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In triangle ABC, AB= c = 3, AC= b = 6, BC= a = 8, and D lies on BC such that AD bisects \(\angle BAC \). Find  \(\cos\angle BAC=cos( \alpha) \)
 

\(cosine:\\ a^2=b^2+c^2-2bc\cdot cos(\alpha)\\ cos(\alpha)= \frac{b^2+c^2-a^2}{2bc}=\frac{36+9-64}{2\cdot 6\cdot 3}\)

\(cos(\alpha)= \cos\angle BAC=-0,52\overline 7\)

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so:

\(cos(\alpha)= \cos\angle BAC=-0,52\overline 7 \)

\(\angle BAD=\frac{1}{2} \angle BAC=\frac{1}{2}arccos (-0.52\overline 7)\\ \color{blue}\angle BAD= 60.928° \)

\(cos \angle BAD=0.48591\)

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