One corner of this cuboid has been sliced off along the plane QTU. WU = 4 cm. a) Calculate the length of the three sides of the triangle QTU b) Calculate the three angles P, Q, and T in triangle PQT.

Since A, B, and X are points on a straight line, $\angle ABC$ and $\angle CBX$ are supplementary angles.

The angle CBX is $\boxed{\text{30}^\circ}$.

The Cosine Rule states $b^2 = a^2 + c^2 - 2ac \cos B$. Here, $a = BC = 6$ cm, $c = AB = 4$ cm, and $B = \angle ABC = 150^\circ$.

We know $\cos(150^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}$.

The length AC is approximately $\boxed{\text{9.67 cm}}$.

Since CX is perpendicular to ABX, $\triangle CBX$ is a right-angled triangle with the right angle at X ($\angle CXB = 90^\circ$). We know BC = 6 cm (hypotenuse) and $\angle CBX = 30^\circ$ from part i).

In a right-angled triangle, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$.

We know $\sin(30^\circ) = \frac{1}{2}$.

The length CX is $\boxed{\text{3 cm}}$.

The problem statement explicitly gives the length of BC.

The length BC is $\boxed{\text{6 cm}}$.