If in a triangle ABC,2cos A = sin B csc C,the | vedclass

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(C) Given the equation: $2 \cos A = \sin B \csc C$Using the sine rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,so $\frac{\sin B}{\sin C} = \frac{b}{c}$.S

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$c = a$

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(C) Given the equation: $2 \cos A = \sin B \csc C$Using the sine rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,so $\frac{\sin B}{\sin C} = \frac{b}{c}$.Substituting this into the given equation: $2 \cos A = \frac{b}{c}$.Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting $\cos A$ into the equation: $2 \left( \frac{b^2 + c^2 - a^2}{2bc} \right) = \frac{b}{c}$.$\frac{b^2 + c^2 - a^2}{bc} = \frac{b}{c}$.Multiplying both sides by $bc$: $b^2 + c^2 - a^2 = b^2$.$c^2 - a^2 = 0 \Rightarrow c^2 = a^2$.Since $a$ and $c$ are side lengths of a triangle,$c = a$.

If in a triangle $ABC$,$2\cos A = \sin B \csc C$,then

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