In a Delta ABC,a, b, A are given and c_1, c_2 | vedclass

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

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$\frac{1}{2}b^2 \sin 2A$

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(A) Using the Law of Cosines,we have $\cos A = \frac{c^2 + b^2 - a^2}{2bc}$.This rearranges to the quadratic equation in $c$: $c^2 - (2b \cos A)c + (b^2 - a^2) = 0$.Since $c_1$ and $c_2$ are the roots of this equation,we have $c_1 + c_2 = 2b \cos A$ and $c_1 c_2 = b^2 - a^2$.The area of a triangle with sides $a, b$ and included angle $C$ is $\frac{1}{2}ab \sin C$.The sum of the areas of the two triangles is $S = \frac{1}{2}ab \sin C_1 + \frac{1}{2}ab \sin C_2 = \frac{1}{2}ab(\sin C_1 + \sin C_2)$.From the Law of Sines,$\frac{c}{\sin C} = \frac{b}{\sin B} = \frac{a}{\sin A} = 2R$,so $\sin C = \frac{c \sin A}{a}$.Thus,$\sin C_1 + \sin C_2 = \frac{(c_1 + c_2) \sin A}{a} = \frac{(2b \cos A) \sin A}{a} = \frac{b \sin 2A}{a}$.Substituting this into the sum of areas: $S = \frac{1}{2}ab \left( \frac{b \sin 2A}{a} \right) = \frac{1}{2}b^2 \sin 2A$.

In a $\Delta ABC$,$a, b, A$ are given and $c_1, c_2$ are two values of the third side $c$. The sum of the areas of the two triangles with sides $(a, b, c_1)$ and $(a, b, c_2)$ is

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