91Ó°ÊÓ

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

Then, the sum of angles will be:

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

Now, we have:

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

From equations (1) and (2), we get:

$f\left(\mathrm{Î\pm },\mathrm{β},\mathrm{γ}\right)=\mathrm{Î\pm }+\mathrm{β}+{\cos }^{-1}\sqrt{{\sin }^2\mathrm{Î\pm }-{\cos }^2\mathrm{β}}$

To find the minimum or maximum angle, we have:

$\frac{∂f}{∂\mathrm{Î\pm }}=0\text{and}\frac{∂f}{∂\mathrm{β}}=0$

For, $\frac{∂f}{∂\mathrm{Î\pm }}=0$:

$$\begin{gathered} \frac{∂f}{∂\mathrm{Î\pm }}=0 \\ 1-\frac{1}{\sqrt{1-{\sin }^2\mathrm{Î\pm }+{\cos }^2\mathrm{β}}}\frac{2\sin \mathrm{Î\pm }\cos \mathrm{Î\pm }}{2\sqrt{{\sin }^2\mathrm{Î\pm }-{\cos }^2\mathrm{β}}}=0 \\ \left({\cos }^2\mathrm{Î\pm }+{\cos }^2\mathrm{β}\right)\left({\sin }^2\mathrm{Î\pm }-{\cos }^2\mathrm{β}\right)={\sin }^2\mathrm{Î\pm }{\cos }^2\mathrm{Î\pm } \\ {\cos }^2\mathrm{β}\left({\sin }^2\mathrm{β}-2{\cos }^2\mathrm{Î\pm }\right)=0 \end{gathered}$$

On further evaluation, we get:

$$\begin{gathered} {\cos }^2\mathrm{β}=0\text{or}{\sin }^2\mathrm{β}=2{\cos }^2\mathrm{Î\pm } \\ ⇒\mathrm{β}=90°\text{or}{\sin }^2\mathrm{β}=2{\cos }^2\mathrm{Î\pm } \end{gathered}$$

Similarly for, $\frac{∂f}{∂\mathrm{β}}=0$:

$$\begin{gathered} \frac{∂f}{∂\mathrm{β}}=0 \\ 1-\frac{1}{\sqrt{1-{\sin }^2\mathrm{Î\pm }+{\cos }^2\mathrm{β}}}\frac{-2\cos \mathrm{β}\left(-\sin \mathrm{β}\right)}{2\sqrt{{\sin }^2\mathrm{Î\pm }-{\cos }^2\mathrm{β}}}=0 \\ \left({\cos }^2\mathrm{Î\pm }+{\cos }^2\mathrm{β}\right)\left({\sin }^2\mathrm{Î\pm }-{\cos }^2\mathrm{β}\right)={\cos }^2\mathrm{β}{\sin }^2\mathrm{β} \end{gathered}$$

As we have $\mathrm{β}=90°$, we get:

$$\begin{gathered} \left({\cos }^2\mathrm{Î\pm }+{\cos }^{2}90°\right)\left({\sin }^2\mathrm{Î\pm }-{\cos }^{2}90°\right)={\cos }^{2}90°{\sin }^{2}90° \\ {\cos }^2\mathrm{Î\pm }{\sin }^2\mathrm{Î\pm }=0 \\ 4{\cos }^2\mathrm{Î\pm }{\sin }^2\mathrm{Î\pm }=0 \\ {\sin }^{2}2\mathrm{Î\pm }=0 \end{gathered}$$

This Implies:

$$\begin{gathered} s\mathrm{in}2\mathrm{Î\pm }=0 \\ 2\mathrm{Î\pm }=0 \\ \mathrm{Î\pm }=0° \end{gathered}$$

Now, from equation (2), we get:

$$\begin{gathered} \mathrm{γ}={\cos }^{-1}\sqrt{{\sin }^{2}0°-{\cos }^{2}90°}  \\ ={\cos }^{-1}0 \\ =90° \end{gathered}$$

Here, the sum of angles will be: $\mathrm{Î\pm }+\mathrm{β}+\mathrm{γ}=0°+90°+90°=180°$.

Now, if ${\sin }^2\mathrm{β}=2{\cos }^2\mathrm{Î\pm }$. Then,

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

Evaluate as:

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

This Implies:

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

Now, from equation (2), we get:

$$\begin{gathered} \frac{1}{3}+\frac{1}{3}+{\cos }^2\mathrm{γ}=1 \\ \mathrm{γ}={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{gathered}$$

Here, the sum of angles will be: $\mathrm{Î\pm }+\mathrm{β}+\mathrm{γ}=3\left({\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)=164.2°$

Hence, the largest and smallest sum of angles are $180°\text{and}164.2°$ respectively.