M - ЭЭ// Bg?<JNAO JkO a0a.A=X=P=| N6Jf`BnBnBNaa p N1JngAV-H=n0<#N1a>"< <NJg-H<-H@aNN/N/a/N.zLNuH0<?N.zLNuB@=@AD-HpNN1Nu0<`0<`H=n =|A-Hp!N1LNuA$ k"X"`Nu=|BnphN1LHH=|p3N1Nu=|A-HpN1NuJngA-HpN1 n(fpaa0. |g |ga>Jnf`&AL @!@!@C!I!I-I-I"Bh Bh"=h$B,B0B:BNu n,n^A@Rh0<(AQAh AhA@ShAPUhAPhQhAThZh*PhBNuH2IA!2I naL?NuH=@R2IA-p na,L?NuH0.RA pa0.RaLNuI8.L(t@0,$AaNu?:MGNI&K&PPJ4Pg RR`TTR| m.BDCJQgBQ:MG7PRDTT| m8.RL0.Ra0.RAH1NuBngPa=@HA1aJf -nv`H0.aaJf6 nv .r".za28.LGH/3@?<INA\L'v@0.2IK;aa@nRf-nv0.Ra 0.a a .BnNung` Nua>N^JgH?<NNT/?NAXJzfBra .r".z nvNua$Jg?AQ-H=|p4N10NuH@P nv"CXf"(fBLNu=@B.Ad-HCB<.:<PQA-HpZN1JngaJf&=|Aj-Hp4N1 nf`t0.a a8Jfb nv .r".za8.LGH/3@?<INA\L'v@0.2IK;-nv nv0.Ra0.0.Nu=n =|phN1Lnbf=|phN1LHH=|/=|plN1LHLH=n =|piN1NuLaB8.HĈDl4B8.nLDl68Bno0.B8Cnno 2.CnNuaL aaBn=|-LHplN1-n-n=n=|piN1LH$H$fNrL$NJ`,Ha(?A?C26/:/>/ a>?CLNuAnEG0C@n2E6@n6NuvNu=n=| piN1Nuaa=n=| phN1JngJngLL aZJBgdJCg^JgNH@0HB4Ԁ-@ -B -@ -B A-H A-H =| 0<1N.z` BDN0=n=| `ba`H=|=|)HplN1-n-nnnLNuBn=|phN1-n-nLB(C KCJBP@CTAUACCB@axNu=|--n-npdN1=nNunnBn=|--n-nplN1-n-n=npeN1Nu=npfN1=npgN1NuA-H=n=|piN1Nu Output Id?<NATA:Bg/ ?<GNAPJfS\*.BSCBNuAdC0<BAA|mRRQNuLBDN0IAN"<SAf"O /NuNPJ  !z!n!n"1n$NuJng"a$p=@RaAL-H-Ha`J/<?<HNA\Jg-@-@"-A" @@2<CSAf-H @aj></<?<HNA\JgJE%p-@"-A" @@2<CHSAf-H @aXG|(mB@Nu<NuJf.-n-n-n B/<?<HNA\-@-@Nu nJg2< BAfNuI8< H/?<INA\LSDfNuJg/.?<INA\NuH~T.,/?<HNA\JfnB` @ L~NuH?<NNT-@?<NNT|f=|=|=| `2|f=|=|=| `=|@=|=| 0.H=@LNuHA-H -H =| JDjD`D@ASBSCJDjH DDH `H DDH 0<1N.zLNu  02 46@ N f ~   {  "$&,. &%x7'6&H(8*G,.467YIXH@jXZZi\^dfY { k z   j p |   { &Z-H\JgB@BABBJg6 [fRAR@ ]fSA |fRB|g |fPNu|nJAfJ@g|nJf/< ?<HNA\JfpNu-@ n0<r:< n^n:<1A!|1|1EBB BBhRAQ nC0|g 21`0<1|. nn 0< 1| .1n1n,1n1n!n2<F4<1Q"n\ n-H"n\$n [g JfPNuRJfPNu ]g`B%nD-H$n`&nBG [g Jf`` nR ]g$ \g |g`B%n `B<F '`G-H ]f`AR"n#H\A[#HT"n80<a><0<?a>Q"nH0<$i gBAJgPA`3AQ"n$IHB@2<i3@5@Q nh1@D1hL41h$ 1h1hT<1h1hl1h  1h 0<"n a><"n0<?a>Q"n0)bh24<@3AQ"nH0<J f Q n4<2.A0(h@l 6h1CQ nh0<BhQ-n=|pN1B@Nu$I8< n^l8<BEJ g5EDQ3ENuBB$I2BC&i gBDJgPD`Dn6Q5CQNu-------------------- Desk Jg/.?<INA\A-HBnpN1B@Nu` 0 00}    0 @ P`p8ppp p0p@pPp`ppp}T ,(<22 2 F m@ d z    $x  $<Z< n00P<%P< D @ @(@8@X@hd@xxd@xd @xd @ @h `  "" Desk File About Runner-------------------- Load Program Run Program-------------- Show Info Dump Output QuitFast BasicRun-Time InterpreterComputer ConceptsProgram execution stoppeddue to:OKProgram InformationCANCELRUNNERFast ST BASICRun-time InterpreterRUNNER bits: Phil MartinInterpreter: Jeremy RustonGavin TheobaldPhil Martinand Nigel, who doesn't like to be mentioned!Version Number: 1.05c/00 Computer Concepts 1986Gaddesden Place, Hemel Hempstead,Herts. 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G E O M E T R Y Xj" lf* HIGHER GRADES r"lf* BY EVELYN MILLS, COPYRIGHT 1989 jL (Lpar(LrHrJ:LtFr8rrX1p/rX2p/xX1tVY1pdrY2pdxY1VPX1lY1X2lY2rrP0" l f*HELLO THERE08" lf*When you see this *8H "lf*Please press C & RETURN to continueHb"&"lf**tZ$rZ$p*C@D&b8$Z$j*C@D8(&Fr8r(R(( p / r( l/l( R`*/l( /x( l/r/xl/l/x( `8,l/x( lr82." l f*PLANAR SHAPES2b0&"l f**tZ$rZ$p*C@D4b82Z$j*C@D&8(4Fr8r (6~ll(l(rTll(l(r(\p--rV(\llrx8r~ l l2l2rT l l2l2r(\p7 -x(:V(\l lr(x<r~llFlFrTllFlFr(\pK-x4>V(\llrrD4<@H r"lf*THE SQUARE<JBrlrxlxlxJnDHr&"lf**tZ$rZ$p*C@DHn8FZ$j*C@D48RHFr8rrr~lldldRBJ"lf*Lr"l f*BB<L"lf*IN A SQUARE, BOTH SIDES<>N"lf* ARE EQUAL AND PARALLEL.>2P"l f*AREA = L X B2HR"lf*e.g. If L of this field = 12 metresH:T"lf*B is also = 12 metres:DV"lf*AREA = 12 X 12 = 144 SQ.METRESDbZ&"lf**tZ$rZ$p*C@D^b8\Z$j*C@DH8F^F r8rDrrlF>`(`pdr(\p x>.b(\l(`l(`l(\r.jdTlllrVllr&"lf**tZ$j8eZ$p*C@Dh88fZ$j*C@D^8(hFr8r((j~dl lln(Jl"lf*IN A SQUARE, EACH ANGLE IS 90 DEGREESJdn" lf*90r"lf*90r" l f*90dp"l f*90r&"lf**tZ$rZ$p*C@D|8rZ$j*C@Dh8(|Fr8r(4~r l l,lx4(`pr(\pr"(\l(`f*THE RECTANGLEr(`p(`vr>(Lprr<l(>v~ l l<l(rT l l<l(rVPl lrVl lv"Vl l"b&" l f**tZ$rZ$p*C@Db8Z$j*C@D|84Fr8rDr4^6r,ldl*RECTANGLESr,ldl*RECTANGLES^4~llldr64<"lf*ARE SIMILAR TO SQUARES<b"lf*LENGTH & BREADTH AREr" l f*NOT THE SAME.b@"lf*AREA = L X B AS IN A SQUARE@d&"lf**tZ$rZ$p*C@Dd8Z$j*C@D84Fr8rr4X(\p r(`p xX4(\l(8l(8l(`rr4d"lf*CALCULATE THE AREASr"l f*OF THE FOLLOWINGd"l f*RECTANGLESr&"lf**tZ$rZ$p*C@D8Z$j*C@D8Fr88"lf*AREAS OF RECTANGLES8D"lf*(VALUES SHOULD NOT EXCEED 500)D&"lf*LENGTH tA$r(,pdA$fr(,ndA$f@D6(,t@D6&"lf*BREADTH tC$r(pdC$fr(ndC$f@D6(t@D6(p(,z(J"l f*THE AREA IS t(t* SQUARE UNITSJr&"lf*ANOTHER TRY (Y/N)tZ$rZ$p*Y@Dr8Z$j*N@D8DF r8rrrlDp<llrllrllp,l<l,h" lf* A PARALLELOGRAM IS Ar" lf* SKEWED RECTANGLE !hf"lf*OPPOSITE SIDES ARE EQUALr" l f*AND PARALLELf"l f*AREA = BASE*PERPINDICULARr" lf*HEIGHTr(Lpr(Lb&"lf**tZ$rZ$p*C@Db8Z$j*C@D8(Fr8r(X(pd/r(pdxX`(l( l(r(p(vr(p(xrr`nHr&"(lf**tZ$rZ$p*C@Dn8Z$j*C@D84Fr8rDr4p(llrlxl<rxl<l<pzl<(lr(l(l<rHr" lf*30 cmz("lf*2cm(FH r"lf*AREA = 60 CM.SQUAREDF@Tlll<rVldl@nHr&"lf**tZ$rZ$p*C@Dn8Z$j*C@D84Fr8rDr4V(p/2r(p V8(l((lrr8N(pr(llrNlJ:r(Tpr"l(Tf*THE TRAPEZIUMrlb&" lf**tZ$rZ$p*C@Db8Z$j*C@D84Fr8rr4\J:rxllrllP\flP<lPr<lPxlrl(lfr"lf*7cm (a)r"l f*11 cm (b)rxlxlPrh"lf*8cm (h)r"l f* AREA = 1/2 X (11 X 7) X 8 CM.SQ.h8"lf* = 72 CM.SQ.8B"lf*a & b are opposite & parallelB>"lf*h is the VERTICAL height>b&" lf**tZ$rZ$p*C@D b8Z$j*C@D8( Fr8r(V (\p,r(`pV\(\l(`l(`lr(\p(\vr(`p(`vrr\^rllr"lf*IN THE FOLLOWING AND^Z"lf*IN SUBSEQUENTr" lf*CALCULATIONSZV" l f*TRY TO FINDr"l f*THE RESULTVT"lf*BEFORE PRESSINGr"lf*RETURNTb&"lf**tZ$rZ$p*C@Db8Z$j*C@D 8(Fr8rD(v"lf*AREAS OF THE TRAPEZIUMr"lf*(Values should not exceed 500)v &"lf*SIDE A tA$r(,pdA$fr(,ndA$f@D6 (,t@D6$&"lf*SIDE B tB$r(0pdB$fr(0ndB$f@D6&(0t@D6(&"lf*HEIGHT tC$r(pdC$fr(ndC$f@D4*(t@D42,(pdd(,v(0fz(f|2L."l f*THE AREA IS t(t* UNITS SQUAREDLr0&"lf*ANOTHER TRY (Y/N)tZ$rZ$p*Y@Dr82Z$j*N@D48(4Fr8r(p6l l(rl(l(rl(l p.8rll.p:l,l,rl,lJrlJl,p.