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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!".

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