GSPk F2mkrcapmdxt UTo try your angle experiments: 1. Set your script tools directory (Preferences in the Display menu, then More) to the Poincare disk directory. 2. Use the Hyperbolic Segment tool to draw the triangle. 3. Use the Hyperbolic Angle tool to measure angles.tBG BCB tch ACfC tP-Disk RadiusC?CtMRP-Disk CenterC C t CtC C t l*The Poincare disk is a model for hyperbolic geometry. In the model, a straight line through two points is defined as an arc passing through the two points and perpendicular to the circle. Play with the hyperbolic triangle to discover things about angles. How large can the sum of the angles of a hyperbolic triangle be? How small? Bill Finzer, 3/95, with much help from Mike AlexandertMCaHide Experiment ey Chat0HDShow Experiment  tbA oCfCCB? tL 'smet jC CCfC?tDiskC CO BY?} N tL$ sC CCB? tA xCBtC C? th abCfCtC C?t*$azC CCB?t'9buC CtC C?t 'cpC CCfC?tty0 FCC  tؓ' rop kC[C@CqC? t tC"B@CB? t;d@i MCB t+0 Q@C7C t38 APC;TB  t smet AQeCdC  t BLaC0+C  t,h BMtCC  tgl CHPCwC  t3 8% CILC/B  t  q@CC@C%C? t(-  EC@tC  tgl t  KctCB  t zC/C@CA?t!'  ac~CpCCB?t I?AQ'CょC t*bgC;TBeCdC?tI?BM'CG t'9cbaC0+CtCC?tI?CI'z-CC t 'cwPCwCLC/B? t'R lC@tCC C? tLf uctCBC C? tbbaC;TBCょC?#t?bcCょCCB?#t beCょCeCdC?#t؍'%bvaC0+CCG?%t&'bxCGtC C?%tQؿ'bzCGtCC?%t؇'cqPCwCz-CC?'t'csz-CCCfC?'t:'cuz-CCLC/B?'t؏' m )vCbCCՎC?)t4L v~CCjCHSB?* t@Ep= ARCB +t a= BNfC5C .t38"> CJCeKXC 1tm r InvertedOWCC  4t!& L)XC B 5 tbbCBeCdC?6tظ'.bwfC5CtCC?7tX'crCeKXCLC/B?8 tbr nOWCCCfC?9 t G w)XC BCB?:tMRAS!CB ,;tBOxC+C /<t38CKȔCEMC 2=thm w GCyC >t27 N,ChvNB ?tbdC;TB!CB?@t 'SbyaC0+CxC+C?At'ctPCwCȔCEMC?Btب'  p Cl~CpCAtC?>Ct yG]CBCB??DtMRATDCuC -EtdiBPhCM'B 0FtqvCLGCMB 3GtLQ HڋD#AC Ht O6vhC !It  RXCC "HtbfCBDCuC?J6t'scafC5ChCM'B?K7t'#cvCeKXCGCMB?L8 t Qsmet rڋD#AC0Cf/C?Mtb;7@ c2ڋD#AClCF%5?F%5?M9tSJ> c36vhC’UCF%5?F%5?N: t aa6vhCCTFB@?NtBq c4XCC5CF%5?F%5?O9 t!  adXCC0Ck]C?Ot!&AU,XC B $PtBQ>JC&*C &Qtm rCMOWCC (Rtej  I~mCOC STtBsGx  PeCB VUt7<  ScBC(C XW tL bhC C,XC B?Y tb dbiCfC,XC B?Yt  bkCfC,XC B?Y t botC C,XC B?YtebqtC C,XC B?Y tRccC C>JC&*C?Z tAcdCB>JC&*C?Zt icfCB>JC&*C?Z thcjCfC>JC&*C?Zt'clCfC>JC&*C?Z tLrcxC COWCC?[ trcytC COWCC?[t0غ$'datC COWCC?[ tlAdeCBOWCC?[tNdgCBOWCC?[tbD> a1܋D#AClC0z,eLCBCfC\tDF a29vhCz‹UC>tg?tC CCB]te a3XCC-C"?#ECfCtC C^tuVz[AVlC+;B _tXjc11,XC B!KCF%5?F%5?Y`t DbjC`@D&B?`Yt\fc12,XC BFCF%5?F%5?YbtbpCtHaC&B?bYt BROC C dtNc15>JC&*CgGCF%5?F%5?Zet9YcekC eCʩ(CP&%B?eZt=c16>JC&*CO5CF%5?F%5?Zgt*'ckCTCʩyCP&gB?gZt]bCNkCMC itY#c19OWCCT4CF%5?F%5?[jt'&czbCOCLC8AC?j[tB:c20OWCCJCF%5?F%5?[lt YdfCiC C8C?l[t5h@c9lC+;BmBF%5?F%5?qYt AX[C\ srt_[d` AYK]D ׹B srt7w<| BEuC-u utt"  BFKDMC uttHPc13OC CBF%5?F%5?vZt AF BTCuC xwt BUC[R xwti n% CABC zyt', CB[C zytc17kCMCBF%5?F%5?{[t CPlC|C }|t CQjCC }|t6";' CWp D@>C ~t CX~%CuC ~tSX"LAWCJ4A  tBS\JqC/FC  tCOkCgWC  tUnOK@c10,XC B8lBF%5?F%5?Yt=c14>JC&*C0 BF%5?F%5?Zt&Wc18OWCCBF%5?F%5?[thm AZCB at BAC atV [ BBCJdA st6; BCCӨ`B st]GbL BG6CtB ct BH"C! ctw| BIdxC utaf BJ7C.B utlq BV5Ch B ft BWCD C ft BXXcCxUMC xt?D BY$1CB xt  CCC.C ht^c CD1 C|sB ht CEuN#CҴFC ztFK CFE/qC>B zt;@ CRCWC ktAF CS*CTC kt CT`C[C }t7=<B CU?C/iC }tuz CYC DC mtdTiY CZ#CժC mt DAYC;ɊC t$) DBECaC t CblCBK]D ׹B?tbr6CtBKDMC?tscg5Ch BC[R?ty'cmC.C[C?tS'dbCWCjCC?t 0dhC DC~%CuC?t vbmCizBMD=FB?t DbnF)lCPTAC@?tbsCJBD|%C?tRbt43CJ”CB?t9chuClRC_B,?tUciCTCCB?tO'cnaCAx(CnjCW?tq'coިCƓC 2CB?t'dc89CfCFC_tC?tZ'ddXCmChuC#͙C?t ^di{Cp,lCNe=CC?t djD|C2CC?t;@CBDًCsB at/4BK=C]DB ct|BZExdC-B ftCGUdC%'C htY^CVJCwC ktpuDCڸC& sC mt{& m2Angle = Angle (ABC) = t{4 m3Angle = Angle (BCA) = t+{B m4Angle = Angle (CAB) = t?V m5 0(Angle (ABC)) + (Angle (BCA)) + (Angle (CAB)) = Sum of angles of the triangle = tK0 Hide Explanation  pontQ Show Explanation P elspont$Reset t$Try your own!  "Arialb