In triangle ABC,with the usual notations,if s | vedclass

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(A) Key Idea: Use the sine rule,i.e.,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Given,$\sin B \sin C = \frac{bc}{a^2}$.From the sin

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Right angled triangle

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(A) Key Idea: Use the sine rule,i.e.,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Given,$\sin B \sin C = \frac{bc}{a^2}$.From the sine rule,we have $\sin B = \frac{b}{2R}$ and $\sin C = \frac{c}{2R}$.Substituting these into the given equation:$\left(\frac{b}{2R}\right) \left(\frac{c}{2R}\right) = \frac{bc}{a^2}$$\Rightarrow \frac{bc}{4R^2} = \frac{bc}{a^2}$$\Rightarrow 4R^2 = a^2$$\Rightarrow a = 2R$.Since $\frac{a}{\sin A} = 2R$,we have $\frac{2R}{\sin A} = 2R$,which implies $\sin A = 1$.Therefore,$A = 90^{\circ}$.Thus,the triangle is a right-angled triangle.

In $\triangle ABC$,with the usual notations,if $\sin B \sin C = \frac{bc}{a^2}$,then the triangle is $\ldots$.

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