I have a irregular triagle know 2 dimensions 1: 66'0'' long 2:48'6'' long angle betwen is 30 degrees missing 3rd dimenssion to close the triagle an the tow onown angles

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Use general Pitagora's theorem ( if you can not just make the measurements) ...I mean get the long one on the ground ..and after that measure with a protractor the angle ..and draw the second one ..when you draw a line from the ends of the 2 of them you will have the 3rd.
or ..this is the Pitagora's theorem in general cases..
BC2=AB2+AC2-2*AB*AC*cosA
meaning

BC2 =48.6sqre+66sqre -2 x 48.6 x 66 x R(3)/2 whwre R(3) is the square root from 3 and R(3)/2 is the cos 30"
and will be with approximation
BC=24.44

Posted on Jan 03, 2011

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Posted on Jan 02, 2017

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