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In a triangle $\mathrm{ABC}$ if $\sin \mathrm{A} \sin \mathrm{B}=\frac{\mathrm{ab}}{\mathrm{c}^2}$, then the triangle is
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Verified Answer
The correct answer is:
right angled
$$
\begin{aligned}
& \text { Hints }: \sin A \sin B=\frac{a b}{c^2} \\
& \Rightarrow c^2=\frac{a b}{\sin A \sin B}=\left(\frac{a}{\sin A}\right)\left(\frac{b}{\sin B}\right) \\
& \Rightarrow c^2=\left(\frac{c}{\sin C}\right)^2 \Rightarrow \sin ^2 C=1 \Rightarrow \sin C=1 \Rightarrow C=90^{\circ}
\end{aligned}
$$
\begin{aligned}
& \text { Hints }: \sin A \sin B=\frac{a b}{c^2} \\
& \Rightarrow c^2=\frac{a b}{\sin A \sin B}=\left(\frac{a}{\sin A}\right)\left(\frac{b}{\sin B}\right) \\
& \Rightarrow c^2=\left(\frac{c}{\sin C}\right)^2 \Rightarrow \sin ^2 C=1 \Rightarrow \sin C=1 \Rightarrow C=90^{\circ}
\end{aligned}
$$
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