Part A
Using the Pythagorean on the right triangle PQR, with PQ and QR as the legs and PR as the hypotenuse,
[tex]14^2 +6^2 =(PR)^2\\\\(PR)=\sqrt{14^2 +6^2}\\\\PR \approx \boxed{15.23 \text{ ft}}[/tex]
Part B
[tex](QR)^2 +6^2 =16^2\\\\(QR)^2 =16^2 -6^2\\\\QR=\sqrt{16^2 -6^2}\\\\QR \approx \boxed{14.83 \text{ ft}}[/tex]