A surveyor fixes a laser at the origin of a hall, with the axes along two walls and the vertical…

Step 1. The first two cosines. $l = \cos 120^\circ = -\dfrac{1}{2}$ and $m = \cos 60^\circ = \dfrac{1}{2}$.

Step 2. The third cosine. $n^2 = 1 - \dfrac{1}{4} - \dfrac{1}{4} = \dfrac{1}{2}$, so $n = \pm\dfrac{1}{\sqrt{2}}$.

Step 3. The two angles. For $n = \dfrac{1}{\sqrt{2}}$ the angle is $\gamma = 45^\circ$. For $n = -\dfrac{1}{\sqrt{2}}$ it is $\gamma = 135^\circ$.

Step 4. The angle between the beams. The direction cosines are $\left(-\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{\sqrt{2}}\right)$ and $\left(-\dfrac{1}{2}, \dfrac{1}{2}, -\dfrac{1}{\sqrt{2}}\right)$. Their dot product is $\dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{2} = 0$. So the beams are perpendicular.