tag:blogger.com,1999:blog-6933544261975483399.post2210624589660899589..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 42Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-26409598907633595662022-11-28T07:38:56.015-08:002022-11-28T07:38:56.015-08:00[For the sake of typing convenience, I will use &q...[For the sake of typing convenience, I will use "a" instead of alpha for below calculation]<br /><br />Join BD<br /><CBD=<CDB=90-a<br />reflex <ADC=360-x-3a<br /><ADB=360-x-3a-(90-a)=270-x-2a<br /><br />Consider triangle ABD<br />sin(a)/BD=sin(270-x-2a)/AB<br />BD/AB=sin(a)/sin(270-x-2a)-----------(1)<br /><br />Consider triangle BCD<br />sin(2a)/BD=sin(90-a)/CD<br />BD/CD=sin(2a)/cosa=2sina----------(2)<br /><br />As AB=CD, By equating (1) & (2)<br />sina/sin(270-x-2a)=2sina<br />sin(270-x-2a)=1/2<br />270-x-2a=30 or 150<br />x+2a=240 (rej.) or 120<br />x=120-2aMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41855445369354614842019-07-01T08:40:12.625-07:002019-07-01T08:40:12.625-07:00Define B’ symmetric of B by AD => AB=AB’
ang BA...Define B’ symmetric of B by AD => AB=AB’<br />ang BAD = a therefore ang BAB’ = 2a<br />Therefore ΔBAB’ is congruent to ΔBCD, and BB’=BD, BD=DB’<br />Therefore ΔBB’D is equilateral<br />Draw a line passing through B and perpendicular to AC meeting in E<br />AB=BC => BE is angle bisector of ang ABC<br />The drawing is symmetric, therefore BE is also angle bisector of ang DBB’<br />AD is angle bisector of ang BDB’ and CB’ is angle bisector of ang BB’D<br />By symmetry lines AD, CB’ and EB meet in point P<br />P is on angle bisector of ang DBB’, on angle bisector ofang BB’D and on angle bisector of ang BDB’ therefore P is the center of ΔBDB’<br />Therefore PD=PB=PB’<br />Consider now ΔAPB<br />Since ΔBB’D is equilateral, ang APB = 120<br />ang APB +ang PBA +ang BAP = 180<br />120+x/2+a=180 =>x+2a=120 => x=120-2a<br />rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62302262786425441772015-12-24T12:35:28.396-08:002015-12-24T12:35:28.396-08:00Hello, I have seen a very similar problem: B and C...Hello, I have seen a very similar problem: B and C are known angles (ie 110 and 20), the angle to find is the one in A position.<br />Never could solve that problem, I think now it will possible for me if I try to follow the comments here.<br />Thank you!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27106677108870094892015-10-12T11:02:24.761-07:002015-10-12T11:02:24.761-07:00Drop a perpendicular from B to AD say. BE and from...Drop a perpendicular from B to AD say. BE and from C to BD say CF. Tr.s DFC, BFC and ABE are all congruent ASA. So DF = BF = BE, hence Tr. BDF is 30 60 90 with < BDE = 30. <br /><br />So < ABD = 30-@ and since < FBC =90-@, x = 120-2@<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16414360416576891332012-04-12T10:07:04.201-07:002012-04-12T10:07:04.201-07:00Yes, Thanks. Following your solution.Yes, Thanks. Following your solution.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37872091383335978992012-03-29T08:52:53.793-07:002012-03-29T08:52:53.793-07:00Draw a circle having its center B and radius BA,
D...Draw a circle having its center B and radius BA,<br />Draw another circle having its center C and radius CB. <br />Let K be the lower intersection point of two circles.<br />Connect BK and CK<br />So, Tri. BKC is an equlateral.<br />Obviously ang.KBC=60deg<br />Since ang.BCK=60deg, ang.DCK=60-2α.<br />Since, ang.DCK is angle on center and ang.DBK is angle on circumferance subtended by same segment: ang.DBK=30-α<br />similarly BCD=2.BKD<br />ang.BKD=α<br />now connect BD<br />as per the symetrical property in triangles ADB and KBD, ang.ABD also 30-α<br />so, ang.ABC=X=ang.ABD+ang.KBD+ang.KBC<br />therefore X=30-α+30-α+60<br />ie: X=120-2α<br />............Anonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40743107006550498722012-03-29T08:39:06.798-07:002012-03-29T08:39:06.798-07:00Sir, Did u receive my solution for Problem 42?Sir, Did u receive my solution for Problem 42?Anonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33046615108616093342010-01-29T10:30:04.366-08:002010-01-29T10:30:04.366-08:00draw bisector of ang C
draw ang DAE = α, E on bise...draw bisector of ang C<br />draw ang DAE = α, E on bisector of C<br />join B to D, D to E, draw altitude BH, H on DE <br /><br />tr BDE is equilateral<br />ang DBH = ang DAC ( perpendicular sides )<br />ang KDE = ang DAC ( AD meetBE on K, AD = EC => DE//AC)<br />ang FED = ang ECA (CE meet BD on F, DE//AC )<br /><br />BH bisector of DBE, DK bisector of EDB, EF of DEB<br />=> DBE = 6 ∙ DAC = 180º<br />=> DAC = 30º<br /><br />from tr ABC<br />A + C + B =( α + 30 ) + ( 30 + α ) + x = 180<br /><br /> x = 120° - 2α<br />-----------------------------------------------c .t . e. onoreply@blogger.com