Answer
(a) F = 574 N
(b) F = 607 N
Work Step by Step
(a) If the man pushes parallel to the incline, the man's force $F$ is equal in magnitude to the component of the piano's weight directed down the incline.
$F = mg~sin(19.0^{\circ})$
$F = (180~kg)(9.80~m/s^2)~sin(19.0^{\circ})$
$F = 574~N$
(b) If the man pushes parallel to the floor, the component of the man's force $F$, which is directed up the incline, is equal in magnitude to the component of the piano's weight directed down the incline.
$F~cos(19.0^{\circ}) = mg~sin(19.0^{\circ})$
$F = \frac{(180~kg)(9.80~m/s^2)~sin(19.0^{\circ})}{cos(19.0^{\circ})}$
$F = 607~N$
