What happened to the solution I just posted to Q10 at 12:42 ??
Drop a perpendicular from C to PB; call the foot of this perpendicular R
Now note: angle CBR = angle PAB = @
Use this simple fact: in a right-angled triangle, hypotenuse of length 'h' and one acute angle @.
Then the 2 sides, one opp angle @ and the other adjacent to @ have, immediately, resp. lengths:
h sin@ and h cos @
Using this fact: AP = x cos@ and PQ = CR = y sin@
.: AQ = AP + PQ = x cos@+ y sin@
QED