Question

If in a triangle ABC, $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$, then show that $a^2,b^2,c^2$ are in A.P.

Answer 1

$InatriangleABCA+B+C=180$

$\frac{\sin A}{\sin C}=\frac{\sin (180-(B+C\left)\right)}{\sin (180-(A+B\left)\right)}$

$\Rightarrow \frac{\sin (B+C)}{\sin (A+B)}=\frac{\sin (A-B)}{\sin (B-C)}$

$\Rightarrow {\sin }^{2}B-{\sin }^{2}C={\sin }^{2}A-{\sin }^{2}B2{\sin }^{2}B={\sin }^{2}C+{\sin }^{2}A$

$\Rightarrow 2{\left(\frac{b}{2R}\right)}^2={\left(\frac{a}{2R}\right)}^2+{\left(\frac{c}{2R}\right)}^2$

$\Rightarrow {2b}^2=a^2+c^2$

So they are in A.P.