<rl<l.p>jl,lrllHrlHjl,p0@rxl,l0pBll,rl,lJrlJlpzDrl,lrTdlllPrVlPlrVlPlz\F&"lftZ$rZ$p*C@DJ\:HZ$j*C@D4:(JFr8rD(BLH rJ: r"lf*NOWB$NHrJ:$pPdl<lxrlxldrlddl<p^R" lf*Ar"l f*Cr"lf*B^6T"lf*The AREA of this6ZV"lf*triangle can ber"lf*calculatedZfX"lf*given two sidesr"l f*and the enclosed angle.f0Z(Lpr(L0<\" lf*AREA = AB X AC X Sin A<<^" lf*-----------------------<0`" lf* 20>b"lf*Area calculations follow>bd&" lf**tZ$rZ$p*C@Dhb8fZ$j*C@DJ8(hFr8rD(Bj"lf*VALUES SHOULD NOT EXCEED 500BFl"lf*Angle input should not exceed 179Fn&"lf*ENTER LENGTH OF SIDE AtA$r(pdA$fr(ndA$f@Dh2p(t@Dh2r&"lf*ENTER LENGTH OF SIDE BtB$r(pdB$fr(ndB$f@Dh2t(t@Dh2v&"l f*ENTER ENCLOSED ANGLEtE$r(pdE$fr(ndE$f@Dh2x(t@Dh2z(<pہ0|SIpd(z(<|f04~(Tpd(z(zSIf|4L"l f*THE AREA IS t(Tt* UNITS SQUAREDLr&"lf*ANOTHER TRY (Y/N)tZ$rZ$p*Y@Dhr8Z$j*N@D8(Fr8rD(rldlPrdlPlPrlPlrlllPr"lf*8cmr"l f* 16 cmlD"l f*The standard area equation is:DB"lf* AREA = BASE X VERTICAL HEIGHTBB"lf* ----------------------B8"lf* 286"lf* = 128 cm.sq.6b&" lf**tZ$rZ$p*C@Db8Z$j*C@D8Fr8:6r, ll*PYTHAGORAS:h6r, ll*PYTHAGORASrr lFZlh8 lG[lr8XTl<lbldrVZl lrVl<lXHJ:r"lf*THEOREMrJ:Hb&"lf**tZ$rZ$p*C@Db8Z$j*N@D8Fr8TJ: r" lf*PYTHAGORAS THEOREMrJ:TF"lf*THEOREM: In a rt.angled triangleFD"lf*the square on the HYPOTENUSE =D>"lf*the sum of the squares on>:"l f*the other two sides.:pldlrllrlldp^"l f*Ar"lf*Cr"lf*B^2"l f*AC = AB + BC22"l f* 2 2 22b&" lf**tZ$rZ$p*C@Db8Z$j*C@D84Fr8rDr4n l  lPr lPnlPrnlP l ntllr"l f**r" lf*The Hypotenuse is oppositetl" lf*the right angle (*)r" lf*and is the longest side.l&"l f*Length of hypotenusetA$r(pdA$fr(ndA$f@D2(n@D2&"lf*Length of a sidetB$r(pdB$fr(ndB$f@D4(n@D42(t(@D20(Hpd(~x(~f0H"lf*3rd. side is t(Ht* units longHt&"lf* ANOTHER TRY (Y/N)tZ$rZ$p*Y@Dt8Z$j*N@D8(Fr8rD(X(\p d r(\l(\lrX(p br(pr(r(ll(rrJ:n"lf*IN THE PREVIOUS EXAMPLEr"lf*DID YOU NOTICE THAT THEn`"lf*CALCULATION WAS FORr"lf*A SIDE OTHER THAN`d"l f*THE HYPOTENUSE ??rJ:r(Lprd0"l f*H = a + b02"l f* 2 2 220"lf*a = H - b0`"l f* 2 2 2r"lf*Remember to take the`<" lf*SQUARE ROOT of the answer<`&""lf**tZ$rZ$p*C@D`8Z$j*C@D8Fr8(pr(p r"(l(f* OF GREATER IMPORTANCEr(p(vrB"lf*IS -------------------------->B`&"lf**tZ$rZ$p*C@D `8Z$j*C@D8 Fr8p dl lPrlPdlPrdlPdl pzr2l<lr" lf*Ar" l f*Br"l f*Cz"lf*The SINEr"lf*COSINE andr"lf*TAN Rulesr(Lpr(LD"l f*SIN angle C = OPPOSITE side/HYPOTDD"l f*COS angle C = ADJACENT side/HYPOTDH"lf*TAN angle C = OPPOSITE side/ADJACENTH4r~ll,l4pllrllrllplTll(l(r(\p#(rV(\llrl`&" lf**tZ$rZ$p*C@D"`8 Z$j*C@D 8("Fr8r($(\p,r(`pxr(\p(\vr(`p(`xr(\l(`l(`rr&r(\p,xr(`pr(\p(\xr(`p(`vr(\l(`l(`rr(Hr"$lf*EXAMPLE:r&".lf**tZ$rZ$p*C@D,8*Z$j*C@D"8(,Fr8rH(p.l dlPrdlPlPrlPl pX0"lf*Ar"l f*Br" l f*CXD2"l f*30r"lf*8 cm.DB4"l f*TO FIND AB: Sin 30 deg. = AB/8BH6"lf* AB = 8 X sin 30HH8"lf* = 8 X 0.5 cmHB:"lf* = 4 cm.B`<&"lf**tZ$rZ$p*C@D@`8>Z$j*C@D,8F@Fr8rr~ll(l(F<BHr"lf*SOrH<rD"lf*If a boat is 300m from ther"lf*bottom of a vertical cliff,rpF"lf*& the angle of elevation ofr"lf*the top of the cliff fromp<H"lf*the boat is 31 degrees -<BJ"lf*Would you use 1. The sine ruleBDL"lf* 2. The cosine ruleDzN"l f* or 3. The tan ruler"l f*to find the height of the cliff ?zNP&"l f*Type 1,2 or 3 to select answertA$NXR(4pdA$fr(4ndA$f@D@XVT(4pz@"lf*GOOD - CORRECT ANSWERVZV(4jz@"lf*Sorry - Tan rule requiredZ`X&"lf**tZ$rZ$p*C@D\`8ZZ$j*C@D@8\Fr8p^,l,lr,l"lr"l"lp`"l,lrr&llr"llrl"ldb rldlr"lf*31r"#l f*hdLd"lf*300 mr"lf*EXPLANTIONL6f"lf*h/300 = tan 31 deg.68h"lf*so h = 300 x tan 3186j"l f* = 300 X 0.6016l"l f* ~ 180 mr&"lf**tZ$rZ$p*C@Dp8nZ$j*N@D\8pF r8r(<pہrR1p(rR2p#r( pHrANp( z z(<|r(\pR2zdANfr(`pR2zdANfr(\vl(`vPlR1llrltHr&"(lf**tZ$rZ$p*C@Dxl8vZ$j*C@Dp84xFr8rHr4jz( pPxr̀v( l( lffz( llrjR~Hr" lf*THEr" lf*CIRCLERxHrr&"lf**tZ$rZ$p*C@Dx8Z$j*C@Dx8Fr8b(\p(2r(pr(b.