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(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Rearranging the terms,we get the quadratic equation in $b$: $b^2 - (2c \cos A)b +

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$b_1 + b_2 = 2c \cos A$

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(A) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Rearranging the terms,we get the quadratic equation in $b$: $b^2 - (2c \cos A)b + (c^2 - a^2) = 0$.Since $c \sin A According to the properties of quadratic equations,the sum of the roots is given by the negative of the coefficient of $b$ divided by the coefficient of $b^2$.Therefore,$b_1 + b_2 = -(-(2c \cos A)) / 1 = 2c \cos A$.

If in a triangle $ABC$,$a, b, c$ are sides and angle $A$ is given,and $c \sin A < a < c$,and $b_1$ and $b_2$ are two possible values of $b$,then:

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