SOLUTION: 7. P is the centre of the upper face of the rectangular block shown in the figure. Calculate (a) PẬC, (b) PÂB. A 3D cuboid figure is given in which its given that ABCD

PQ = 5.

Let Q be the center of the bottom face of the rectangular block

So PQ is the altitude of triangle PAC.  PQ is the side opposite

angle PAC (aka angle PAQ) in right triangle APQ.  We need to 

find AQ, the side adjacent angle PAC.  AQ is 1/2 the diagonal AC.

We find diagonal AC by the Pythagorean theorem on right triangle

ACD. 












  <--- answer for (a)

-----------------------------------
(b)  Angle PAB.

I'll tell you how, but I'll leave it up to you.

 

We need to draw in three additional line segments, QA, QB, and
PR perpendicular to AB

We know that QA=QB=10, for they are both half the diagonal we
found in part (a).  

So triangles AQB and APB are isosceles.

We can find PA by using the Pythagorean theorem on right
triangle AQP.

We know AR=6 (half of AB=12)

Then find angle PAB by .

Edwin