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在\(\triangle ABC\)中,角\(A,B,C\)的对边分别为\(a,b,c\),已知\(\cos {A}=\frac {3} {5}\).
若\(\Delta ABC\)的面积为\(3\),求\(\overrightarrow {AB} \cdot \overrightarrow {AC} \)的值;
参考答案:\(\overrightarrow {AB} \cdot \overrightarrow {AC} = \frac{9}{2}\)(2)\(\sin \left( {B - 2C} \right) = - \frac{{31\sqrt 2 }}{{50}}\)
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