tag:blogger.com,1999:blog-6933544261975483399.post2589713915921314192..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 727: Triangle, Isosceles, Congruence, Angles, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-7810909440054590262019-10-08T05:34:34.420-07:002019-10-08T05:34:34.420-07:00Solucion enviada por Pedro Miranda de Peru
Ver sol...Solucion enviada por Pedro Miranda de Peru<br /><a href="http://www.gogeometry.com/problem/p727-solucion-p-miranda.jpg" rel="nofollow">Ver solucion aqui</a>Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70306245673734434242016-07-22T14:32:23.216-07:002016-07-22T14:32:23.216-07:00Problem 727
Forms the isosceles trapezoid ABEC (t...Problem 727<br />Forms the isosceles trapezoid ABEC (the point Eis the right of point B and above the point C).<br />(ΒΕ//ΑC,<BAC=88=<ECA). Then AB=BD=EC=BE and BC is bisector <ACE.So ABEC is cyclic. But triangle ABD=triangle CED (AB=BD=CE, AD=DC,<BAD=<ECD).Then BD=DE=BE, so triangle BED is equilateral .Therefore x=<DBC=<DBE-<CBE=60-44=16.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37569070745130035312015-05-16T20:19:09.024-07:002015-05-16T20:19:09.024-07:00Dear Ajit,
sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/...Dear Ajit,<br /><br />sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2) <br /><br />sin(x)*cos(24-x/2)=2sin(24-x/2)*cos(24-x/2)*sin(22+x/2), hence cos(24-x/2) get cancelled.<br /><br />sin(x)=2sin(24-x/2)*sin(22+x/2). Using product to sum formula,<br /><br />cos(90-x)=cos(2-x)-cos46<br /><br />cos(2-x)-cos(90-x)=cos46. After using sum to product formula,<br /><br />-2sin(46-x)*sin(-44)=sin44<br /><br />-2sin(46-x)*[-sin44]=2sin30*sin44<br /><br />2sin44*sin(46-x)=2sin30*sin44, hence 2sin44 get cancelled.<br /><br />Finally sin(46-x)=sin30, then x=16.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48324931176469982082012-02-25T23:44:46.003-08:002012-02-25T23:44:46.003-08:00Thanks, Pravin.
My email: ajitathle@gmail.com
Need...Thanks, Pravin.<br />My email: ajitathle@gmail.com<br />Need ur email address.<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27522440928955978112012-02-12T22:27:53.410-08:002012-02-12T22:27:53.410-08:00Dear Ajit:
sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2...Dear Ajit:<br />sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)<br />=2sin(24-x/2)*cos(24-x/2)*sin(22+x/2)<br />Observe sin(66+x/2)and cos(24-x/2)get cancelled <br />since the angles are complementary<br />We now have<br />sin(x)= 2sin(24-x/2)*sin(22+x/2)= cos(2-x)-cos(46)<br />cos(46)=cos(2-x)-sin(x)=sin(88+x)-sin(x)=2cos(44+x)sin(44)<br />Again cos(46)and sin(44)get cancelled <br />since the angles are complimentary<br />So we are left with <br />1=2cos(44+x)<br />cos(44+x)= cos(60)<br />x=16 deg<br />PravinPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67020076249665043672012-02-12T18:33:01.355-08:002012-02-12T18:33:01.355-08:00A trigonometric Proof:
B = 48°
∠ABD = 48° - x
ΔABD...A trigonometric Proof:<br />B = 48°<br />∠ABD = 48° - x<br />ΔABD is isosceles<br />AD = 2 AB sin(∠ABD/2)= 2 c sin(24° - x/2)<br />∠BAD = ∠BDA = 66° + x/2<br />∠DAC = ∠DCA = 22° - x/2<br />ΔABD is isosceles<br />b = AC = 2 AD cos (22° - x/2)<br />b = 4c sin(24° - x/2) cos(22° - x/2)<br />sin B = 4 sin C sin(24° - x/2) cos(22° - x/2)<br />sin 48° = 4 sin 44° sin(24° - x/2) cos(22° - x/2)<br />sin 48° = (2 sin 44°)[2 sin(24° - x/2) cos(22° - x/2)]<br />sin 48° = (2 sin 44°)[sin(46° - x) + sin 2°]<br />cos 42° = 2 cos 46° sin(46° - x) + 2 sin 44°sin 2°<br />cos 42° = 2 cos 46° sin(46° - x) + cos 42° - cos 46°<br />sin(46° - x) = 1/2<br />46° - x = 30°<br />x = 16°Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41349551737285569402012-02-12T17:45:11.064-08:002012-02-12T17:45:11.064-08:00Nice proof, Peter. I applied Ceva's Trigonomet...Nice proof, Peter. I applied Ceva's Trigonometric Theorem to Tr. ABC and landed up with: sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)<br />Can anyone tell how this is solved the answer, of course, being x=16 deg.? <br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86662427845836007922012-02-11T19:06:29.091-08:002012-02-11T19:06:29.091-08:00http://img197.imageshack.us/img197/8941/problem727...http://img197.imageshack.us/img197/8941/problem727.png<br /><br />Let DD’ is the perpendicular bisector of AC and DD’ cut BC at M<br />AM is the angle bisector of angle BAC ( see picture)<br />Draw BK//AC and locate point L on BK such that BL=BA=BD<br />Note that m(BAM)=m(MAC)=44<br />Triangle ABL is isosceles => m(BLA)=1/2*(180-92)=44<br />So m(BAM)=m(BAL)=44 => A, M ,L are collinear <br />Triangle BKM congruence to LKM ( case ASA)<br />so BK=KL => triangle BDL is equilateral<br />and m(DBL)=60 => x=60-44=16Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com