9514 1404 393
Answer:
85°
Step-by-step explanation:
Each of the isosceles triangles PAB and QBC has interior base angles of ...
(180° -80°)/2 = 50°
Then angle PBC is 90° -50° = 40°.
Segments PB and QB are the same length, so triangle PBQ is an isosceles right triangle. This means angle BPQ is 45°. Angle QEB is the sum of the remote interior angles of ∆BEP, so is 45° +40° = 85°.
The acute angle between PQ and BC is 85°.