Answer
(a) $25^{\circ}$
(b) $F_y=34.8N$
Work Step by Step
(a) From trigonometry, we know that
$cos\theta=\frac{F_x}{F}$
$\implies \theta=cos^{-1}\frac{F_x}{F}$
We plug in the known values to obtain:
$\theta=cos^{-1}\frac{74.6}{82.3}$
This simplifies to:
$\theta=25^{\circ}$
(b) Now, we can determine the magnitude of the force as follows:
$F_y=\sqrt{F^2-F_x^2}$
We plug in the known values to obtain:
$F_y=\sqrt{(82.3)^2-(74.6)^2}$
This simplifies to:
$F_y=34.8N$
