tag:blogger.com,1999:blog-6500033298667240354.post3616000125953999129..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 2638foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6500033298667240354.post-80942267119320088942010-03-14T03:08:15.488-07:002010-03-14T03:08:15.488-07:00Let the bigger isosceles triangle be ABC with BC a...Let the bigger isosceles triangle be ABC with BC as the base and let the smaller isosceles triangle be DBC with DC as the base. Let AD & BC intersect in M. Now /_DAC =/_DBC = α and if AB=a then BC=2asin(α/2) while DC=4a(sin(α/2))^2<br />Since Tr.ABM and DMC are similar, we've A/D =(AB/DC)^2 or 9 = a^2/16a^2(sin(α/2))^4 which gives us: (sin(α/2))^4 =1/144 or sin(α/2)=√(1/12) and cos(α/2)=√(11/12). Finally, <br />cos(α)=(cos(α/2))^2 (sin(α/2))^2 = 11/12-1/12 =10/12 = 5/6 or Option D.<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com