nitter
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Aaron Parecki
@aaronpk
1 Oct 2011
Replying to
@andormaybe
@amarkor
It gets better! No sin/cos/tan needed! d = (b^2 + c^2 - a^2) / 2c Where d is the missing edge segment and a,b,c are known edges (!)
Oct 1, 2011 · 2:52 AM UTC