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प्रश्न
In \[∆ ABC \text{ if } \cos C = \frac{\sin A}{2 \sin B}\] prove that the triangle is isosceles.
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उत्तर
Let \[∆ ABC\] be any triangle.
Suppose \[\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k\]
If \[\cos C = \frac{\sin A}{2\sin B}\] then
\[\frac{b^2 + a^2 - c^2}{2ab} = \frac{ka}{2kb} \left( \because \cos C = \frac{b^2 + a^2 - c^2}{2ab} \right)\]
\[\Rightarrow b^2 + a^2 - c^2 = a^2 \]
\[ \Rightarrow b^2 - c^2 = 0\]
\[ \Rightarrow \left( b - c \right)\left( b + c \right) = 0\]
\[ \Rightarrow b - c = 0\]
\[ \Rightarrow b = c \left( \because b, c > 0 \right)\]
Thus, the lengths of two sides of the \[∆ ABC\] are equal.
Hence, \[∆ ABC\] is an isosceles triangle.
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