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November 20th, 2012, 07:11 PM   #1
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a rhombus

[attachment=0:3nqwwm2w]rhombus.JPG[/attachment:3nqwwm2w]
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 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 = 180-30 = 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
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Re: a rhombus

Quote:
 Originally Posted by Hoempa angle BCE = angle DCE + Angle BCD = 90 + 45 = 135.
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 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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