Explanation

(a)
Given that, $2\sin A = \sqrt{3}\sin B$
$\Rightarrow 2\sqrt{5}\sin{A = \sqrt{15}\sin B}$ …(i)
and$\ 2\cos{A = \sqrt{5}\cos B}$
$\Rightarrow 2\sqrt{3}\cos{A = \sqrt{15}\cos B}$ …(ii)
On squaring and adding Eqs. (i) and (ii), we get
$20\sin^{2}{A + 12\cos^{2}{A = 15}}$
$\Rightarrow 8\sin^{2}{A = 3\ \Rightarrow \sin^{2}{A = \frac{3}{8}}}$
$\Rightarrow \cos^{2}{A =}\frac{5}{8}$
$\therefore\frac{\sin^{2}A}{\cos^{2}A} = \frac{3}{5}\ \Rightarrow \tan{A = \sqrt{\frac{3}{5}}}$