Answer
$\sin\theta=D\sin\theta_i-\dfrac{m\lambda}{D}\tag{$m=\pm1,\pm2,..$}$
Work Step by Step
First of all, to find the central maxima, we need to find the net path difference which is given by
$$\Delta r=D\sin\theta_i-D\sin\theta$$
where $D\sin\theta_i$ is the path difference between the top and the bottom ends of the slit for the incident wave and $D\sin\theta$ is the path difference between the top and the bottom ends of the slit for the diffracted wave.
When $\theta_i=\theta$, the net path difference is zero which means that a constructive interference is created there.
Now we can be sure that the central maximum is laying at the position in which $\theta=28^\circ$.
Thus, the path difference equal to
$$\Delta r=D\sin\theta_i-D\sin\theta=m\lambda$$
where $m=\pm1,\pm2,\pm3,...$
And hence, the angle of minima $\theta$ is given by
$$\sin\theta=D\sin\theta_i-\dfrac{m\lambda}{D}\tag{$m=\pm1,\pm2,..$}$$
Therefore, and from all the above, at an initial angle of $\theta_i=28^\circ$, the first minimum fringe will be surrounding the central maximum fringe symmetrically at $28^\circ$.
