In triangle ABC,with usual notations,if cos A | vedclass

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

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right angled triangle

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(D) Given $\cos A = \frac{\sin B}{\sin C}$.Using the Sine Rule,$\frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$.Substituting these into the given equation: $\cos A = \frac{b/2R}{c/2R} = \frac{b}{c}$.Using the Cosine Rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Equating the two expressions: $\frac{b^2 + c^2 - a^2}{2bc} = \frac{b}{c}$.Multiplying both sides by $2bc$: $b^2 + c^2 - a^2 = 2b^2$.Rearranging the terms: $c^2 = a^2 + b^2$.Since the square of one side is equal to the sum of the squares of the other two sides,the triangle satisfies the Pythagorean theorem.Therefore,$	riangle ABC$ is a right-angled triangle.

In $\triangle ABC$,with usual notations,if $\cos A = \frac{\sin B}{\sin C}$,then the triangle is $......$

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