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In $\triangle \mathrm{ABC}$, if $\sin ^2 \mathrm{~B}=\sin \mathrm{C}$ and $3 \cos ^2 \mathrm{~B}=2 \cos ^2 \mathrm{C}$, then $\triangle \mathrm{ABC}$ is
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Verified Answer
The correct answer is:
a scalene triangle
Given $2 \cos ^2 C=3 \cos ^2 B$
$\Rightarrow 2-2 \sin ^2 C=3\left(1-\sin ^2 B\right)$
$\begin{aligned}
& \Rightarrow 2-2 \sin ^2 C=3-3 \sin C \\
& \Rightarrow 2 \sin ^2 C-3 \sin C+1=0 \\
& \Rightarrow(2 \sin C-1)(\sin C-1) \\
& \Rightarrow \sin C=\frac{1}{2}, 1 \Rightarrow C=\frac{\pi}{6} \text { or } \frac{\pi}{2}
\end{aligned}$
So $\triangle A B C$ is a scalene triangle.
$\Rightarrow 2-2 \sin ^2 C=3\left(1-\sin ^2 B\right)$
$\begin{aligned}
& \Rightarrow 2-2 \sin ^2 C=3-3 \sin C \\
& \Rightarrow 2 \sin ^2 C-3 \sin C+1=0 \\
& \Rightarrow(2 \sin C-1)(\sin C-1) \\
& \Rightarrow \sin C=\frac{1}{2}, 1 \Rightarrow C=\frac{\pi}{6} \text { or } \frac{\pi}{2}
\end{aligned}$
So $\triangle A B C$ is a scalene triangle.
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