In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $5y + 8 = 0$.

(A) The given equation of the plane is $5y + 8 = 0$.
We can rewrite this in the form $Ax + By + Cz = D$ as $0x + 5y + 0z = -8$.
To convert this to the normal form $lx + my + nz = d$,we divide the equation by $\sqrt{A^2 + B^2 + C^2} = \sqrt{0^2 + 5^2 + 0^2} = 5$.
Dividing by $5$,we get $\frac{0}{5}x + \frac{5}{5}y + \frac{0}{5}z = -\frac{8}{5}$,which simplifies to $0x + 1y + 0z = -\frac{8}{5}$.
Since the distance $d$ must be positive,we multiply by $-1$: $0x - 1y + 0z = \frac{8}{5}$.
Comparing this with the normal form $lx + my + nz = d$,the direction cosines of the normal are $(0, -1, 0)$ and the distance from the origin is $\frac{8}{5}$ units.