tag:blogger.com,1999:blog-6933544261975483399.post293338685993538446..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 819: Quadrilateral, Triangle, Angles, 30 degreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger23125tag:blogger.com,1999:blog-6933544261975483399.post-6394338911479363792024-02-09T20:07:25.333-08:002024-02-09T20:07:25.333-08:00<BCA=40
<ABC=100
sin100/AC=sin40/BC
BC=ACsin...<BCA=40<br /><ABC=100<br />sin100/AC=sin40/BC<br />BC=ACsin40/sin100<br />CD=BC=ACsin40/sin100<br />sin30/CD=sinx/AC<br />CD=ACsin30/sinx<br />sin40/sin100=sin30/sinx<br />sin40sinx=sin100sin30<br />2sin40sinx=sin100<br />sinx=cos40<br />x=50Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22234340955343900692021-02-07T20:00:17.994-08:002021-02-07T20:00:17.994-08:00Draw BE as perpendicular to AC and CF perpendicula...Draw BE as perpendicular to AC and CF perpendicular to AD. Join EF.<br />BE bisects AC,since AB = BC<br />E being the midpoint of the hypotenuse AC of the right triangle AFC,<br />EA = EC = EF and consequently ΔEFC is equilateral.<br />Now consider the right triangles BEC and DFC.<br />By Pythagoras, DF^2 = CD^2 - CF^2 = BC^2 - CE^2 = BE*2.<br />So DF = BE and consequently the right triangles CDF and CBE are congruent.<br />Follows x = ∠FBC.<br />Note that ΔABC is isosceles (with base angles FCB, FAB each 40°),<br />Hence x = ∠FBC = 90° - ∠FCB = 90° - 40° = 50°<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41897531549784627232015-11-06T11:52:45.569-08:002015-11-06T11:52:45.569-08:00Let D' on AD so that CD'=CD and <AD'...Let D' on AD so that CD'=CD and <AD'C is obtuse and let B' be the circumcenter of tr. AD'C. Clearly tr CB'D' is equilateral and CB'=AB'=CD'=CB=AB, so B'=B, thus <AD'C=180-m(<ABC)/2=130 degs. and, as constructed, <ADC=<CD'D=180-m(<AD'C)=50 degs.<br /><br />Best regardsStan Fulgerhttps://www.facebook.com/stan.fulgernoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28539505795195429542015-07-01T09:13:50.928-07:002015-07-01T09:13:50.928-07:00Another proof. Find E on AD such that AB = BE. The...Another proof. Find E on AD such that AB = BE. Then < EBC = 60 and Tr. EBC is equilateral, Tr. ECD is isoceles and x = 50. <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88735146528147719652015-06-24T11:30:15.345-07:002015-06-24T11:30:15.345-07:00Draw altitudes BH and CG. Then by 30-60-90 Triangl...Draw altitudes BH and CG. Then by 30-60-90 Triangle and congruence x=50<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74730720523015906852013-10-10T06:52:38.901-07:002013-10-10T06:52:38.901-07:00http://www.mathematica.gr/forum/viewtopic.php?f=20...http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32307&p=149499Μιχάλης Νάννοςhttps://www.blogger.com/profile/02379101429577964881noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21587122022311194802013-02-03T13:55:18.009-08:002013-02-03T13:55:18.009-08:00En la figura adjunta resuelvo el problema sin usar...En la figura adjunta resuelvo el problema sin usar CD...:D<br />http://www.subirimagenes.com/otros-ahorasi-8277920.html<br /><br />By Tony García.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72284219393561461682012-12-10T20:34:29.740-08:002012-12-10T20:34:29.740-08:00Sorry for the extra zeros in the solution, they al...Sorry for the extra zeros in the solution, they all refers degrees.<br />Thank You.Anonymoushttps://www.blogger.com/profile/11413383907798175910noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23970690793623454582012-12-10T06:37:06.307-08:002012-12-10T06:37:06.307-08:00Let E be the point on AD such that AB= BE.
