Given,
$\mathit{A}=\left\{1,2,3,4,5,6\right\}$

(i) $\left(2,4\right)\in \mathit{R}$    $\{\because 4$ is divisible by 2 $\}$
But $\left(4,2\right)$∉ R $\{\because 2$ is not divisible by 4$\}$
$\therefore \mathit{R}$ is not symmetric.
 

(ii) Let
$\Rightarrow \mathit{b}=\mathit{\lambda }\mathit{a}$ and $\mathit{c}=\mathit{\mu }\mathit{b}$
Now, $\mathit{c}=\mathit{\mu }\mathit{b}=\mathit{\mu }\left(\mathit{\lambda }\mathit{a}\right)\Rightarrow \left(\mathit{a},\mathit{c}\right)\in \mathrm{R}$
$\Rightarrow \mathit{c}$ is divisible by $\mathit{a}$
$\therefore \mathbf{R}$ is transitive.
 
OR

$=\frac{9}{4}\left(\frac{\mathit{\pi }}{2}-\mathit{s}\mathit{i}\mathit{n}{}^{-1}\frac{1}{3}\right)$
$=\frac{9}{4}\left({\cos }^{-1}\frac{1}{3}\right).....\left(1\right)$
Now, let ${\cos }^{-1}\frac{1}{3}=\mathrm{x}$ .
Then, $\cos \mathit{x}=\frac{1}{3}\Rightarrow \mathit{s}\mathit{i}\mathit{n}\mathit{x}=\sqrt{1-{\left(\frac{1}{3}\right)}^2}=\frac{2\sqrt{2}}{3}$
$\therefore \mathit{x}=\mathit{s}\mathit{i}\mathit{n}{}^{-1}\frac{2\sqrt{2}}{3}\Rightarrow {\cos }^{-1}\frac{1}{3}=\mathit{s}\mathit{i}\mathit{n}{}^{-1}\frac{2\sqrt{2}}{3}$