If a=3, b=5, c=7 are the sides of a triangle | vedclass

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(C) Using the Law of Cosines,we find the values of $\cos A, \cos B, \cos C$:$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{25+49-9}{2(5)(7)} = \frac{65}{70

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$\frac{83}{15 \sqrt{3}}$

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(C) Using the Law of Cosines,we find the values of $\cos A, \cos B, \cos C$:$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{25+49-9}{2(5)(7)} = \frac{65}{70} = \frac{13}{14}$$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{9+49-25}{2(3)(7)} = \frac{33}{42} = \frac{11}{14}$$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9+25-49}{2(3)(5)} = \frac{-15}{30} = -\frac{1}{2}$Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{3+5+7}{2} = 7.5 = \frac{15}{2}$.$\Delta = \sqrt{\frac{15}{2}(\frac{15}{2}-3)(\frac{15}{2}-5)(\frac{15}{2}-7)} = \sqrt{\frac{15}{2} \cdot \frac{9}{2} \cdot \frac{5}{2} \cdot \frac{1}{2}} = \sqrt{\frac{675}{16}} = \frac{15\sqrt{3}}{4}$.Since $\cot A = \frac{\cos A}{\sin A} = \frac{\cos A}{\Delta / (\frac{1}{2}bc)} = \frac{2bc \cos A}{4\Delta} = \frac{b^2+c^2-a^2}{4\Delta}$,we have:$\cot A + \cot B + \cot C = \frac{b^2+c^2-a^2 + a^2+c^2-b^2 + a^2+b^2-c^2}{4\Delta} = \frac{a^2+b^2+c^2}{4\Delta}$$= \frac{9+25+49}{4(\frac{15\sqrt{3}}{4})} = \frac{83}{15\sqrt{3}}$.

If $a=3, b=5, c=7$ are the sides of a triangle $ABC$,then $\cot A+\cot B+\cot C=$

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