In a triangle ABC,with usual notations,if fra | vedclass

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(C) Using the Sine Rule,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the given equation: $\frac{2 \cos A}{2R \

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$\frac{\pi}{2}$

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(C) Using the Sine Rule,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the given equation: $\frac{2 \cos A}{2R \sin A} + \frac{\cos B}{2R \sin B} + \frac{2 \cos C}{2R \sin C} = \frac{2R \sin A}{(2R \sin B)(2R \sin C)} + \frac{2R \sin B}{(2R \sin C)(2R \sin A)}$ $\frac{1}{R} (\cot A + \frac{1}{2} \cot B + \cot C) = \frac{1}{2R} (\frac{\sin A}{\sin B \sin C} + \frac{\sin B}{\sin C \sin A})$ Multiplying by $2R$: $2 \cot A + \cot B + 2 \cot C = \frac{\sin^2 A + \sin^2 B}{\sin A \sin B \sin C}$ Using $\sin^2 A + \sin^2 B = \sin^2(A+B) = \sin^2 C$,the $RHS$ becomes $\frac{\sin^2 C}{\sin A \sin B \sin C} = \frac{\sin C}{\sin A \sin B} = \frac{\sin(A+B)}{\sin A \sin B} = \frac{\sin A \cos B + \cos A \sin B}{\sin A \sin B} = \cot B + \cot A$. Substituting back: $2 \cot A + \cot B + 2 \cot C = \cot B + \cot A$ $\cot A + 2 \cot C = 0$. Since $\cot A = -2 \cot C$,and using the property of triangles,this leads to $\angle A = \frac{\pi}{2}$.

In a triangle $ABC$,with usual notations,if $\frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$,then $\angle A = $

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