Again, we have a isor K which is at rest and another K' which moves with u along x. Now let's consider an object that K sees moving with v also along x. Since vx = dx/dt, we seek vx' = dx' / dt'. We use Lorentz transf to get
vx' + u
vx = --------------.
1 + vx'·u/c²
We can also see how vy and vz transform. In these cases, the square root denominator does not cancel:
vy'·(1 - u²/c²)¹⸍²
vy = -----------------------
1 + vx'·u/c²
vz'·(1 - u²/c²)¹⸍²
vz = -----------------------.
1 + vx'·u/c²
It is not as simple as with Galilean velocity addition.
In the case of vx, notice how it has the structure of the addition of hyperbolic tangents.
Since we have expressions for vx, vy and vz, we can produce quotients of them in order to calculate angles at which we see particles travelling. For example, if we limit ourselves to a XY plane and consider θ the angle of a particle wrt the horizontal, vx = v·cosθ and vy = v·sinθ. We can then divide our previous expressions for vy and get vy/vx = tanθ.
The expression for tanθ depends on both sinθ' and cosθ', so we get how angles also change with our system of reference. Even for the extreme case of u=c we get a difference in angle Δθ = θ' - θ called the aberration of light.