In a triangle ABC,if a=5, b=3, c=7,then sqrt{ | vedclass

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(A) Using the Law of Cosines:$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3^2 + 7^2 - 5^2}{2 	imes 3 	imes 7} = \frac{9 + 49 - 25}{42} = \frac{33}{

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$\frac{4}{7}$

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(A) Using the Law of Cosines:$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{3^2 + 7^2 - 5^2}{2 	imes 3 	imes 7} = \frac{9 + 49 - 25}{42} = \frac{33}{42} = \frac{11}{14}$$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{5^2 + 7^2 - 3^2}{2 	imes 5 	imes 7} = \frac{25 + 49 - 9}{70} = \frac{65}{70} = \frac{13}{14}$Using the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$,so $\sin A = ak, \sin B = bk, \sin C = ck$.Since $\sin(A+B) = \sin(180^\circ - C) = \sin C$,we have:$\frac{\sin(A-B)}{\sin(A+B)} = \frac{\sin A \cos B - \cos A \sin B}{\sin C} = \frac{ak \cos B - \cos A bk}{ck} = \frac{a \cos B - b \cos A}{c}$Substituting the values:$= \frac{5 	imes \frac{13}{14} - 3 	imes \frac{11}{14}}{7} = \frac{\frac{65-33}{14}}{7} = \frac{32}{14 	imes 7} = \frac{32}{98} = \frac{16}{49}$Therefore,$\sqrt{\frac{\sin(A-B)}{\sin(A+B)}} = \sqrt{\frac{16}{49}} = \frac{4}{7}$.

In a triangle $ABC$,if $a=5, b=3, c=7$,then $\sqrt{\frac{\sin(A-B)}{\sin(A+B)}}=$

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