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In a Δ ABC, if cos C = \(\frac{sin\,A}{2sin\,B},\) Prove that triangle is isosceles. |
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Answer» \(\frac{sin\,A}{2sin\,{B}}\) = cos C Sin A = 2 sin B cos C ka= 2kb \(\frac{a^2+b^2-c^2}{2ab}\) a = \(\frac{a^2+b^2-c^2}{2ab}\) a2 = a2 + b2 – c2 ⇒ b2 – c2 = 0 ⇒ b2 = c2 ⇒ b = c Hence, the triangle is isosceles. |
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