To represent \(\vec{F}\) in spherical coordinates, we must find the radial distance \(r\text{,}\) the azimuthal angle \(\theta\text{,}\) and the polar angle \(\phi\text{.}\)
Coordinate \(r\) is simply the magnitude of force \(\vec{F}.\) Since the three components of \(\vec{F}\) are perpendicular, we can apply the Pythagorean Theorem to find it.
\begin{align*}
F = |\vec{F}| = r \amp= \sqrt{F_x^2+F_y^2+F_z^2}\\
\amp = \lbf{\sqrt{20^2 + \left( -30 \right) ^2 + 60^2}}\\
\amp = \lbf{70}
\end{align*}
Azimuthal angle \(\theta\) measures the angle between the \(x\) axis and the projection of \(\vec{F}\) onto the \(xy\) plane, \(F_{zy}\text{.}\)
Using a right triangle with sides \(F_x\text{,}\) \(F_y\text{,}\) and \(F_{xy} \text{,}\) we can find \(\theta\) using the inverse tangent of the ratio of the opposite to adjacent sides.
\begin{equation*}
\theta = \tan^{-1} \frac{F_y}{F_x} = \tan^{-1}\left( \frac{-30}{20}\right) = \ang{-56.31}
\end{equation*}
This angle is negative because it is measured clockwise from the positive \(x\) axis, opposite the standard CCW direction.
The polar angle \(\phi\) is measured down from the \(+z\) axis to the vector \(\vec{F}\text{.}\) We can find it using a right triangle with sides \(F\text{,}\) \(F_z\text{,}\) and \(F_{xy}\text{.}\) Note that \(\phi\) is the same as the direction cosine angle \(\theta_z\text{.}\)
\begin{equation*}
\phi = \theta_z =\cos^{-1} \frac{F_z}{|\vec{F}|} = \cos^{-1}\left( \frac{60}{70} \right) = \ang{31.0}
\end{equation*}
Also notice that the azimuthal angle \(\theta\) is smaller than the direction cosine angle \(\theta_x\text{,}\) since \(\theta\) is in the \(xy\) plane, but \(\theta_x\) is a 3D angle from the \(x\) axis to the vector \(\vec{F}.\)