Answer
The tension in each cable has a magnitude of $4.26kN$.
Work Step by Step
As shown in the below photo, two tension forces in two cables, when translated into the vertical component, each has a magnitude of $T\sin\theta$. Opposing the tension forces is the weight of the I-beam.
Since the I-beam is lifted at a constant velocity, these forces balance each other. $$mg=T\sin\theta+T\sin\theta=2T\sin\theta$$ $$T\sin70=\frac{mg}{2}=\frac{8kN}{2}=4kN$$ $$T=4.26kN$$