(\l(lrr.R"l f*AREA = PI X (RADIUS)r"l f*2R"l f*CIRCUMFERENCE = 2 X PI X RADIUSr&"lf**tZ$rZ$p*C@D8Z$j*C@D8Fr8NJ:r"lf*AT THIS POINT, 3 DIMENSIONALNz"lf*FIGURES ALLOW THE USE OF MOREr"lf*ADVANCED GEOMETRICAL CALCULATIONSzX(`prl(`6l(`rXX(`pdrl(`6l(`rX`&"l f**tZ$rZ$p*C@D`8Z$j*C@D86Fr8rJ:r6pdllrllZrlZlpplnlrnl<lZr<lZdlppnlnlxrnlxlxrlxlppnlx<lZrnlxl(rl(lxpplxlZrl(dlrl(lpZrnllrllZplnlBrnlBdlrnlB<lZpNnlBlBrlBlNNlBlrlBlZN\&" lftZ$rZ$p*C@D\8Z$j*C@D8(Fr8r(p lFl rFl l rl ZlpZl lr l lFr lFZlFrZlFZlNZlFl2rl2l Nb"lf* THE CUBOIDr" lf*hr"l f*bbZ"lf*lr"l f*VOUME = l x b x h units cubedZ&"l f*LengthtA$r(,pdA$fr(,ndA$f@D&"lf*BreadthtB$r(pdB$fr(ndB$f@D&"lf*HeighttC$r(pdC$fr(ndC$f@D`(Tp(,z(z(r"lf*Volume is t(Tt* units cubed``&" lf**tZ$rZ$p*C@D`8Z$j*C@D8Fr8(p,2r(px r(l((lrrD"l f*THEr"l f*PRISMD`&" lf**tZ$rZ$p*C@D`8Z$j*C@D8Fr8p2l lZr lZZlZrZlZ2lp@r lr2lFl@p2ll rl lDrlDZlZp@r lrxl(l@XJ:r"lf*THIS IS Ar"lf*PRISMXNlxxlrxlxlNpxllrllrllxpplxldrld,lxr,lxlp,lx,lr,llrr l rllvr lrllr lr rllv0"lf*SO IS THIS ->0`&"lf**tZ$rZ$p*C@D`8Z$j*C@D8*Fr8rJ:*pl l2r l2 lFr lFlZpplZ2lZr2lZFlFrFlFFl2ppFl22lr2llrll ppl l rl lrll2pl2lFrlF2lZr2ll rFl2lrFlFl2Z rl lrll2Zpl2lFrlFlFr>l*lp8Fl<l2r88" lf*This is also a PRISM8`&"lf**tZ$rZ$p*C@D `8 Z$j*C@D8 Fr8pl l2r l2 lFr lFlZpplZ2lZr2lZFlFrFlFFl2ppFl22lr2llrll ppl l rl lrll2pl2lFrlF2lZr2ll rFl2l,FlFl2,B"l f*The VOLUME of this unsharpenedBF"lf*pencil is found by multiplying theFT" lf*AREA by LENGTH (L)r"lf*AREATz " lf*Lr&"lf**tZ$rZ$p*C@D$z8"Z$j*C@D 8($F r8r(V&(`pr(\p V`((\l(`(`l(\rrrH r"lf*NOW`^*Hr"lf*Calculater"l f*the following^`,&" lf**tZ$rZ$p*C@D0`8.Z$j*C@D$8(0Fr8r(N2l<6l<rld6ldNp4xldl(rdl(dl<rdl<l<pp6l<l(rl(xlrxllpp8ll(rl(l<rl(l(pX:r l rxl2lrl2lXR< r l rllr lRp>l lr ll(rl(l$p@l$2l2r2l2<l0r<l0Hl<rl2l"Br "pDl<l4rl4l1rl1l(pFl(?l$rl2lr rllrrll @H"lf*Breadth of house = 2.1 units@6J"lf*Length = 2.8 units6:L"l f*Height of wall = 1 unit:N"l f*Total height = 2.4 unitsr"l f*Find the volume of the house.rr lFP"l f*Is the volume 1) 8.250 units cubedFFR"lf* or 2) 9.996 units cubedFT&"lf*Type 1 or 2 to select answertA$r(4pdA$fr(4ndA$f@D0VV(4pz@"lf*GOOD - CORRECT ANSWERVXX(4jz@"lf*SORRY - WRONG CHOICE !X2Z&" lf**tZ$26rZ$p*C@D^68\Z$j*C@D08^Fr8p`2l lrllFrlFPlFp\bPlFPlrPl2l rJ:\fd" lf*TO CALCULATE THE VOLUMEr" lf* OF THE HOUSE SHOWNfjf" lf*ON THE PREVIOUS SCREEN,r" l f*IT IS NECCESARY TO FINDj|h"l f*THE AREA OF THE END SECTION.r"l f*1. FIND AREA OF RECTAGLE AS SHOWN.|xjlPlr r2l2lr(Lpr(Lxdl"lf*2. NOW FIND AREA OF TRIANGLE.rr2lldn"lf*ADD BOTH AREAS & MULTIPLY BY LENGTH.r&"lf**tZ$rZ$p*C@Dr8pZ$j*C@D^8*rFr8rJ:*jt(\p  r(`pr(`p(`xjhv(\ll(`rrr" lf*AND NOW TO THE PYRAMIDh`x&"lf**tZ$rZ$p*C@D|`8zZ$j*C@Dr8|Fr8p~l PlrPllrllPpplPl rl lnrl lppPl}lPr}lPlPr}lPl p`"lf*Hrr}llrlZl`^"lf*Volume of pyramidr"lf*= 1/3(base area)^0"lf*X Height (H)0`&"l f**tZ$rZ$p*C@D`8Z$j*C@D|8(Fr8r (X(\p r(`p xX|(pr(r(\l(`lr(`p(`xr(\p(\v|rrlHr&"*lf**tZ$rZ$p*C@Dl8Z$j*C@D8(Fr8rH(N" lf*THE CYLINDERrr lN~dlPlrl<lrdl>l*rdlblN~r"lf*lrdlPxlPr lr" l f*rrh"lf*For a cylinder of radius 'r'r" lf*and length 'l'h8" lf*VOLUME = pi x r x l84" lf* 24`&" lf**tZ$rZ$p*C@D`8Z$j*C@D8Fr8PJ:r" lf*CYLINDER VOLUMESrJ:PB"lf*(Values should not exceed 500)B&"lf*Enter radius of basetA$r(pdA$fr(ndA$f@D2(t@D2&"lf*Enter length of cylindertB$r(pdB$fr(ndB$f@D2(t@D2(<pہ.(Tpd(<z(z(z(f.J"l f*The VOLUME is t(Tt* units cubedJp&"lf*Another try (Y/N)tR$rR$p*Y@Dp8R$j*R@D8Fr8z(`p r(Dp 2rl(`l(Dllr(`p(`vrzbJ:rHr"lf*THE r"lf* CONE blHr&"lf**tZ$rZ$p*C@Dl8Z$j*C@D8*Fr8rJ:*xdldl(lllrdlldrdl<ldxhdldldr"l f*rrdlddlhx" lf*hr"lf*For a cone of radius rr"lf*& height h:x8"lf*VOLUME = 1/3 x pi x h8r(`p<Z rl(`l(`rrr&"lf**tZ$rZ$p*C@Dƀ8Z$j*C@D8(Fr8r (llrT l l2l2r(\p2(rV(\llrr(Lpr(LrllrT l l2l<r(\p2,rV(\l lrHH rJ:r"lf*THE SPHEREHrlPlPrHr&"lf**tZ$rZ$p*C@D8Z$j*C@D8*Fr8rJ: *^" lf*VOLUME OF A SPHEREr(Lpr(L^h(prD(r" lf*= 4/3 X PI X Rrh2J:r"lf*32r(\prll(\llrrTldl<lN(\p<,<rV(\ldlrNr(\prll(\llrrTldl<l(\p<@rV(\ldlrr&" lf**tZ$rZ$p*C@D8Z$j*C@D8*Fr8rJ:*z"lf*6 SPHERES WITH RADIUS 5 CM AREr"lf*ARE PACKED IN A RECTANGULAR BOXzt"lf*MEASURING 30 X 20 X 10 CM.r"lf*THE SPACE AROUND THE SPHEREStF"lf*IS FILLED WITH SAWDUST FOR PACKING.FH"lf*WHAT IS THE VOLUME OF THE SPACE TO BEH8 "l f*FILLED WITH SAWDUST ?8b "l f*1) 2660 CM CUBED ?r"l f*2) 2280 CM CUBED ?bb"l f*3) 2860 CM CUBED ?r"lf*4) 2000 CM CUBED ?b>&"l f*1,2,3 OR 4 tA$>Z(4pdA$fr(4ndA$f@DZV(4pz@" lf*GOOD - CORRECT ANSWERVZ(4jz@" lf*SORRY - 3 IS THE ANSWER !Z`&" lf**tZ$rZ$p*C@D`8Z$j*C@D8Fr8^ PlPl<l<llrPlPl(l<ll^4"PlPll<ll4h$rPlPlr r2lPlrnlPlhF&rlPlrlPlFX("lf*THE SURFACE AREA OFr"lf*A SPHEREXJ,"l f*= 4 X PI X Rr" l f*2J`.&" lf**tZ$rZ$p*C@D2`80Z$j*C@D82Fr8V4(\p ,r(`pVZ6(pr(r~ l l(\l(`ZD8(\p(\vr(`p(`vrrrD$:J:rH $D<"lf* THIS CONCLUDES THIS SECTION OF D>>"lf* G.C.S.E. G E O M E T R Y >$@HrJ:$jB"lf*WOULD YOU LIKE A RE-RUN OFr" lf*THE PROGRAM ? 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HIGHER GRADE):x" l f*BYr"lf*EVELYN MILLS Copyright 1989r(Lpu0r(Lx$J:rH$Fr8<"lf*WHEN YOU SEE THIS *, PLEASE<< "l f*PRESS 'P' (RETURN) TO PAGE<6""l f*FROM SCREEN TO SCREEN6V$&"lf**tZ$rZ$p*P@D(V2&Z$j*P@D$2(Fr8V*J:r"lf*PROPORTIONALITY REQUIRES A KNOWLEDGE OFV:," lf*SIMILARITYrJ::<."lf*SIMILAR TRIANGLES RERQUIRE:<60"lf*3 PROPORTIONAL SIDES6J2"l f*2 SIDES PROPORTIONAL AND ENCLOSED ANGLESJ(4"l f*EQUAL.(@8"lf*THE TRIANGLES ARE EQUIANGULAR.@F9rlr~ll@lFh:(pr(\p r(p(vr(p @(ph><(rl(\lr>4Frlr4XP&" lf* * tZ$rZ$p*P@DTX0RZ$j*P@DP0TFr8jVdl lZrlZPlZrPlZdl jfXllPrlPlPrlPlfRZ" lf*Ar"l f*Br" l f*CRR\"lf*Dr"l f*Er"l f*FR^"lf*IF ABC & DEFr"lf*ARE SIMILARr"lf*THENr(Lp'r(L.