In the ambiguous case of a triangle,if a, b a | vedclass

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(B) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.This can be rearranged as a quadratic equation in $c$: $c^2 - (2b\cos A)c + (b^2

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$c_1 - c_2 = 2\sqrt{a^2 - b^2\sin^2 A}$

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(B) Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.This can be rearranged as a quadratic equation in $c$: $c^2 - (2b\cos A)c + (b^2 - a^2) = 0$.Let $c_1$ and $c_2$ be the two roots of this quadratic equation.Then,the sum of roots is $c_1 + c_2 = 2b\cos A$ and the product of roots is $c_1 c_2 = b^2 - a^2$.The difference between the roots is given by $|c_1 - c_2| = \sqrt{(c_1 + c_2)^2 - 4c_1 c_2}$.Substituting the values: $|c_1 - c_2| = \sqrt{(2b\cos A)^2 - 4(b^2 - a^2)}$.$|c_1 - c_2| = \sqrt{4b^2\cos^2 A - 4b^2 + 4a^2} = \sqrt{4a^2 - 4b^2(1 - \cos^2 A)}$.Since $1 - \cos^2 A = \sin^2 A$,we get $|c_1 - c_2| = \sqrt{4a^2 - 4b^2\sin^2 A} = 2\sqrt{a^2 - b^2\sin^2 A}$.

In the ambiguous case of a triangle,if $a, b$ and $A$ are given and there are two possible values for the third side,$c_1$ and $c_2$,then what is the value of $|c_1 - c_2|$?

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