November 20th, 2012, 07:11 PM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  a rhombus
[attachment=0:3nqwwm2w]rhombus.JPG[/attachment:3nqwwm2w]

November 20th, 2012, 08:38 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,988 Thanks: 995  Re: a rhombus
Did you give ALL the info? Like, why can't DF be, say, horizontal? 
November 20th, 2012, 11:27 PM  #3 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: a rhombus
sorry an error corrected 
November 21st, 2012, 06:48 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,988 Thanks: 995  Re: a rhombus
Square sides = 1 ; then rhombus sides = SQRT(2) Draw right triangle CFG (BCG is straight line). Angle CFG = 45 (same as angle BDC), angle DCF = 45 : so angle BCE = 135 Triangle BCE: angle BEC = ASIN[SIN(135) / SQRT(2)] = 30 So angle BEF = 18030 = 150 = angle BDF. Mehopes that's clear nuff! 
November 21st, 2012, 06:55 AM  #5 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: a rhombus
To find angle BCE, you could also use: angle DCE = angle BDC = 45 degrees (BD and CE parallel) angle BCE = angle DCE + Angle BCD = 90 + 45 = 135. 
November 21st, 2012, 11:51 AM  #6  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,988 Thanks: 995  Re: a rhombus Quote:
 
November 22nd, 2012, 01:03 AM  #7 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: a rhombus Nice!


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