If in triangle ABC,cos A = frac{sin B}{2sin C | vedclass

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(B) Given $\cos A = \frac{\sin B}{2\sin C}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin B = \frac{

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Isosceles

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(B) Given $\cos A = \frac{\sin B}{2\sin C}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$.Substituting these into the given equation:$\cos A = \frac{b/2R}{2(c/2R)} = \frac{b}{2c}$.Using the Cosine Rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Equating the two expressions for $\cos A$:$\frac{b^2 + c^2 - a^2}{2bc} = \frac{b}{2c}$.Multiplying both sides by $2bc$:$b^2 + c^2 - a^2 = b^2$.$c^2 - a^2 = 0$ $\Rightarrow c^2 = a^2$ $\Rightarrow c = a$.Since two sides are equal,the triangle is isosceles.

If in triangle $ABC$,$\cos A = \frac{\sin B}{2\sin C}$,then the triangle is

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