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If in a $\Delta A B C, \cos B=(\sin A) /(2 \sin C)$, then the triangle is
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The correct answer is:
Isosceles triangle
Consider $\cos B=\frac{\sin A}{2 \sin C}$
$\Rightarrow \frac{c^{2}+a^{2}-b^{2}}{2 a c}=\frac{a}{2 c}$
$\quad\left(\because \frac{a}{\sin A}=\frac{c}{\sin C} \Rightarrow \frac{\sin A}{\sin C}=\frac{a}{c}\right)$
$\Rightarrow c^{2}+a^{2}-b^{2}=\frac{2 a^{2} c}{2 c}$
$\Rightarrow c^{2}+a^{2}-b^{2}=a^{2}$
$\Rightarrow c^{2}-b^{2}=0$
$\Rightarrow c=b$
Hence, $\Delta \mathrm{ABC}$ is isosceles triangle.
$\Rightarrow \frac{c^{2}+a^{2}-b^{2}}{2 a c}=\frac{a}{2 c}$
$\quad\left(\because \frac{a}{\sin A}=\frac{c}{\sin C} \Rightarrow \frac{\sin A}{\sin C}=\frac{a}{c}\right)$
$\Rightarrow c^{2}+a^{2}-b^{2}=\frac{2 a^{2} c}{2 c}$
$\Rightarrow c^{2}+a^{2}-b^{2}=a^{2}$
$\Rightarrow c^{2}-b^{2}=0$
$\Rightarrow c=b$
Hence, $\Delta \mathrm{ABC}$ is isosceles triangle.
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