The point O lies inside the triangle EFG, and ∠OEF = ∠OFG = ∠OGE.
Given that EF = 13, FG = 14, and GE =15, determine tan ∠OEF

Name the lengths OE, OF, OG by a,b,c respectively.  Then use the law of cosines once for each triangle and ∠OEF which gives three equations for the three unknowns:
cos( ∠OEF)=(13^2+a^2-b^2)/(2*13*a)= etc.
which is solvable in theory. 
The algebra got horrid quick so I turned to wolfram alpha which tried to give me multiple solutions for b that were roots of a sixth degree polynomial but still in terms of a.  Nope.

So I drew the figure in Geometers Sketchpad.
∠OEF = 29.66117
tan(∠OEF)=.56949
Which doesn't appear to be a nice rational number or the square root of a nice rational.

Posted by Jer on 2016-01-10 20:07:19