Answer
$$\dfrac{\pi}{2}$$
Work Step by Step
Let $\theta$ be the angle between $r$ and $a$
So, $\theta =\cos^{-1} [\dfrac{r \cdot a}{|r| |a|}]$
Now, $v(t)=\dfrac{dr}{dt}=i[e^t(\cos t -\sin t) ]+j [e^t(\sin t +\cos t) ]$
$\implies a(t)=\dfrac{dv(t)}{dt}=-2 e^t ( \sin t i- \cos t j)$
Next, $\theta =\cos^{-1} [\dfrac{ e^t \cos t \times (-2 e^t \sin t)
+(e^t \sin t) (2e^t \cos t)}{\sqrt {(e^t \cos t )^2+(e^t \cos t )^2 \cdot \sqrt {(-2 e^t \sin t)^2 +( e^t \cos t)^2}}}]$
$\implies \theta=\cos^{-1} (0)=\dfrac{\pi}{2}$
