1.

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

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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