Question

if 2 cos a = Sin B / sin C then show that triangle is isosceles

Answer 1

perpendicular is drawn from A and B to meet at G and H ay opposite sides respectively.

Answer 2

AnsWer:

Let

$\sf\dfrac{sin\:A}{a}=$$\sf\dfrac{sin\:B}{b}=$$\sf\dfrac{sin\:C}{c}=k.$

Then , sin A = ka, sin B = kb, sin C = kc.

$\sf \:Now, \: \: \: \: \qquad \: \cos \: A \: = \frac{ \sin \:B }{2 \sin \: C} \\ \\ \implies \sf \: 2 \cos \: B \sin \: C = \sin \: B \\ \\ \implies \sf \: 2( \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} )kc = kb \\ \\ \implies \sf \: {b}^{2} + {c}^{2} - {a}^{2} = {b}^{2} \\ \\ \implies \sf \: {c}^{2} = {a}^{2} \\ \\ \implies \sf \: c = a \\ \\ \implies \sf \triangle \: ABC \: is \: isosceles.$

Therefore, triangle ABC is isosceles.