Yes, imagining it would be difficult. At least this 3D illustration makes things simpler, but in theory the triangle is right angled.

Because the triangle still involves two perpendicular planes of the cuboid, the angle there is still 90 degrees.

The point directly below X (let’s call it Y) will make triangle QRY right angled. But lines XR and YR are on the same plane, so if triangle QRY is right angled, then so does triangle QRX.

Another way to see it :

Extend RV and name the end of the extended line as Z. Let RZ = √45 cm

(√(6² + 3²) = √45)


Draw triangle QRZ, which is right angled.

Now, treating QR as a hinge, rotate the whole QRZ anticlockwise until Z coincides with X.

(Notice the circular rotation, where RZ and RX are actually radii of this circle)

Now you have triangle QRX. But notice the right angle QRZ (Which is now QRX) didn't change.

Technically if someone views the picture from a very large height (seeing triangle QRX as the cross sectional surface, something like an aerial view of an NDP parade), the person would observe a right angled triangle.

Or you draw the rectangle QRXY, where Y is a point 3cm from U and along TU.

Then XQR is just this rectangle's diagonal.
It follows that triangle XQR is right angled since QR is perpendicular to RX, meaning XRQ is a right angle (90°)