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November 20th, 2012, 06:11 PM   #1
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a rhombus

[attachment=0:3nqwwm2w]rhombus.JPG[/attachment:3nqwwm2w]
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November 20th, 2012, 07:38 PM   #2
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Re: a rhombus

Did you give ALL the info?
Like, why can't DF be, say, horizontal?
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November 20th, 2012, 10:27 PM   #3
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Re: a rhombus

sorry an error corrected
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November 21st, 2012, 05:48 AM   #4
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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!
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November 21st, 2012, 05:55 AM   #5
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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.
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November 21st, 2012, 10:51 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, 12:03 AM   #7
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Re: a rhombus

Nice!
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