We use the relation:
\[
\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1
\]
Given:
\[
\cos 90^\circ = 0, \quad \cos 60^\circ = \dfrac{1}{2}, \quad \cos^2 \theta = 1 - 0^2 - \left(\dfrac{1}{2}\right)^2
\]
\[
\cos^2 \theta = 1 - \dfrac{1}{4} = \dfrac{3}{4}
\]
\[
\cos \theta = \dfrac{\sqrt{3}}{2}
\]
Thus, \( \theta = \dfrac{\pi}{6} \).