The direction of change in velocity $\left(\delta v\right)$ is given by
$\varphi =\frac{{180}^o-\theta }{2}={90}^o-\frac{\theta }{2}$ …. (i)
This can be shown graphically as

For small interval i.e. $\delta t\to 0$ , then angle between $v_1$ and $v_2$ is very small i.e., $\theta \approx 0^o$ .
So, from Eq. (i) we get
$\varphi ={90}^o$