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For a ΔABC\Delta ABCΔABC,
c=3cmc=3cmc=3cm
b=6cmb=6cmb=6cm
cos(∠BAC)=18\cos(\angle BAC)=\frac {1}{8}cos(∠BAC)=81
Find the length of ADADAD, the angle bisector of ∠BAC\angle BAC∠BAC, such that DDD lies on BC‾\overline{BC}BC.
This problem is part of the set Trigonometry.
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