tag:blogger.com,1999:blog-6933544261975483399.post5915392432633486130..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 848: Special Right triangle, Catheti or legs ratio 1:2, 26.5 Degrees. Double angleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-38268865578769733662013-06-23T07:34:49.839-07:002013-06-23T07:34:49.839-07:00Problem 848 Hints
1 Draw the cevian AD (D on AC) ...Problem 848 Hints <br />1 Draw the cevian AD (D on AC) so that AD = DC<br />2 Prove that triangle ABD is the special right triangle 3k-4k-5k (angles aprox. 37 and 53 degrees).<br />3 Apply 180 degrees in isosceles triangle ADC<br />Good luckAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63267301730168359872013-06-23T03:00:42.915-07:002013-06-23T03:00:42.915-07:00i did not get the solution
please helpi did not get the solution<br />please helpAnonymoushttps://www.blogger.com/profile/00908049546546731829noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19897459848961631182013-06-21T17:33:05.517-07:002013-06-21T17:33:05.517-07:00Dear Pranav,
I apologize for the problem (Site Tem...Dear Pranav,<br />I apologize for the problem (Site Temporarily Unavailable), I am looking for a solution. In the meantime you can see this problem at http://agutie.homestead.com/files/geometry/p848-special-right-triangle-cathetus-1-2-angle-26.5-degrees.htmAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40588991960346871122013-06-21T09:44:42.506-07:002013-06-21T09:44:42.506-07:00I did not get this problem
dear antonio gutierrez ...I did not get this problem<br />dear antonio gutierrez can u post this soln<br />my teachers the problem is immpossible<br />Anonymoushttps://www.blogger.com/profile/00908049546546731829noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12668726450752500862013-01-28T18:48:22.392-08:002013-01-28T18:48:22.392-08:00Draw a perpendicular BD from B to AC and the Let t...Draw a perpendicular BD from B to AC and the Let the two circles intersect in E.<br />Now, AC = sqrt(5)*c while BD * AC = AB * BC or BD = (c * 2c)/(sqrt(5)*c) = 2c/sqrt(5)<br />We can say that /_ACB//_ECB ~= BD/BE = 2/sqrt(5) or /_ACB = /_ECB * 2/sqrt(5) = 60/sqrt(5)<br />or /_ ACB ~= 26.83281Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65534731555876826012013-01-27T04:41:26.741-08:002013-01-27T04:41:26.741-08:00 we have tan c=c/2c
then tanc =0.5
... we have tan c=c/2c<br /> then tanc =0.5<br /> c = shift tan(0.5)= 26.56Anonymoushttps://www.blogger.com/profile/01933091843322188770noreply@blogger.com