Answer
$\alpha=\cos^{-1}\Big(\frac{45.93}{75}\Big)\approx 52.24^\circ$
$\beta=\cos^{-1}\Big(\frac{45.93}{75}\Big)\approx 52.24^\circ$
$\gamma=\cos^{-1}\Big(\frac{37.5}{75}\Big)=120^\circ$
Work Step by Step
The direction of a vector can be specified using two angles, namely, a transverse angle $(\theta)$ and an azmuth angle $(\phi)$. As shown in Figure F2–13,
$\theta=(90^{\circ}-45^{\circ})=45^{\circ}$
$\phi=(90^{\circ}+30^{\circ})=120^{\circ}$
The components of $\vec F$ can then be determined by applying trigonometry.
$F_z = F \cos\phi$
$\implies F_z = 75 \cos120^\circ\;lb=-37.5\;lb$
$F_x = F \sin\phi\cos\theta$
$\implies F_x = 75 \sin120^\circ\cos45^\circ\;lb\approx 45.93\;lb$
and
$F_y = F \sin\phi\sin\theta$
$\implies F_x = 75 \sin120^\circ\sin45^\circ\;lb\approx 45.93\;lb$
Having determined the magnitudes and directions of the components of each force, we can express the force as a Cartesian vector.
$\vec F=\Big\{45.93\hat i+45.93\hat j-37.5\hat k\Big\}\;lb$
The magnitude of $\vec F$ is given by
$F=\sqrt {(45.93)^2+(45.93)^2+(37.5)^2}\;lb$
or, $F\approx 75\;lb$
Now, the unit vector acting in the direction of $\vec F$ is given by
$\vec u=\frac{45.93}{75}\hat i+\frac{45.93}{75}\hat j-\frac{37.5}{75}\hat i$
$u\approx 0.61\hat i+\hat j-37.5\hat k$
Now, the coordinate direction angles of $\vec F$ can be determined from the components of the unit vector $\vec u$
$\alpha=\cos^{-1}\Big(\frac{45.93}{75}\Big)\approx 52.24^\circ$
$\beta=\cos^{-1}\Big(\frac{45.93}{75}\Big)\approx 52.24^\circ$
$\gamma=\cos^{-1}\Big(\frac{37.5}{75}\Big)=120^\circ$
