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February 1st, 2010, 04:27 AM  #1 
Member Joined: May 2009 Posts: 90 Thanks: 0  Length of sides and angle calculations
Hi, Can someone tell me how I would find a) the length of side BC and b) the angle B,on the sketch, Thanks....[attachment=0:3nsc7msn]01022010 124526_0063.jpg[/attachment:3nsc7msn] 
February 1st, 2010, 11:38 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,034 Thanks: 2271 
Is AC perpendicular to BD? You need some extra piece of information such as that.

February 1st, 2010, 01:18 PM  #3 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Length of sides and angle calculations Hello, manich44! I explained this at another site. We are given three sides and one angle of a quadrilateral. [color=beige]. . [/color]This does not determine a unique quadrilateral. We find that:[color=beige] .[/color] Then we have , but we know only two of its sides. If , we have a chance . . . ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Hello, skipjack! I thought of that, too. But if , we'd have a kite and 
February 2nd, 2010, 01:23 AM  #4 
Member Joined: May 2009 Posts: 90 Thanks: 0  Re: Length of sides and angle calculations
Sorry Guys, Forgot to include total area = 1276m2 
February 2nd, 2010, 07:19 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,034 Thanks: 2271 
Area triangle ADB = (1/2)(34.37)(41.45)sin(105.24°)m² = 687.26831m² Area triangle CDB = 1276m²  687.26831m² = 588.73169m² By the cosine rule, BD² = (34.37² + 41.45²  2(34.37)(41.45)cos(105.24°))m² = 3648.3673m², so BD = 60.4017164m Area triangle CDB = (1/2)(BD)(CD)sin(angle BDC), so sin(angle BDC) = 2(588.73169)/((60.4017164)(37.62)) = 0.51817844, and so angle BDC = 31.210144° or 148.789856°. Now you can use the cosine rule to find the two possible values for the length of BC. You can then use the sine rule to find angles ABD and CBD, and hence angle ABC. 
February 3rd, 2010, 02:43 AM  #6 
Member Joined: May 2009 Posts: 90 Thanks: 0  Re: Length of sides and angle calculations
Hi, Thanks for the help. I get length BC = 34.30 and angle B 74.6 deg. Can you explain 31.210144° or 148.789856°. though Thanks... 
February 4th, 2010, 03:07 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 21,034 Thanks: 2271 
Nothing in the problem requires that angle BDC is acute. How did you calculate 74.6° for angle ABC?


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