`"l f*AB AC BC..b"lf*-- = -- = --.Xc"lf*DE DF EFr"l f*CORRESPONDING SIDESX2d"lf*ARE PROPORTIONAL2BfHr"lf*SO ~~~~~~rHBVh&"lf**tZ$rZ$p*P@DlV0jZ$j*P@Dh0lFr8jjdl lZrlZPlZrPlZdl jflllPrlPlPrlPlfRn" lf*Ar"l f*Br" l f*CRRp"lf*Dr"l f*Er"l f*FRpr"lf*IF AB=12cm.r"lf* BC= 6cm.r"lf* DE= 8cm.p.t"l f*WHAT IS EF ?.8v"l f*AB BC 12 6 88x"lf*-- = -- => ---- = ---88z"lf*DE EF 8 EF8X|"lf*12xEF = 48 i.e. EF = 4r(Lp@r(LXH~"lf*NOTE THAT THE RATIOS ARE 12/8 (=> 3/2)HV&"lf**tZ$rZ$p*P@DV0Z$j*P@D0>Fr8rli@li>jdl lZrlZPlZrPlZdl jllPrlPlPrlPlrrdlPlR" lf*Ar"l f*Br" l f*CRR"lf*Dr"l f*Er"l f*FRt"lf*THE LINE RATIOr"lf*IS 3/2r"lf*THE AREA RATIOtL"l f*IS SQUAREDr(Lp@r(LL>"lf*i.e. AREA ABC/AREA DEF = 9/4>H"lf*If the area of ABC = 36 cm sq. then theHJ"lf*area of DEF must be 4/9 of 36 = 16 cm.sqJX&"lf**tZ$rZ$p*P@DX0Z$j*P@D0(Fr8r(D(\p((r(`p<xD(p r(r(\l(`lr(`p(`xr(\p(\vrrrPH rJ: r"lf*PROPORTION AND VOLUMEP$J:rH$X&"lf**tZ$rZ$p*P@DX0Z$j*P@D0Fr8L"lf* VOLUME RATIO IS ALWAYS CUBED. rL^Pl<llllrPldllll^J<l<<ldrdl<dldJ^l(ll llrldll ll^Ll(ldrl(ldLVrPlddldrldldVD"l f*1Ocm.r"l f*12 cm.DB" lf*5cm.r"lf*7cm.BP&"lf* Are the cylinders similar (Y/N) tA$PLA$j*YhA$j*N@DL^A$p*N@"lf* GOOD - THE REASON IS AS FOLLOWS ^VA$p*Y@"lf* WRONG - EXPLANTION FOLLOWS VV&"lf**tZ$rZ$p*P@DV0Z$j*P@D0&Fr8r&^Pl<llllrPldllll^J<l<<ldrdl<dldJ^l(ll llrldll ll^Ll(ldrl(ldLVrPlddldrldldVDHr"lf*A BrHDD"l f*1Ocm.r"l f*12 cm.DB" lf*5cm.r"lf*7cm.BH"lf*RATIO OF HEIGHT/BASE IN A = 10/25 [*PI]HH"lf*RATIO OF HEIGHT/BASE IN B = 12/49 [*PI]HB"lf*These ratios are NOT prportionalBV&""lf**tZ$rZ$p*P@DV0Z$j*P@D0:Fr8rrl:D(p (r(p DZ(rl(,l(r(p(vrrZ$J:rH $N"lf*SOME TYPICALr"lf*G.C.S.E.NP"l f*QUESTIONSrrJ:rHPV&" lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8J"lf*TWO CONTAINERS ARE SIMILAR IN SHAPE; ONEJH"lf*IS 12cm. HIGH AND THE OTHER 16cm. HIGH.HD"lf*IF THE SMALLER HOLDS 1.35 litres OFDJ"lf*OF LIQUID. HOW MUCH DOES THE LARGER HOLDJ&"l f*Answer tA$r(pdA$fr(ndA$f@D< (j́@"l f*NO....<^(ṕ@"l f*CORRECTr(Lp'r(L^>"l f*Think of the line ratio (3/4)>H"l f*For volume ratios this is cubed (27/64)HD"lf*HENCE ANSWER IS 1.35 x 64/27 litresDJ"lf*The ratio is INVERTED to increase volumeJX&"lf**tZ$rZ$p*P@DX0Z$j*P@D0:Fr8rrl:l <l(lZr(lZl rl lPl,"lP<l,R$"lf*Ar"l f*Br"lf*DR:&"l f*Er" lf*C:H(" lf*4 6r" l f*7 1O.5HV*"lf*TRIANGLES ABC &r"lf*CDE ARE SIMILARVJ,"lf*BECAUSE:r(Lp@r(LJF."lf*4 7 2 14 2 2FF0"lf*- = --- -> - = -- i.e. - = -FF2"lf*6 1O.5 3 21 3 3F*4(Lp@r(L*@6"lf*IF THE AREA OF ABC IS 20cm.sq.@8&"lf*WHAT IS AREA OF CDE tA$r(pdA$fr(ndA$f@D::(j@"lf*..NO:<<(p@"lf*CORRECT<0@&"lf*Area = 1O x 9/40VB&" lf**tZ$rZ$p*P@DFV0DZ$j*P@DB0*FFr8r*JH(`p rll6l(`rJLJrr~ll"lrLDLJ:r"lf*IF LINE RATIO = 3/4DBN"lf*AREA RATIO = 9/16 (SQUARED)B@P"lf*VOLUME RATIO = 27/64 (CUBED)@LRJ:r"l f*QUESTIONS OF THIS TYPE APPEARL4T" l f*YEAR AFTER YEAR !!4VV&"lf**tZ$rZ$p*P@DZV0XZ$j*P@DV0(ZFr8rH(@\r(l(llll@0h"lf*TWO EXAMPLES OF0Vj"lf*SIMPLE PROPORTIONr"lf*FOLLOW NEXT.VHl"lf*WHILE SUCH QUESTIONS DEAL WITH DIRECTHFn"lf* PROPORTION, THEY COLLECT A FEW MARKSF@p"lf*WHICH CAN BE EXTREMELY USEFUL.@Vr&"lf**tZ$rZ$p*P@DvV0tZ$j*P@Dr0vFr8Hx"lf*A SHADE OF PAINT IS MADE UP OF 3 PARTSHHz"lf*BLUE & 4 PARTS YELLOW. HOW MANY LITRESHB|"lf*OF BLUE ARE REQUIRED TO MAKE 1O.5BX~"l f*LITRES OF THIS PAINT ?r(Lp@r(LX~&"l f*ANSWER tA$r(pdA$fr(ndA$f@Dv~B(j@"l f*....NO!!B>(p@"l f*CORRECT><"lf*ANSWER IS SIMPLY 1O.5 X 3/7<V&"lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8Rr(llrr5l l RRrl!lrrl!l R4" lf*THIS ONE IS SIMPLE4`Tll6lPr(`prVl(`lr`F"lf*A CONCRETE MIXTURE IS NADE BY MIXINGFD"lf*CEMENT SAND AND GRAVEL BY VOLUME INDD"lf* THE RATIO 1:2:4. HOW MUCH SAND ANDD@"lf*AND GRAVEL MUST BE ADDED TO O.5@Z"l f* CUBIC METERS OF CEMENT ?r(Lp@r(LZ>"l f*SAND = 2 x 0.5 cubic meters>>"l f*CEMENT = 4 x 0.5 cubic meters>X&"lf**tZ$rZ$p*P@DX0Z$j*P@D0(Fr8r(d(`pxr(\pr(`pd(\|fz d6(\l(`(`l(\rr6d"lf*QUESTIONS ON PROPORTIONr"l f*ARE MOSTLY RELATED TOd>" l f*GEOMETRY (AREA, VOLUME ETC.)>X&"lf**tZ$rZ$p*P@DX0Z$j*P@D0Fr8hTldl#lr(p@r(`pdz(|fz2hRV(l(`lrrr~lPl@lxRf"l f*S P E E D - D I S T A N C E - T I M Er(LpN r(LfFFr8rr~ll@l7F4" lf* DISTANCE 44" lf* SPEED = -------- 44" lf* TIME 4h"lf*BY CROSS MULTIPLICATION IT FOLLOWS THAT:r~lPl@lh:"l f* DISTANCE = SPEED x TIME :2" lf* DISTANCE 22" lf* TIME = -------- 22" lf* SPEED 2>"lf*THESE FORMULAE MUST BE KNOWN>V&" lf**tZ$rZ$p*P@DV0Z$j*P@D0(Fr8r(F"lf*CONSIDER NOW THE FOLLOWING QUESTION:FH"lf*A car travels 45 km at an average speedHH"lf*of 3O km/hour & then travels 175 km atHH"lf*7O km/hour. What is the average speed HX"lf*for the whole journey ?r(Lp'r(LXF"l f*S=D/T TOTAL DISTANCE = 22O KM.FH"l f* TIME TAKEN FOR STAGE 1 = 1.5 HRS.HH"lf* TIME TAKEN FOR STAGE 2 = 2.5 HRS.Hf"lf* TOTAL TIME = 4 HRS.r(Lp'r(LfJ"lf*AVERAGE SPEED IS 22O/4 KM/HR = 55 KM/HR.J@"lf*Did you note the TWO stages ??