ang. AE...Let E be the point on AD such that AB= BE.<br />ang. AEB= 700 (∆ ABE is isosceles) <br />ang ABC= 1000= ang. ABE+ ang. EBC → ang EBC= 600<br />Since BE=BC (by our construction) so, <br />ang. BEC= ang. ECB = 1800-600/2= 600<br />This implies ∆ BEC is equilateral. → CD= CE.<br />so ang. X= ang. CED = 1800-(700+600) = 500<br />□<br />Anonymoushttps://www.blogger.com/profile/11413383907798175910noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84912804266430238982012-11-20T23:01:26.581-08:002012-11-20T23:01:26.581-08:00Let BA=BC=CD=a and AC=d
Using sine rule,
In triang...Let BA=BC=CD=a and AC=d<br />Using sine rule,<br />In triangle ABC, a/sin 40=d/sin 100…(1)<br />In triangle ACD, a/sin 30=d/sin x…(2)<br />Using (1),(2), we get, sin 100/sin40=sin x/sin 30<br /> sin 80/sin40=sin x/sin 30<br /> 2(sin 40)(cos 40)/sin40=sin x/(1/2)<br /> cos 40=sin x<br /> sin (90-40)=sin x<br /> sin 50=sin x<br /> 50=x.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49971042860841197812012-11-04T08:48:39.851-08:002012-11-04T08:48:39.851-08:00x=50deg if AD>AC; or x=130deg IF AD<AC.x=50deg if AD>AC; or x=130deg IF AD<AC.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76065747942595933532012-11-02T03:39:06.411-07:002012-11-02T03:39:06.411-07:00Draw BE perpendicular to AC and CF perpendicular t...Draw BE perpendicular to AC and CF perpendicular to AD. Join EF.<br />BE bisects AC, since AB = BC<br />E being the midpoint of the hypotenuse AC of the right triangle AFC, <br />EA = EC = EF and consequently ΔEFC is equilateral.<br />Now consider the right triangles BEC and DFC.<br />By Pythagoras, DF^2 = CD^2 - CF^2 = BC^2 - CE^2 = BE*2.<br />So DF = BE and consequently the right triangles CDF and CBE are congruent.<br />Follows x = ∠FBC.<br />Note that ΔABC is isosceles (with base angles FCB, FAB each 40°),<br />Hence x = ∠FBC = 90° - ∠FCB = 90° - 40° = 50°<br /><br /><br /> Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20381910627057536712012-11-01T02:53:55.483-07:002012-11-01T02:53:55.483-07:00To Anonymous: I think the problem has two solution...To Anonymous: I think the problem has two solutions. First, according to the figure, when the angle x is acute and the second, when the angle x is obtuse. <br />Thanks.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34261849687392332822012-11-01T02:18:01.682-07:002012-11-01T02:18:01.682-07:00Antonio, did you judge the answer according to fig...Antonio, did you judge the answer according to figure?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75084863637127922092012-10-29T10:08:09.317-07:002012-10-29T10:08:09.317-07:00Costruind simetricul punctului C fata de dreapta A...Costruind simetricul punctului C fata de dreapta AD si notandu-lcu P obtinem :<br />ΔACP echilateral si ΔACP=ΔPCD(LLL);∠CDP = ∠ABC = 100°,AD mediatoarea seg AD =>AD bisectoarea ∠CDP => ∠CDA = 50° <br />Prof Radu Ion,Sc.Gim.Bozioru,BuzauProf Radu Ion,Sc.Gim.Bozioru,Buzaunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67991547423188560292012-10-29T08:01:36.144-07:002012-10-29T08:01:36.144-07:00draw the altitudes BM of triangle ABC,CH of triang...draw the altitudes BM of triangle ABC,CH of triangle ACD<br />CM=AC/2 (ABC is isoscele)<br />CH=AC/2 (triangle 30-60-90)<br />the right triangles BMC and CHD are congruent<br />x=50Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88019027972782253662012-10-29T03:21:59.471-07:002012-10-29T03:21:59.471-07:00Let E be the circumcenter of ΔACD.
Then
∠CED = 2...Let E be the circumcenter of ΔACD. <br /><br />Then<br />∠CED = 2×∠CAD = 60°, CD = CE<br />⇒ ΔCDE is equilateral<br />⇒ AB = BC = CD = DE = CE = AE<br />⇒ ABCE is a rhombus<br />⇒ ∠AEC = ∠ABC = 100°<br />⇒ ∠ADC = 1/2×∠AEC = 50°Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8326981339018009242012-10-28T19:28:39.778-07:002012-10-28T19:28:39.778-07:00Problem 819: Try to use elementary geometry (Eucli...Problem 819: Try to use elementary geometry (Euclid's Elements).Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56188133752581891672012-10-28T19:28:12.143-07:002012-10-28T19:28:12.143-07:00To Anonymous: according to the figure angle x is a...To Anonymous: according to the figure angle x is acute, therefore x=50. If D' is on AD so that AB=BC=CD', then angle x is obtuse (x=130).Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73294533630961491962012-10-28T15:20:00.896-07:002012-10-28T15:20:00.896-07:00Why x=50 only?
I believe x=50 or 130.Why x=50 only?<br />I believe x=50 or 130.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2004669450237856712012-10-28T14:59:03.899-07:002012-10-28T14:59:03.899-07:00why x not= 130?
I believe x=50 or 130.why x not= 130?<br />I believe x=50 or 130.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43722224371017656332012-10-27T23:39:38.284-07:002012-10-27T23:39:38.284-07:00In triangle ABC:
AC = 2 AB cos(40) ---- (1)
In ...In triangle ABC: <br />AC = 2 AB cos(40) ---- (1)<br />In triangle ACD:<br />AC = CD sin(x)/sin(30) = 2 CD sin(x) ---- (2)<br />AB = CD;<br />Then by (1) and (2):<br />sin(x) = cos(40) = sin(50)<br />x = 50Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58087376701950288432012-10-27T17:31:49.214-07:002012-10-27T17:31:49.214-07:00Triangle ABC is isosceles, AC = 2ABcos40
Using sin...Triangle ABC is isosceles, AC = 2ABcos40<br />Using sine rule on triangle ACD, <br />AC/sinx = CD/sin30<br />2ABcos40/sinx = CD/sin30<br />cos40 = sinx<br />x = 50W Fungnoreply@blogger.com