In a triangle ABC, if sin A sin B = frac{ab}{ | vedclass

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(C) Given the relation: $\sin A \sin B = \frac{ab}{c^2}$ Using the Sine Rule, we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2

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right angled

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(C) Given the relation: $\sin A \sin B = \frac{ab}{c^2}$ Using the Sine Rule, we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. This implies $\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, and $\sin C = \frac{c}{2R}$. Substituting these into the given equation: $\left(\frac{a}{2R}ight) \left(\frac{b}{2R}ight) = \frac{ab}{c^2}$ $\frac{ab}{4R^2} = \frac{ab}{c^2}$ Since $a, b 
eq 0$, we can cancel $ab$ from both sides: $\frac{1}{4R^2} = \frac{1}{c^2}$ $\Rightarrow c^2 = 4R^2$ $\Rightarrow c = 2R$. Since $c = 2R$, we have $\frac{c}{\sin C} = 2R \Rightarrow \sin C = \frac{c}{2R} = \frac{2R}{2R} = 1$. Therefore, $C = 90^{\circ}$. Thus, the triangle is a right-angled triangle.

In a triangle $ABC$, if $\sin A \sin B = \frac{ab}{c^2}$, then the triangle is

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