March 7th, 2019, 06:36 AM  #1 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024  Geometryagogo Code: A b B a a D b C c E 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: 6,754 Thanks: 695 
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. 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: 14,412 Thanks: 1024 
Agree...which results in angleBDC = 20.989984...degrees. I guess that eliminates a possible nontrig solution: if result was exactly 21 degrees instead, then a nontrig solution would at least "have a chance!". 

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