In a Delta ABC,a, c, A are given and b_1, b_2 | vedclass

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(B) Using the Law of Cosines,we have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.This rearranges to the quadratic equation in $b$: $b^2 - (2c \cos A)b + (c

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$\sqrt{\frac{9a^2 - c^2}{8c^2}}$

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(B) Using the Law of Cosines,we have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.This rearranges to the quadratic equation in $b$: $b^2 - (2c \cos A)b + (c^2 - a^2) = 0$.Given that $b_1$ and $b_2$ are the roots of this equation,we have $b_1 + b_2 = 2c \cos A$ and $b_1 b_2 = c^2 - a^2$.Since $b_2 = 2b_1$,we get $3b_1 = 2c \cos A \Rightarrow b_1 = \frac{2c \cos A}{3}$ and $2b_1^2 = c^2 - a^2$.Substituting $b_1$ into the second equation: $2(\frac{2c \cos A}{3})^2 = c^2 - a^2$.$2(\frac{4c^2 \cos^2 A}{9}) = c^2 - a^2 \Rightarrow 8c^2 \cos^2 A = 9(c^2 - a^2)$.Using $\cos^2 A = 1 - \sin^2 A$,we get $8c^2(1 - \sin^2 A) = 9c^2 - 9a^2$.$8c^2 - 8c^2 \sin^2 A = 9c^2 - 9a^2 \Rightarrow 8c^2 \sin^2 A = 9a^2 - c^2$.Therefore,$\sin A = \sqrt{\frac{9a^2 - c^2}{8c^2}}$.

In a $\Delta ABC$,$a, c, A$ are given and $b_1, b_2$ are two values of the third side $b$ such that $b_2 = 2b_1$. Then $\sin A = $

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