91Ó°ÊÓ

Let the line through the origin makes angle $\mathrm{Î\pm },\mathrm{β}\text{and}\mathrm{γ}$ with the three coordinate planes, respectively.

Then, the sum of angles will be:

$S=\mathrm{Î\pm }+\mathrm{β}+\mathrm{γ}                                 \left(1\right)$

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

Therefore, we have:

$$\begin{gathered} {\cos }^2\left(90°-\mathrm{Î\pm }\right)+{\cos }^2\left(90°-\mathrm{β}\right)+{\cos }^2\left(90°-\mathrm{γ}\right)=1 \\ {\sin }^2\mathrm{Î\pm }+{\sin }^2\mathrm{β}+{\sin }^2\mathrm{γ}=1 \end{gathered}$$

By using Lagrange's multipliers method, we get:

$$\begin{gathered} f\left(\mathrm{Î\pm },\mathrm{β},\mathrm{γ}\right)=S+\mathrm{λ}\mathrm{θ} \\ =\mathrm{Î\pm }+\mathrm{β}+\mathrm{γ}+\mathrm{λ}\left({\sin }^2\mathrm{Î\pm }+{\sin }^2\mathrm{β}+{\sin }^2\mathrm{γ}\right) \end{gathered}$$

To find the minimum or maximum angle, we have:

$$\begin{gathered} \frac{∂f}{∂\mathrm{Î\pm }}=1+2\mathrm{λ}\sin \mathrm{Î\pm }\cos \mathrm{Î\pm }=1+\mathrm{λ}\sin 2\mathrm{Î\pm } \\ \frac{∂f}{∂\mathrm{β}}=1+\mathrm{λ}\sin 2\mathrm{β} \\ \frac{∂f}{∂\mathrm{γ}}=1+\mathrm{λ}\sin 2\mathrm{γ} \end{gathered}$$

Put each partial derivative equal to zero, and we get: $\mathrm{Î\pm }=\mathrm{β}=\mathrm{γ}$. So,

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

And,

$$\begin{gathered} S=\mathrm{Î\pm }+\mathrm{β}+\mathrm{γ} \\ =3\mathrm{Î\pm } \\ =3{\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ =105.8° \end{gathered}$$

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