# Question #3419a

Jul 6, 2016

(a) $460.0 N$, rounded to one decimal place.
(b) $372.6 N$, rounded to one decimal place.

#### Explanation:

The figure above shown the position of the force vector whose $x$ and $y$ components are to be resolved.
(a) the $x$ scalar component ${F}_{x} = F \cos {39}^{\circ}$
Inserting value of $\cos$ from tables
$\implies {F}_{x} = F \times 0.7771 = 592 \times 0.7771$
$= 460.0 N$, rounded to one decimal place.

(b) $y$ scalar component of the vector ${F}_{y} = - F \sin {39}^{\circ}$
($- v e$ sign has been placed in front as the angle is in the fourth quandrant or ${F}_{y}$ is in the $- y$ direction).
Inserting value of $\sin$ from tables
$\implies {F}_{y} = F \times 0.6293 = 592 \times 0.6293$
$= 372.6 N$, rounded to one decimal place.