In a triangle ABC,with usual notations a=2, b | vedclass

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(A) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{c^2+a^2-b^2}{2ca}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting th

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$\frac{19}{30}$

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(A) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{c^2+a^2-b^2}{2ca}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting these into the expression:$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc} + \frac{c^2+a^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc}$$= \frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2abc}$$= \frac{a^2+b^2+c^2}{2abc}$Given $a=2, b=3, c=5$,note that $a+b=c$ $(2+3=5)$,which means the triangle is degenerate (a straight line).However,applying the formula:$= \frac{2^2+3^2+5^2}{2(2)(3)(5)} = \frac{4+9+25}{60} = \frac{38}{60} = \frac{19}{30}$.

In a triangle $ABC$,with usual notations $a=2, b=3, c=5$,then $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=$

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