MHB Triangle Challenge: Evaluate $\cos \angle B$

[anemone]

In a triangle $ABC$ with side lengths $a,\,b$ and $c$, it's given that $17a^2+b^2+9c^2=2ab+24ac$.

Evaluate $\cos \angle B$.

 

[kaliprasad]

In a triangle $ABC$ with side lengths $a,\,b$ and $c$, it's given that $17a^2+b^2+9c^2=2ab+24ac$.

Evaluate $\cos \angle B$.

Spoiler

we have $a^2 - 2ab + b^2 + 16a^2- 24ac + 9c^2 = (a-b)^2 + (4a-3c)^2 = 0$
hence $ a = b $ and $4a= 3c=>\frac{c}{a} = \frac{4}{3}$
hence $\cos \angle B = \frac{a^2+c^2-b^2}{2ac} = \frac{c^2}{2ac} = \frac{c}{2a} =\frac{2}{3} $

 

[anemone]

Spoiler

we have $a^2 - 2ab + b^2 + 16a^2- 24ac + 9c^2 = (a-b)^2 + (4a-3c)^2 = 0$
hence $ a = b $ and $4a= 3c=>\frac{c}{a} = \frac{4}{3}$
hence $\cos \angle B = \frac{a^2+c^2-b^2}{2ac} = \frac{c^2}{2ac} = \frac{c}{2a} =\frac{2}{3} $

Well done, kaliprasad and thanks for participating!