@V&" lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8NlldrldldN^"lf*(B)30r"lf* 20r"lf* 10^V"l f*(A)Or"l f*noon 1pm 2pm 3pmV~ldFlFrFlF_lFr_lFlr"lf*km~bZlldr"l f*Pamr" lf*Ruthbv"lf*Pam cycles from Ar"lf*to B as shown.r(Lp'r(Lvv"lf*Ruth cycles from Br"lf*to A as shownr(Lp'r(Lv8"l f*Intermediate times are8T"l f*1.30 pm & 2.30 pm.r(Lp'r(LT&"lf*P's av.speed (km.hr)-NO rest tP$r(<pdP$fr(<ndP$f@D8 (<p @""lf*O.K.8> (<j @" lf*D/T = 10>H&"lf*R's av.speed from B to A tR$HP(DpdR$fr(DndR$f@DP8(Dp@" lf*O.K.8<(Dj@" lf*D/T = 30<D"lf*From B to A is a negative gradientDD"lf*showing a return journey to LINE ADV&"#lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8X J:r"lf*Average speed is related to the GRADIENTX:"" lf*of the linerJ::J$lldrlddldJ~&"lf*Yr"l f*XrldZlFrZlFZld~R0"l f*Or" lf*Ar" l f*BRP2"lf*The GRADIENT of OAr"lf* ABPn4"lf*= --r"l f* OBr"l f*This is true for ANYn.6"lf*straight line.8rlr llrr#llrTlldl`:(\pPrV(\llrrrl`V<&"lf**tZ$rZ$p*P@D@V0>Z$j*P@D<0(@Fr8r(HB"lf*I WONDER IF YOU CAN FIND THE ALGEBRAICHhD"lf*EXPRESSION FOR THE FOLLOWING QUESTION ?r(Lp'r(LhF"lf*What is the average speed of a boatr"lf*which travels for the first x hours atpH"lf*1O km/hr and for the next y hours at 15 km/hr ?r(Lp'r(LpHJ"l f*ANSWER ?: 1) 1O + 15 2) 1Ox + 15yHFL"l f* ------- or --------FFN"l f* x + y x + yFP&"lf*YOUR CHOICE 1 OR 2 tA$r(pdA$fr(ndA$f@D@>R(p@"lf*CORRECT>BT(j@"lf*NOT SO !!BJV"lf*DISTANCE = SPEED X TIME i.e. 1Ox and 15yJXX&"lf**tZ$rZ$p*P@D\X2ZZ$j*P@DX2\Fr8P^X1p/xrX2pX1vPL`Y1pdrY2pdxY1LPaX1lY1X2lY2rrPLbHr"l f* V A R I A T I O N rHLVd&"lf**tZ$rZ$p*P@DhV2fZ$j*P@Dd2hFr8hj"lf*QUESTIONS OF THE FOLLOWING TYPEr" lf*APPEAR FREQUENTLYhHl"lf*If W is directly proportional to D andHn"lf*W = 24 when D = 6, find the value of Wr"lf*when D = 7.r(Lp'r(LHp"l f*Here a constant k is involved, and W=kDHBr"l f*ALWAYS find the value of k first.B\t"lf*HENCE: 24 = k x 6 => k = 4r(Lp@r(L\@v"lf*Using the new value of 7 for D@2x" lf*W = 4 x 7 => 282Fz"lf*NOTE THE FOLLOWING TYPES OF VARIATIONFZ|&"lf**tZ$rZ$p*P@DZ0~Z$j*P@D|0Fr8B"lf*DIRECT PROPORTION where y = kxBB"lf* 2BB"lf*The SQUARE LAW where y = kxBB"lf* 3BD"lf*The CUBE LAW where y = kxDB"lf* kBB"lf*INVERSE VARIATION WHERE y = ---BB"l f* XBB"l f* kBB"l f* OR y = ---BB"l f* 2Bj"lf* X etc.r(Lp'r(Ljf"lf*COMPUTER CALCULATIONS ON PROPORTIONr"lf*NOW FOLLOWfZ&" lf**tZ$rZ$p*P@DZ4Z$j*P@D4Fr8bJ:r"lf*DIRECT PROPORTION (use values < 1OO)rJ:b&"lf*FIRST VALUE OF Y tF$r(pdF$fr(ndF$f@D2(td@D22(n@D2&"lf*FIRST VALUE OF X tG$r(pdG$fr(ndG$f@D2(td@D2.(n@D.((p(|(F"lf*Value of constant k is t((t* FX&"l f*NEW VALUE OF X tN$r(4pdN$fX0(4td@D0.(4n@D.((pd(|(fz(4(N"l f*The required value of Y is now t(t* Nj&" lf*ANOTHER RUN (Y/N) tR$rR$p*Y@Dj6R$j*Y@D6Fr8\J:r"lf*NEXT: THE SQUARE LAW (Y = kX )rJ:\>"lf* 2>&"lf*FIRST VALUE OF Y tF$r(pdF$fr(ndF$f@D2(td@D22(n@D2&"lf*FIRST VALUE OF X tG$r(pdG$fr(ndG$f@D2(td@D2.(n@D.(((p(|d(~f(F"lf*Value of constant k is t((t* FX&"l f*NEW VALUE OF X tN$r(4pdN$fX0(4td@D0.(4n@D.<(pdd(|(~fzd(4~ff<N"l f*The required value of Y is now t(t* Nj&" lf*ANOTHER RUN (Y/N) tR$rR$p*Y@Dj6R$j*Y@D6Fr8F"lf*IT SHOULD BE REMEMBERED THAT, IN ALLFD"lf*INSTANCES, THERE IS NO NECESSITY TODf"lf*CONTINUALLY USE THE TERMS 'X' AND 'Y'r(Lp@r(LfJ"lf*In the previous example using the squareJf"lf*law, the question could have beenr"lf*rephrased as:fD"l f* 2DD"l f*A variable A is proportional to r .DF"l f*If A = 2O when r = 2, find the valueF0" lf*of A when r = 50J"lf*In the computer calculation, the A inputJJ"lf*would be that of 'Y' and the r input = XJV&"lf**tZ$rZ$p*P@DV2Z$j*P@D2Fr8v(pr(`pr(\p@r(p(vr(p@(pv\(r(\l(`l(\r(`p(`xrr\t"lf* INVERSE r"lf* PROPORTION r"lf* CALCULTIONS tX&"lf**tZ$rZ$p*P@DX2Z$j*P@D2(Fr8r(> "lf* k>> "lf*INVERSE PROPORTION y = --->`"lf* Xr$ll`&"lf*FIRST VALUE OF Y tF$r(pdF$fr(ndF$f@D2 (td@D22"(n@D2$&"lf*FIRST VALUE OF X tG$r(pdG$fr(ndG$f@D2&(td@D2.((n@D.*((p(z(F,"lf*Value of constant k is t((t* FX.&"l f*NEW VALUE OF X tN$r(4pdN$fX00(4td@D0.2(4n@D.(4(pd(z(f|(4(N6"l f*The required value of Y is now t(t* Nj8&" lf*ANOTHER RUN (Y/N) tR$rR$p*Y@Dj0:R$j*Y@D<0<Fr8B>(p@r(pBt@(pr(r~(l(l(l(r(p(vrrrtLBJ:r"lf* THIS CONCLUDES THIS SECTION L>D"lf* OF THE G.C.S.E. CURRICULUM >$FJ:r$lH&" lf* ANOTHER RUN (Y/N) tA$rA$p*Y@Dl*RA$j*Y@*'J SET SCREEN$'lll$$'lll$"'r8r4"'                             ,PROG1.BSCPROG2.BSCPROG3.BSCPROG4.BSCPROG5.BSCPROG6.BSCPROG7.BSCPROG8.BSCPROG9.BSCPROG10.BSC<<<<<<<<<<A:\*.BSCab Hz/:NuH z0(2(@g4"P&1::g*g2g:ftgl`2g\`NgL`g<`.E\@f#`@Y@j0(Y@1@`,E0Y@j0(Y@XBhmBBhf1@LNu,X.6H*2( 6^VjTzVb&$N Hf($*  n| 60X2 `\V"8"&Z,NNN  $ L& " V ,>v" 2^nP ZV 4n  $ rT*0 0D>L  " >: RR8N".$B.d Dbn^jX2 8*  v2`ll|:D.D&|"(8BBBP&&LHJHNLHJH$., THJH*&.ZR PFHH@ (0\HxT0&TDL R$6.* $`T&` l 4MJy f2*O*m-M - ЭЭ// Bg?<JNAO JkO a0a.A=X=P=| N6Jf`BnBnBNaa p N1JngAV-H=n0<#N1a>"< <NJg-H<-H@aNN/N/a/N.zLNuH0<?N.zLNuB@=@AD-HpNN1Nu0<`0<`H=n =|A-Hp!N1LNuA$ k"X"`Nu=|BnphN1LHH=|p3N1Nu=|A-HpN1NuJngA-HpN1 n(fpaa0. |g |ga>Jnf`&AL @!@!@C!I!I-I-I"Bh Bh"=h$B,B0B:BNu n,n^A@Rh0<(AQAh AhA@ShAPUhAPhQhAThZh*PhBNuH2IA!2I naL?NuH=@R2IA-p na,L?NuH0.RA pa0.RaLNuI8.L(t@0,$AaNu?:MGNI&K&PPJ4Pg RR`TTR| m.BDCJQgBQ:MG7PRDTT| m8.RL0.Ra0.RAH1NuBngPa=@HA1aJf -nv`H0.aaJf6 nv .r".za28.LGH/3@?