In a triangle ABC , with usual notations if $a=4, \mathrm{~b}=8, \angle \mathrm{C}=60^{\circ}$, then the value of $\angle \mathrm{B}$ and the ratio $\cos \mathrm{A}: \cos \mathrm{C}$ respectively are,

Given triangle sides a = 4, b = 8, and included angle C = $60^{\circ}$, determine angle B and the ratio $\cos A : \cos C$.

Using the Law of Cosines, side c is found by:

$c^2 = a^2 + b^2 - 2ab\cos C = 4^2 + 8^2 - 2\cdot4\cdot8\cdot\cos 60^{\circ} = 16 + 64 - 64\cdot\frac{1}{2} = 48$

Thus, $c = \sqrt{48} = 4\sqrt{3}$.

To find angle B, apply the Law of Sines:

$\frac{b}{\sin B} = \frac{c}{\sin C}$
$\frac{8}{\sin B} = \frac{4\sqrt{3}}{\sin 60^{\circ}} = \frac{4\sqrt{3}}{\sqrt{3}/2} = 8$

So $\sin B = 1$ and $\angle B = 90^{\circ} = \frac{\pi}{2}$.

$\angle A = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$

Now compute the cosine ratio:

$\cos A = \cos 30^{\circ} = \frac{\sqrt{3}}{2}$
$\cos C = \cos 60^{\circ} = \frac{1}{2}$

The ratio simplifies to $\frac{\sqrt{3}}{2} : \frac{1}{2} = \sqrt{3} : 1$

So $\angle B = \frac{\pi}{2}$ and $\cos A : \cos C = \sqrt{3} : 1$, matching option B.