I have a wuick question.

Well let the speed of light be c, the speed of the bus be b the distance to the mirror m.

Than the time t for the light to start at your body and come back to you is t = 2m/c.

Further the bus driven in this time a distance of a = b*t

So the angle you watch yourself is

θ = atan(m/a) = atan(m/(b*t/2)) = atan( m/(b*2m/(2c)) ) = atan( c/b )
At the distance of 5 m, due to the size of your eyes, there will be no difference whether you assume light moves instantaneously or at the speed of light.
It takes time t for light to travel from the bus to the mirror and back again. What you see in the mirror is light that left the bus time t ago.

In this case t is the time for light to go 10 meters, t = 10/c.

Your image appears to be 10 meters away and a distance vt behind you. So the angle relative to directly perpendicular satisfies tan θ = (vt) / 10 = (10v/c) / 10 = v/c.

θ = tan\^(-1) (v/c)
by
In your reference frame it's the mirror that's moving, not you. So the answer is 0&deg; from the normal (or 90&deg; to the direction of travel).