A particle is subjected to two mutually perpendicular Simple...

The position of the particle is given by $x = A \sin(\omega t)$ and $y = A \cos(\omega t)$.

Squaring and adding the equations, we get:
$x^2 + y^2 = A^2 \sin^2(\omega t) + A^2 \cos^2(\omega t) = A^2$
This represents the equation of a circle. Thus, the trajectory of the particle is circular.

The position vector is $\vec{r} = A \sin(\omega t) \hat{i} + A \cos(\omega t) \hat{j}$.

The velocity vector is obtained by differentiating the position vector with respect to time:
$\vec{v} = \dfrac{d\vec{r}}{dt} = A\omega \cos(\omega t) \hat{i} - A\omega \sin(\omega t) \hat{j}$

Taking the dot product of $\vec{r}$ and $\vec{v}$:
$\vec{r} \cdot \vec{v} = (A \sin(\omega t))(A\omega \cos(\omega t)) + (A \cos(\omega t))(-A\omega \sin(\omega t)) = 0$

Since the dot product is zero, the position vector and velocity vector are always perpendicular to each other. Therefore, the angle between them is $90^{\circ}$.

Answer: Circle, $90^{\circ}$