<INA\L'v@0.2IK;aa@nRf-nv0.Ra 0.a a .BnNung` Nua>N^JgH?<NNT/?NAXJzfBra .r".z nvNua$Jg?AQ-H=|p4N10NuH@P nv"CXf"(fBLNu=@B.Ad-HCB<.:<PQA-HpZN1JngaJf&=|Aj-Hp4N1 nf`t0.a a8Jfb nv .r".za8.LGH/3@?<INA\L'v@0.2IK;-nv nv0.Ra0.0.Nu=n =|phN1Lnbf=|phN1LHH=|/=|plN1LHLH=n =|piN1NuLaB8.HĈDl4B8.nLDl68Bno0.B8Cnno 2.CnNuaL aaBn=|-LHplN1-n-n=n=|piN1LH$H$fNrL$NJ`,Ha(?A?C26/:/>/ a>?CLNuAnEG0C@n2E6@n6NuvNu=n=| piN1Nuaa=n=| phN1JngJngLL aZJBgdJCg^JgNH@0HB4Ԁ-@ -B -@ -B A-H A-H =| 0<1N.z` BDN0=n=| `ba`H=|=|)HplN1-n-nnnLNuBn=|phN1-n-nLB(C KCJBP@CTAUACCB@axNu=|--n-npdN1=nNunnBn=|--n-nplN1-n-n=npeN1Nu=npfN1=npgN1NuA-H=n=|piN1Nu Output Id?<NATA:Bg/ ?<GNAPJfS\*.BSCBNuAdC0<BAA|mRRQNuLBDN0IAN"<SAf"O /NuNPJ  !z!n!n"1n$NuJng"a$p=@RaAL-H-Ha`J/<?<HNA\Jg-@-@"-A" @@2<CSAf-H @aj></<?<HNA\JgJE%p-@"-A" @@2<CHSAf-H @aXG|(mB@Nu<NuJf.-n-n-n B/<?<HNA\-@-@Nu nJg2< BAfNuI8< H/?<INA\LSDfNuJg/.?<INA\NuH~T.,/?<HNA\JfnB` @ L~NuH?<NNT-@?<NNT|f=|=|=| `2|f=|=|=| `=|@=|=| 0.H=@LNuHA-H -H =| JDjD`D@ASBSCJDjH DDH `H DDH 0<1N.zLNu  02 46@ N f ~   {  "$&,. &%x7'6&H(8*G,.467YIXH@jXZZi\^dfY { k z   j p |   { &Z-H\JgB@BABBJg6 [fRAR@ ]fSA |fRB|g |fPNu|nJAfJ@g|nJf/< ?<HNA\JfpNu-@ n0<r:< n^n:<1A!|1|1EBB BBhRAQ nC0|g 21`0<1|. nn 0< 1| .1n1n,1n1n!n2<F4<1Q"n\ n-H"n\$n [g JfPNuRJfPNu ]g`B%nD-H$n`&nBG [g Jf`` nR ]g$ \g |g`B%n `B<F '`G-H ]f`AR"n#H\A[#HT"n80<a><0<?a>Q"nH0<$i gBAJgPA`3AQ"n$IHB@2<i3@5@Q nh1@D1hL41h$ 1h1hT<1h1hl1h  1h 0<"n a><"n0<?a>Q"n0)bh24<@3AQ"nH0<J f Q n4<2.A0(h@l 6h1CQ nh0<BhQ-n=|pN1B@Nu$I8< n^l8<BEJ g5EDQ3ENuBB$I2BC&i gBDJgPD`Dn6Q5CQNu-------------------- Desk Jg/.?<INA\A-HBnpN1B@Nu` 0 00}    0 @ P`p8ppp p0p@pPp`ppp}T ,(<22 2 F m@ d z    $x  $<Z< n00P<%P< D @ @(@8@X@hd@xxd@xd @xd @ @h `  "" Desk File About Runner-------------------- Load Program Run Program-------------- Show Info Dump Output QuitFast BasicRun-Time InterpreterComputer ConceptsProgram execution stoppeddue to:OKProgram InformationCANCELRUNNERFast ST BASICRun-time InterpreterRUNNER bits: Phil MartinInterpreter: Jeremy RustonGavin TheobaldPhil Martinand Nigel, who doesn't like to be mentioned!Version Number: 1.05c/00 Computer Concepts 1986Gaddesden Place, Hemel Hempstead,Herts. 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sǵk}1ۏt=.p{?2 k( ` h&4^TmԳ1ne}I<" 6FŏŏLG`g0VGf͖tЮbⰏs=%mecV,57SF€f/-Z\*x*n{PIЋ e|! sLK:Nװ:i2)2}*)Қ ?N󈈈L?Ӿzp e%7Xx}  XÀp T%0 FF X)}/Հ ` ` vH;`[FF$;`Nu@P$ k(C@ԥ* _2ɿ/ k:#x !$'+.158;?BBBaJAfNuJAgfSANua< g<@f J@gRS@Nu<Xf `@<Nf `4<|fa`(<gθ<|faa`<CfJ@gBS@RBn Jb Nu<{fFJ@g(H@@ }gaz`J@gL`RSANuH }gaR`LRSANu<1eҸ<9bD 0e 9bSAE E` *f"RSABc8:g@DEDa^SDPQNu $fRSA`@ LfRSABcDa0SD QNu Rf,RSA:BcDaSDkPQSE PQNuH@@?aSWgL` Nu<(f )ga\`<Af J@gS@<ae<zb`<af J@gS@<Ae<Zbx `p<*f JBg``<BfVJBgSSBNuH"`H 2E)8BBBa|JAfBaLHBBBHBJk4EFgFA)8 SBf`8Hav(<<g&<xza"aRF>LxaaL0A)8BBaJAfNuaH< g<Dg(<Eg <.g<dg <Zg<zf<DRB|(bNu<^fPNu<|g<g"<|fa`<+fJj H a|LNu<-fJjNu<1eH<9b@D 0e 9bSAE E`H@@?a0SWgL` Nu<(f )ga `aN< g<Xgj<Ngn<|gr<BgB<Zg<.g<JFf>Ek$aQg 5eR 9c0`>A)8E)a 0gSRGNuJFfT0gJGkGg k S@fJEg0jD@Ec00S@f0GkgS@f0g"H !0f S@fNu0gS@f0gS@f"H0 !0f S@fD0jD@H|e0B@H@d|e0B@H@ |e0H@0DF 00`f  `RFfNuNPJ  pzpZ#,#{ϐmӐܐ$lll$$lll$,r8r4r, Fr8` (phr(p@xr( p`(pr(r~(ll(l( r(p(vr(p(xr( p( vrrrrFHrJ: r"lf*TRIGONOMETRYFH"lf*PLUS BEARINGSrJ:rHH$"lf*BY$t"l f*EVELYN MILLSr"l f*Copyright 1989r(Lpr(Lt(Fr8r(B"lf*WHEN YOU SEE THIS STAR *, PLEASEBh"l f*PRESS 'P' (RETURN) TO PAGEr" l f*FROM SCREEN TO SCREENhV&"lf**tZ$rZ$p*P@D V0Z$j*P@D0 Fr8>""lf*AMY STUDY INVOLVING BEARINGS><$"lf*REQUIRES A KNOWLEDGE OF THE<>&"lf*SIN/COS/TAN RULES AS WELL AS>d("lf*THE COSINE RULE AS SUCH.rr~ll@l<d,R1p<rR2p#r(<pGr( p2xrANp( zz(<|V.(\pR2zdANfr(`pR2zdANfVd0(\vl(`vlR2llrR2pR2xrdJ2Hr"lf* SO rHrJV4&" lf**tZ$rZ$p*P@D8V06Z$j*P@D40(8Fr8r(B:(p=r(TpSr(,p2r(pBH<ANphr(pAN|H>V1p(Tx(,zd(frH1p(,zd(fv(r(p(vr(p@(pR@(r(Tl(V1lH1rrR\B"lf* CONDIDER NOW THE r"lf* GENERAL USE OF \@DJ:r"lf* THE SIN/COS/TAN @^F"l f* RULES BEFORE r"l f* APPLYING SAME TO THE ^FH"lf* THE STUDY OF BEARINGS rJ:F>Jrlllr>ZL&"lf**tZ$rZ$p*P@DPZ4NZ$j*P@DL4(PFr8r(`Rl lPrlPlPrlPl `RT"lf*Ar"l f*Br"l f*CR>V"l f*SIN ANGLE C = OPPOSITE (AB)>:X"f* --------------:>Z"l f* HYPOTENUSE (AC)>>\"lf*COS ANGLE C = ADJACENT (BC)>>^"lf* -------------->>`"lf* HYPOTENUSE (AC)>>b"lf*TAN ANGLE C = OPPOSITE (AB)>>d"lf* --------------->>f"lf* ADJACENT (BC)>Zh&"lf**tZ$rZ$p*P@DlZ2jZ$j*P@Dh2lFr8fn(p r(pxr(l((l(f*p(p(xrr*Xr"lf* REMEMBER THAT r"lf* THE HYPOTENUSE XTt"lf* IS ALWAYS r"l f* OPPOSITE THE T,v"l f* RT. ANGLE ,Vx&"lf**tZ$rZ$p*P@D|V0zZ$j*P@Dx0|Fr8@~J:r" lf*THE S I N E RULE@JJ:r"lf*(FINDING THE UNKNOWN ANGLE)J>"lf*VALUES SHOULD NOT EXCEED 9OO>h&"lf*LENGTH OF SIDE OPPOSITE ANGLE tA$r(pdA$fh6(ndA$f@D|6*(t@D|*,(n@D|,^&"l f*LENGTH OF HYPOTENUSE tB$r(pdB$f^6(ndB$f@D|60(n(@D|0,(n@D|,ANp($(pd(|(f$( pd(fB"l f*THE ANGLE IS t( t* DEGREESBD"lf*The HYPOTENUSE is the longest sideDh&"lf*ANOTHER TRY (Y/N) tZ$rZ$p*Y@D|h0Z$j*Y@D0Fr8F(pxJ r(pd(|fzF"r"P(l((l(rr"lf* USING THE PN"lf* SINE RULE r"l f* SOLVE THE NX"l f* FOLLOWING r"lf*PROBLEM rXV&"lf**tZ$rZ$p*P@DV0Z$j*P@D0(Fr8r(B"lf*When a kite is flying, the stringBB"lf*makes an angle of 22 degrees withB@"lf*the horizontal & the string is @H"lf*200 metres long. How high is the kite ?H*(Lp r(L*h&"l f*ANSWER TO NEAREST WHOLE NUMBER tA$r(pdA$fh6(ndA$f@D6<(pK@"l f*CORRECT<N(jK@"l f*SORRY ANSWER IS 75 metres.N:"lf*1. ALWAYS MAKE A DRAWING.::"lf*2. REMEMBER YOUR ALGEBRA.:8"lf* EXPLANATION FOLLOWS8V&"lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8`lZl<rZl<l<rl<l`B"lf*22r" lf*200 m.BR"lf*Kr"lf*Gr" lf*BR."l f*SIN 22 KG.0"l f*------ = ----0."l f* 1 2OO.8"l f*By CROSS MULTIPLICATION8@"lf*KG = SIN 22 X 200 = 74.921319@8"lf* = 758V&"lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8`lZl<rZl<l<rl<l`D"lf*75 M.r" lf*200 m.DR"lf*Kr"lf*Gr" lf*BR>"l f*ASSUMING THAT KG IS KNOWN AND>H"l f*ANGLE B IS RERQUIRED (ELEVATION ANGLE),HD"lf*THE SAME EQUATION IS USED TO GIVE:-DB"lf*SIN B = 75/200 = 0.375 = 22 deg.BH"lf*USE THE INV SIN ON YOUR CALCULTOR HEREHV&"lf**tZ$rZ$p*P@DV0Z$j*P@D0&Fr8r&D(p,|r(pDR(\pr( pxr(pR(r(l(\(l( r(p(vr(p(vr(\p(\vr( p( xrrrr\r"lf*NOW TO THEr"lf*COSINE RULE\X&"lf**tZ$rZ$p*P@DX0Z$j*P@D0Fr8> J:r" lf*THE COSINE RULE>N J:r"lf*(VALUES SHOULD NOT EXCEED 900)Nb&"lf*LENGTH OF ADJACENT SIDE tA$r(pdA$fb6(ndA$f@D6*(t@D**(n@D*^&"l f*LENGTH OF HYPOTENUSE tB$r(pdB$f^6(ndB$f@D60(n(@D0*(t@D**(n@D* ANp($"(pd(|(f$&( pd(fB*"l f*THE ANGLE IS t( t* DEGREESB@,&"lf*ANOTHER TRY (Y/N)tZ$@2.Z$p*Y@D200Z$j*Y@D20:2Fr8rrl:L4X1p,rX2p,xX1LL6Y1p-rY2pxY1L&7(p&b8(rX1lY1X2lY2rrrb4:rrl4:<"lf*A COSINE PROBLEM FOR YOU:V>&"lf**tZ$rZ$p*P@DBV0@Z$j*P@D>0BFr8NDlldrlddldNvF"lf*Yr" l f*XrldPlrPlPldvVH" lf*Ar"lf*12r"l f*60V\J&"lf*X COORDINATE OF A tA$r(pdA$f\6L(ndA$f@DB6@N(p@"lf*CORRECT@HP(j@"lf*VALUE SHOULD BE 6HRR"lf******************r(Lp@r(LR4T"lf* ADJACENT42V"lf*COS 66= --------2.X"lf* 12.DZ"lf*ADJACENT = COS 66 X 12 = 0.5 X 12DV\&"lf**tZ$rZ$p*P@D`V0^Z$j*P@D\0`Fr8@b(`pr(pr(@zd(\pr(`l(\(`l(`r(`p(`xr(\p(\vrrzTfrJ:rH r"lf*THE TAN RULET$hJ:rH$Vj&"lf**tZ$rZ$p*P@DnV0lZ$j*P@Dj0nFr8>pJ:r" lf*THE T A N RULE>NrJ:r"lf*(VALUES SHOULD NOT EXCEED 900)Nbt&"lf*LENGTH OF ADJACENT SIDE tA$r(pdA$fb6v(ndA$f@D6*x(t@D**z(n@D*^|&"l f*LENGTH OF HYPOTENUSE tB$r(pdB$f^6~(ndB$f@D60(n(@D0*(t@D**(n@D*ANp($(pd(|(f$( pd(fB"l f*THE ANGLE IS t( t* DEGREESB@&"lf*ANOTHER TRY (Y/N)tZ$@2Z$p*Y@Dn24Z$j*Y@D4(Fr8r(Lrlrldldldl L:rlilPldl :FJ:r" lf* A TAN PROBLEM FOR YOU FBJ:rrlrBX&"lf**tZ$rZ$p*P@DX0Z$j*P@D0Fr8J"lf*A SURVEYOR IS ESTIMATING THE HEIGHT OF AJD"lf*BUILDING 70 M HIGH BY TAKING ANGLEDF"lf*MEASUREMENTS FROM THE ROAD. FIND THEFB"lf*ANGLE OF ELEVATION FROM A GROUNDB`"l f*POINT 5OO M. FROM THE BUILDING.r(Lp@r(L`b&"l f*ANGLE TO NEAREST DEGREE tA$r(pdA$fb<(p@"lf*CORRECT<h(j@" lf*HARDLY TRUE !!r(Lppr(LhnldlrdldlrdllnR" lf*Ar" lf*Br"lf*CR4"lf*ANS = INV TAN AB/BC48"lf* = ~ E I G H T deg.8V&"lf**tZ$rZ$p*P@DV0Z$j*P@D08Fr8rrl8`(pr(Dp<x r(pr(`Bldl(DlPll rrB4rlr4Z(\pr(\l(\lrrZX"lf*THE SIN/COS/TANr"lf*RULES SHOULD NOWXN"lf*BE APPLIED TOr"lf*BEARINGS.NZ"l f*PROBLEMS OF THISr"l f*TYPE CONSISTENTLYZT"l f*APPEAR IN THEr"lf*G.C.S.E. EXAM.T2"lf*GRADES A,B, AND C2X&"lf**tZ$rZ$p*P@DX0Z$j*P@D0Fr8J"lf*ALL BEARINGS ARE TAKEN FROM THE NORTH INJ6"lf*CLOCKWISE DIRECTION.6H"(8lf*AN IMPORTANT POINT TO REMEMBER IS THATHH"lf*THE INTERIOR ANGLES OF 2 PARALLEL LINESH@"l f*ADD UP TO 180 deg. AS FOLLOWS:@nlrllZrlrlPlnNlrPlPlZrlNh"lf*60r"lf*120r"lf*OrlhR" lf*Ar" l f*If O=60 deg. then B MUSTRL"lf*be 120 deg.r(Lp@r(LL@" lf*THE BEARING OF A FROM 0 IS O6O@@" lf*THE BEARING OF O FROM A IS 240@V&"lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8nlrllZrlrlPlnNlrPlPlZrlNh"lf*60r"lf*120r"lf*Orlh"" lf*A"D"lf*THERE IS NO NEED FOR CONFUSION HEREDH "lf*ALWAYS DRAW A North LINE AND CALCULATEHH "lf*YOUR BEARINGS clockwise FROM THIS LINEH@ lrZllll@Vlr"lf*THE BEARING OF A TO O (or O FROM A)V:"l f*MUST BE TAKEN AS SHOWN BY:<"l f*THE ARROW i.e. 360-120 deg<,"lf* = 240 deg.,:"lf*O TO A IS, OF COURSE 060:Z&"lf**tZ$rZ$p*P@DZ0Z$j*P@D0Fr8hTlllr(Lp zr(\pdd(Lfzfv(hJ (`pdd(LfzfvPrV(\l(`l rJF"rrPlllFlFr$FllrrJ:rHr"lf* A SAMPLE QUESTION rL&" lf* AS PER G.C.S.E. rJ:rHLV0&"lf**tZ$rZ$p*P@D4V02Z$j*P@D004Fr8D6"lf*A SHIP SAILS 14 KM. OM A BEARING OFDJ8"lf*136 deg. HOW FAR HAS IT TRAVELLED SOUTHJh:"lf*SOUTH AND EAST OF ITS STARTING POINT ?r(Lp@r(LhH<"lf*Here it is essential to draw a diagramHJ>"l f*to create the right angles necessary forJJ@"l f*a sin/cos or tan calculation as requiredJHA"l f*ANSWER INPUTS SHOULD BE GREATER THAN 4HhB&"lf*DISTANCE SOUTH (2 dec.places) tS$r(HpdS$fh8D(HndS$f@D48<F(Hp@"lf*CORRECT<LH(Hn@"lf*////OFF COURSE////LNJ(Ht@"lf*////OFF COURSE////NhL&"lf*DISTANCE EAST (2 dec. places) tE$r(pdE$fh6N(ndE$f@D46<P(p@"lf*CORRECT<NR(t@"lf*//OFF COURSE//NLT(n@"lf*//OFF COURSE//LdV&"lf*METHOD -> *tZ$rZ$p*P@DZd0XZ$j*P@DV0ZFr8^\lr ld l r lddld^>^lr lddl>L` ld lr ldlLRb"lf*Nr" l f*Er"l f*OR:d"lf*Br" lf*A:RfJ: r"lf*YOUR DIAGRAM SHOULD LOOK LIKE THISRlhJ:r"lf*NOW INSERT BEARING & DISTANCEr(Lp@r(LlZjlrlZll lr"l f*136Z:jlr"lf*14 km.:@l"lf*ANGLE AOB IS OBVIOUSLY 44 deg.