Find the largest and smallest values of the acute angles a line through the origin makes with the three coordinate planes.

Let the line through the origin makes angle $\alpha ,\beta \text{and}\gamma$ with the three coordinate planes, respectively.

Then, the sum of angles will be:

$S=\alpha +\beta +\gamma \left(1\right)$

Since the angle between the line and plane is the complement of the angle between the normal plane and the line.

Therefore, we have:

$$\begin{gathered} {\cos }^2\left(90°-\alpha \right)+{\cos }^2\left(90°-\beta \right)+{\cos }^2\left(90°-\gamma \right)=1 \\ {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =1 \end{gathered}$$

By using Lagrange's multipliers method, we get:

$$\begin{gathered} f\left(\alpha ,\beta ,\gamma \right)=S+\lambda \theta \\ =\alpha +\beta +\gamma +\lambda \left({\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma \right) \end{gathered}$$

To find the minimum or maximum angle, we have:

$$\begin{gathered} \frac{\partial f}{\partial \alpha }=1+2\lambda \sin \alpha \cos \alpha =1+\lambda \sin 2\alpha \\ \frac{\partial f}{\partial \beta }=1+\lambda \sin 2\beta \\ \frac{\partial f}{\partial \gamma }=1+\lambda \sin 2\gamma \end{gathered}$$

Put each partial derivative equal to zero, and we get: $\alpha =\beta =\gamma$. So,

$$\begin{gathered} 3{\sin }^2\alpha =1 \\ \sin \alpha =\frac{1}{\sqrt{3}} \\ \alpha ={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$

And,

$$\begin{gathered} S=\alpha +\beta +\gamma \\ =3\alpha \\ =3{\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =105.8° \end{gathered}$$

Hence, the largest and smallest sum of angles are $105.8°\text{and}90°$ respectively.