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 March 7th, 2019, 06:36 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Geometry-a-go-go Code: A b B a a D b C c E Rectangle ABCD, dimensions a by b. BC is extended to E, CE = c, angle CED = 75 degrees. DE = 2a + c. Angle BDC = ? You're welcome  March 7th, 2019, 01:17 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Basic equations: $\frac{b}{c}=\tan(75^\circ)$, $(2a+c)^2=b^2+c^2$. Solve and get $r^2+r\cot(75^\cdot)-\frac{1}{4}=0$, where $r=\frac{a}{b}=\tan(\angle BDC)$. I'll leave the algebra to others. Thanks from topsquark Last edited by skipjack; March 7th, 2019 at 03:26 PM. March 7th, 2019, 02:08 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Agree...which results in angleBDC = 20.989984...degrees. I guess that eliminates a possible non-trig solution: if result was exactly 21 degrees instead, then a non-trig solution would at least "have a chance!". Tags geometryagogo Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post whsvin Geometry 0 February 1st, 2017 07:07 AM becko Math Books 2 December 29th, 2010 08:47 PM Sara so Geometry 4 December 17th, 2010 12:27 PM TungLHang Geometry 2 December 10th, 2010 08:31 PM cursed_mask Geometry 0 July 22nd, 2008 11:52 PM

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