If M is the midpoint of BC, then find AM^2.
\(sin(22.2^\circ)=\frac{\overline{BM}}{4}\\\overline{BM}=4\:sin(22.5^\circ)\\\overline{AM}^2=4^2-\overline{BM}^2=16-16\:sin^2(22.5^\circ) =16(1-sin^2(22.5^\circ)\\\textcolor{red}{\overline{AM}^2=13.657}\)
Credits to asinus!
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