​In a Δ ABC, if cos C = ​sin A/ 2sin B Prove that triangle is isosceles.

Question

In a Δ ABC, if cos C = \(\frac{sin\,A}{2sin\,B},\) Prove that triangle is isosceles.

Answer 1

\(\frac{sin\,A}{2sin\,{B}}\) = 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.