For the curve A
$\begin{align} 16x^{2}+9y^{2}&=144\\ \Rightarrow 16(2x)+9(2y)\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\ \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} &=-\frac{16x}{9y}\\ \therefore \left.\begin{matrix} \frac{\mathrm{d}y}{\mathrm{d}x} \end{matrix}\right|_{(3,0)} &=-\frac{48}{0}\to -\infty\end{align}$
$\therefore \mathrm{\theta}_L=\tan^{-1}(-\frac{48}{0})=90^{\circ}$
Now for the curve B
$\begin{align} x^{2}+y^{2}-10x &=-21\\ \Rightarrow 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}-10&=0\\ \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} &=-\frac{5-x}{y}\\ \therefore \left.\begin{matrix} \frac{\mathrm{d}y}{\mathrm{d}x} \end{matrix}\right|_{(3,0)} &=-\frac{2}{0}\to -\infty\end{align}$
$\therefore \mathrm{\theta}_M=\tan^{-1}(-\frac{2}{0})=90^{\circ}$
$\therefore$ The angle between the tangents $L$ and $M$ is $\mathrm{\theta}_L-\mathrm{\theta}_M=90^{\circ}-90^{\circ}=0^{\circ}$
So the correct answer is A.