Determine the angles formed by the given line with the coordinates axis $ (D): { x-2y-5=0} / {x-3z+8=0 } $ ( this is a system of 2 equations, I don't know how to format it properly )
I know the directionals are ( 1, -2 , 0) and (1,0,-3) but I don't really know how to continue it.
On
The system of equations describes the line as the intersection of two planes. It must therefore be perpendicular to the normals of both planes, so its direction vector is $(1,0,-3)\times(1,-2,0)=(6,-3,2)$. The corresponding unit vector is $\left(\frac67,-\frac37,\frac27\right)$. Since $v\cdot w = \|v\|\,\|w\|\cos\theta$, the components of this vector are the cosines of the angles it makes with the coordinate axes.
just a hint to finish
The line $(D) $ can be defined by following parametric equations
$$x=x $$ $$y=(x-5)/2$$ $$z=(x+8)/3$$
thus the vector director of $(D) $ is $$\vec {u}=(1,1/2,1/3) $$ or $$(6,3,2) $$
use $$\cos (\vec{u},\vec{v})=\frac {\vec {u}\cdot \vec{v}}{||\vec {u}||||\vec {v}||} $$