@4n"lf*OB = DISTANCE SOUTH4,p" lf*COS 54 = OB,.r" l f* ----.,t" l f* 14,2v"l f*OB = COS 44 x 142Px"lf* = 1O.O7 km.r(Lp@r(LP8z"lf*SIMILARLY, SIN 44 = BA88|"lf* ----88~"lf* 148>"lf*DISTANCE EAST (BA) = 9.73 km>V&" lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8|"lf*A CAR WHICH IS INITIALLY 20 KM.r"lf*DUE NORTH OF A TOWN T, TRAVELS 3O KM.|d"lf*DUE EAST. WHAT IS ITS FINAL BEARINGr"lf*FROM T ?dD(Lp'r(LrJ:rHDz"lf*THE ANSWER TO THIS IS O56 deg.rJ:rHr(Lp@r(Lzlr l lr lPlrlrPl lL"lf*Tr"lf*ANGLE T REQUIRED !!LV&"lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8@J:r" lf* SOPHISTICATION @^Tlll2r(`prV l(`l r^D(\p$r"(\lf**rDD(`pr"l(`f*/rDD(`pr"#l(`f*/rDJ:X&"lf**tZ$rZ$p*P@DX0Z$j*P@D0Fr8F"lf*A MORE SPOHISTICATED COSINE RULE FORFh"lf*ANY TRIANGLE (NO RT.ANGLE REQUIRED)r"lf*IS AS FOLLOWShnl2lnrlnlZrlZl2nR"lf*Ar"lf*Br"$l f*CRR"lf*cr"lf*br"l f*aR6"lf* 2 2 2 6X"l f*a = b + c -(2bcxCosA)r(Lp@r(LX."l f*REARRANGING: .4"lf* 2 2 242"lf* b + c - a22"lf*Cos A = ---------20"lf* 2bc0`"lf*ANY UNKNOWN SIDE ORr"lf*OR ANGLE MAY BE FOUND`>"lf*USING THIS RULE.r>V&"lf**tZ$rZ$p*P@DV2Z$j*OP@D2&Fr8r&D(p(r(p<xDL(l(l(lrrrrLZr"lf*CONSIDER THEr"l f*FOLLOWINGZV&" lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8H"lf*TWO BOATS SAIL FROM PORT A ON BEARINGSHF"lf*OF O47 deg. AND 325 deg. AT SPEEDS OFFB"lf*4 KNOTS AND 5 KNOTS RESPECTIVELY.Bj"lf*FIND THEIR DIFFERENCE APART AFTER 1 HOURr(Lp@r(LjPlrllZrlPbll}rl}AlZrAlZlb>"lf*35r"lf*47>R"lf*Ar"lf*Br"l f*CRF"lf*4 n.m.r"lf*5 n.m.F6" l f*n.m. = Nautical miles6V&" lf**tZ$rZ$p*P@DV2Z$j*P@D2Fr8PlrllZrlPbll}rl}AlZrAlZlb>"lf*35r"lf*47>R"lf*Ar"lf*Br"l f*CRF "lf*4 n.m.r"lf*5 n.m.F> "lf*The distance required is BC.>0"lf* 2 2 20H"lf* BC = 5 + 4 - (2 x 5 x 4 cos82 deg.)H>"(8lf* = 41 - (4O x O.139)><"lf* = 41 - 5.56 = 35.44<2"l f*BC=sq.root 35.442."l f* = 5.95 n.m..V&" lf**tZ$rZ$p*P@DV0Z$j*P@D0(Fr8r(p (pr(\p  r(`p( r(pd(`|xfz<pD!(p(vr(p@(pDD"(r(\l(`(l(rrD^$J:r"lf* COMPUTER r"lf* CALCULATION ^Z&"l f* USING r"l f* COSINE RULE rJ:ZX(&" lf**tZ$rZ$p*P@D,X2*Z$j*P@D(2(,Fr8r(L.J:r" lf*THE COSINE RULErJ:Lh0&"lf*FIRST SIDE [less than 1000] tA$r(pdA$fh82(ndA$f@D0864(t@D,606(n@D,0h8&"lf*SECOND SIDE [less than 1000] tB$r(pdB$fh8:(ndB$f@D,8.<(t@D,.0>(n@D,0h@&"l f*ENCLOSED ANGLE [< 90 deg.] tD$r( pdD$fh6B( ndD$f@D,62D( tZ@D,22F( n@D,2H(pd( f\J(Dpdd(~v(~fxdz(z(z(ff\HL"lf*OPPOSITE SIDE IS t(Dt* UNITS LONGHhN&"lf*ANOTHER RUN (Y/N) tZ$rZ$p*Y@D,h0PZ$j*Y@DR0RFr8FT"lf*SHOULD YOU WISH TO FIND THE ENCLOSEDFBV"lf*ANGLE BETWEEN TWO SIDES, THEN THEB.X"lf*FORMULA IS: .4\"lf* 2 2 244^"l f* a + c - b44`"l f*Cos B = ----------42b"l f* 2ac2jdllrllxrlxljRf"lf*Ar" lf*Br"lf*CRRh"lf*br" lf*ar"lf*cRVj&"lf**tZ$rZ$p*P@DnV2lZ$j*P@Dj2(nFr8r(&p(Dpr(p&2(Lpz,X|(D2B(p(vr(p@(pBT(rl(Dzd(Lf(Dzd(LflrTD"l f*NOW TRY A SCALE DRAWINGrDV&"lf**tZ$rZ$p*P@DV0Z$j*P@D0Fr8D"lf*THERE ARE FOUR TOWNS A, B, C AND D.Dv"lf*B IS 1OO km. NORTH OF Ar"lf*C IS 9O km. ON A BEARING OF 14O FROM Avj"lf*D IS 12O km. ON A BEARING OF 26O FROM A.r(Lp'r(LjF"l f*FIND THE DISTANCES BETWEEN THE TOWNSFJ"l f*B & C, C & D, D & B WITH A SCALE DRAWINGJLxlxlnrxllL\xlFlr"l f*Br"lf*C\n"lf*Dr"lf*Arlrll}n4"lf*Nrl4V&" lf**tZ$rZ$p*P@DV0Z$j*P@D08Fr8rlr8^(DpPxr(p(vr(p@(p^F(rldl(DlPll (rF8"lf*B TO C = 179 METRES.86"l f*C TO D = 182 METRES.6T"l f*D TO B = 169 METRES.rrlTX&" lf**tZ$rZ$p*P@DX0Z$j*P@D0F(8r8F"lf*IN GENERAL QUESTIONS ON BEARINGS MAYFB"lf*BE ANSWERED USING THE SIN/COS/TANB@"lf*METHOD OF CALCULATION INVOLVING@X"lf*RIGHT ANGLED TRIANGLES.r(Lp'r(LXJ"l f*RE THE ALTERNATIVE COSINE RULE AS SHOWN,JH"l f*- CHECK YOUR SYLLABUS FOR REQUIREMENTS.HV&"lf**tZ$rZ$p*P@DV0Z$j*P@D0&Fr8r&Pr(p(r(p@(PL(l((l(rrrrL@r~(l(llr@LJ:r"lf* THIS CONCLUDES THE SECTION L@" l f* ON TRIGONOMETRY rJ:@8"lf* DO YOU WISH A RE-RUN 8l&"lf* OF THE PROGRAM (Y/N) tA$rA$p*Y@Dl(A$j*Y@(,PROG1.BSCPROG2.BSCPROG3.BSCPROG4.BSCPROG5.BSCPROG6.BSCPROG7.BSCPROG8.BSCPROG9.BSCPROG10.BSC<<<<<<<<<<A:\*.BSCab Hz/:NuH z0(2(@g4"P&1::g*g2g:ftgl`2g\`NgL`g<`.E\@f#`@Y@j0(Y@1@`,E0Y@j0(Y@XBhmBBhf1@LNu,X.6H*2( 6^VjTzVb&$N Hf($*  n| 60X2 `\V"8"&Z,NNN  $ L& " V ,>v" 2^nP ZV 4n  $ rT*0 0D>L  " >: RR8N".$B.d Dbn^jX2 8*  v2`ll|:D.D&|"(8BBBP&&LHJHNLHJH$., THJH*&.ZR PFHH@ (0\HxT0&TDL R$6.* $ To the best of my knowledge these programs are error free. If you have any queries please contact me as follows : Evelyn Mills, B.Sc.,Ph.D, 43 Chancel Close, Nailsea, BRISTOL BS19 2NE Educational programs written on request. #a000000 #b000000 #c7770007000600070055200505552220770557075055507703111103 #d #E 98 11 #W 00 00 0E 03 0F 0B 08 A:\*.*@ #W 00 00 0D 08 15 0B 00 @ #W 00 00 0E 09 15 0B 00 @ #W 00 00 0F 0A 15 0B 00 @ #M 00 00 00 FF A FLOPPY DISK@ @ #M 00 01 00 FF B FLOPPY DISK@ @ #T 00 03 02 FF TRASH@ @ #F FF 04 @ *.*@ #D FF 01 @ *.*@ #G 03 FF *.APP@ @ #G 03 FF *.PRG@ @ #F 03 04 *.TOS@ @ #P 03 04 *.TTP@ @