tag:blogger.com,1999:blog-6933544261975483399.post7604228291277879216..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1074: Quadrilateral, Right Triangle, Angles, 15, 22,5, 30, 90 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-60567450193779806392024-03-13T13:56:04.505-07:002024-03-13T13:56:04.505-07:00There are more general forms of this question.
The...There are more general forms of this question.<br />The first one is, BAD = 120-4t=90, BAC=30-t=22.5, BCA=30, ACD=2t=15 where t=7.5 and the result is 3t=22.5.<br /><br />The second one is, BAD = 60+4t=90, BAC=3t=22.5, BCA=60-4t, ACD=30-2t=15 where t=7.5 and the result is 30-t=22.5<br />I have not investigated yet that the above solutions are also the solutions of the general form. One can also try to give the solution for the general cases.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62010028859915173972016-01-06T09:30:29.421-08:002016-01-06T09:30:29.421-08:00Let O be the circumcentre of ABC and let CO and AD...Let O be the circumcentre of ABC and let CO and AD meet at E.<br /><br />< AOB = 60 so Tr. OAB is equilateral. Also < BOC = 45 so < AOE = 75. Hence <OAC = OCA = 37.5. Hence < DCE = 22.5 and < OAE = 30 so it follows that AO = AE<br /><br />Therefore Tr. BAE is right isoceles and so < AEB = 45 = < DCB implying that BCED is cyclic. <br /><br />So < DBE = < DCE = 22.5 and so x = 22.5<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2268553898912866012015-04-06T13:26:48.643-07:002015-04-06T13:26:48.643-07:00Let O be the center of a circle through B, D and C...Let O be the center of a circle through B, D and C where cuts AC at P so we have triangle BPO equilateral (I) ( Because m∠BOP = 2m∠BCP=30º.Now have m∠BOD = 2.m∠BCD = 90º so #BODA is cyclic because m∠BAD + m∠BOD = 90º+90º= 180º then m∠BAO = m∠BDO = 45º but m∠BAO = m∠BAC+m∠CAO then m∠CAO = 22.5º (II). Circunference ABOD cuts AC at Q but AC bissects ∠BAO then we have BQ = QO then using (I), triangles BPQ and QPO are congruents by equal sides then m∠BPQ = 30º = m∠BAP + m∠ABP ( ∠BPQ external angle) .: m∠ABP = 30º - 22.5º = 7.5º (III) but m∠PBD = m∠PCD = 15º (inscritive angles), finaly using (III) we have m∠ABD = m∠ABP + m∠PBD = 7.5º + 15º .: m∠ABD = 22.5 ºsoloqueuehttps://www.blogger.com/profile/02702247656742384375noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23285305165695796712015-02-05T21:38:48.699-08:002015-02-05T21:38:48.699-08:00With out loss of generality assume AB = 1.
(all a...With out loss of generality assume AB = 1.<br />(all angles are measured in degrees)<br />First note CAD = 67.5,<br />ADC = (180 - 82.5),<br />ABC = (180 - 57.5) <br />By Sine Rule applied to Δ ABC:<br />AC / sin 57.5 = AB / sin 30 = 2.<br />So AC = 2 sin 57.5 = 2 cos 37.5 ..... (i).<br />By Sine Rule applied to ΔADC:<br />AD / sin 15 = AC / sin 82.5 = AC / cos 7.5<br />So AD = AC. [sin 15 / cos 7.5]<br />= 2 [cos 37.5].[2 sin 7.5]<br />= 2[2 cos 37.5sin 7.5] <br />= 2[sin 45 – sin 30] <br />= 2[1/(sqrt 2) - 1/2]<br />= (sqrt 2) - 1 = tan 22.5 <br /> So tan x = AD / AB = (tan 22.5) / 1 = tan 22.5<br />Hence x = 22.5 degrees<br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59482309647207680042015-01-26T05:18:20.649-08:002015-01-26T05:18:20.649-08:00To Peter Tran
I am sory because there was an error...To Peter Tran<br />I am sory because there was an error in my solution in line 5.<br />Thanks for your corection. But I am sure that x = 22.5 degree.<br />I still working to find that solution.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9410237511884477512015-01-24T21:57:48.834-08:002015-01-24T21:57:48.834-08:00To Anonymous
Refer to line 5 "so that m∠DBC ...To Anonymous<br />Refer to line 5 "so that m∠DBC = 105 degree"<br />Please provide more explanation how do you get 105 degree.<br />http://s17.postimg.org/mmbwojrcv/pro_1074.png<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87395610000965516312015-01-23T06:30:53.412-08:002015-01-23T06:30:53.412-08:00Let O is a circle through A, B and D where cuts AC...Let O is a circle through A, B and D where cuts AC and DC at E and F respectively.<br />Let AD and BC intersect at G<br />Note that A, B, E, F and D concyclic<br />Because ∠GBA an exterior angle of triangle ABC then m ∠GBA = 52.5 degree <br />so that m∠DBC = 105 degree. This implies ∠DBC = 30 degree<br />Because BADF cyclic then m∠BAF = 30 degree so that m∠EAF = 7.5 degree = m∠EDF (A, B, E, F and D concyclic)<br />Because ∠AED an exterior angle of triangle ECD then m ∠AED = (7.5 + 15) degree = 22.5 degree = x <br />(From wisna email: putu_wisna_ariawan@yahoo.com)<br />Anonymousnoreply@blogger.com