The reflection matrix, to reflect a point $(x,y)$ about the line $y=mx$ where $m=tan\theta$ is given by,
$\begin{bmatrix} cos2\theta & sin2\theta\\ sin2\theta&-cos2\theta \end{bmatrix}$
So, if we have a point $(x,y)$ and after reflection, new point is $(x',y')$ then we can write it as :
$\begin{bmatrix} x'\\y' \end{bmatrix}$ = $\begin{bmatrix} cos2\theta & sin2\theta\\ sin2\theta&-cos2\theta \end{bmatrix}$$\begin{bmatrix} x\\y \end{bmatrix}$
Now, here in the given question, given point is $(vcos\phi,vsin\phi)$
After reflecting about $x-$axis, It becomes, $(vcos\phi,-vsin\phi)$
Now, according to given information for $\theta = 5^{\circ}$ and $\phi = 10^{\circ}$,
$O_{u}(v)= v\begin{bmatrix} cos10 &sin10 \\ sin10 &-cos10 \end{bmatrix}\begin{bmatrix} cos10\\-sin10 \end{bmatrix}= v\begin{bmatrix} cos20\\sin20 \end{bmatrix}$
$O_{u}^{2}(v)= v\begin{bmatrix} cos10 &sin10 \\ sin10 &-cos10 \end{bmatrix}\begin{bmatrix} cos20\\-sin20 \end{bmatrix}= v\begin{bmatrix} cos30\\sin30 \end{bmatrix}$
$O_{u}^{3}(v)= v\begin{bmatrix} cos10 &sin10 \\ sin10 &-cos10 \end{bmatrix}\begin{bmatrix} cos30\\-sin30 \end{bmatrix}= v\begin{bmatrix} cos40\\sin40 \end{bmatrix}$
Similarly,
$O_{u}^{8}(v)= v\begin{bmatrix} cos10 &sin10 \\ sin10 &-cos10 \end{bmatrix}\begin{bmatrix} cos80\\-sin80 \end{bmatrix}= v\begin{bmatrix} cos90\\sin90 \end{bmatrix}$
It means new point is $w= (vcos90^{\circ}, vsin90^{\circ})$.
So, angle subtended by $w$ and $x-$axis at origin is $90^{\circ}.$