January 4th, 2014, 12:56 PM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  Rhombus.
ABCD is Rhombus. Measuring the angle (A)=40 Point m belongs to the segment AD Measuring the angle (m BD)=30 find Measuring the angle mCD 
January 4th, 2014, 06:42 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,065 Thanks: 1621 
It's easy to show that it's 10° by using trigonometry. A geometrical proof is achievable, based on the diagram below. [attachment=0:28nbddm5]Rhombus.gif[/attachment:28nbddm5] 
January 4th, 2014, 09:26 PM  #3 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  Re: Rhombus.
can you send the by using trigonometry. solve

January 4th, 2014, 10:00 PM  #4  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,621 Thanks: 845  Re: Rhombus.
Nice one Skip! I got to angleDCM = angleBMC  40, but tires started spinning at that point... Quote:
I can supply a hint if YOU're willing to do some work: let the side lengths = 1 then, since triangle ABM is isosceles and angleAMB = 100, then AM = BM = .5 / SIN(50)  
January 5th, 2014, 04:41 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,065 Thanks: 1621 
Further hint: using .5 = sin(30°), MD = 1  BM = (sin(50°)  sin(30°))/sin(50°) = 2cos(40°)sin(10°)/sin(50°) = 2sin(10°). 

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