Geometry

If M is the midpoint of BC, then find AM^2.

$sin\ ( 22.2^{\circ})=\frac{\overline {BM}}{4}\\ \overline {BM}=4\ sin\ ( 22.5^{\circ})\\ \overline {AM}^2 =4^2-\overline {BM}^2=16- 16\ sin^2\ (22.5^{\circ })=16\ (1-sin^2\ (22.5^{\circ }))\\ \color{blue}\overline {AM}^2=13.657$

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