Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$\sqrt{3}\cos x+\frac{2b}{a}\sin x=\frac{c}{a}$
Now, $\sqrt{3}\cos \alpha +\frac{2b}{a}\sin \alpha =\frac{c}{a} \dots \left(\mathrm{i}\right)$   
$\sqrt{3}\cos \beta +\frac{2b}{a}\sin \beta =\frac{c}{a} \dots \left(\mathrm{ii}\right)$ 

From equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$\Rightarrow \sqrt{3}\left[\cos \alpha -\cos \beta \right]+\frac{2b}{a}\left(\sin \alpha -\sin \beta \right)=0$
$\Rightarrow \sqrt{3}\left[-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)\right]+\frac{2b}{a}\left[2\cos \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)\right]=0$

$\Rightarrow -\sqrt{3}+2\sqrt{3}·\frac{b}{a}=0$
$\Rightarrow \frac{b}{a}=\frac{1}{2}=0.5$