Consider another symmetric field line below line joining centres of charges $q_1$ and $q_2$

In given situation, flux (field lines) leaving charge $q_1$ at a solid angle $2\alpha =$ flux terminating over charge $q_2$ at a solid angle $2\beta$.

Clearly, in given situation

We can say that, flux leaving charge $q_1$ through a cone of semi-vertical angle $\alpha =$ flux terminating on charge $q_2$ through a cone of semi-vertical angle $\beta$.

To calculate flux, we first find flux through an elemental ring of base of cone and then we integrate to get total flux.

Area of elemental ring,

$dA=2\pi rds=2\pi R\sin \alpha ·Rd\alpha$

Flux through elemental ring $\left(E\Vert dA\right)$

$d\varphi =\frac{kQ}{R^2}·2\pi R^2\sin \alpha d\alpha$

Total flux through base of cone

$\varphi ={\int }_0^{\alpha }\frac{kQ}{R^2}·2\pi R^2\sin \alpha d\alpha$

$=kQ2\pi {\int }_0^{\alpha }\sin \alpha d\alpha$

$\varphi =\frac{Q}{2{\epsilon }_0}·\left(1-\cos \alpha \right)$

So, equating flux of both cones, we get

$\frac{q_1}{2{\epsilon }_0}\left(1-\cos \alpha \right)=\frac{q_2}{2{\epsilon }_0}\left(1-\cos \beta \right)$

$\Rightarrow q_1\left(1-\cos \alpha \right)=\frac{3}{2}{q}_1\left(1-\cos \beta \right)$

Substituting $\alpha =30°$ in above equation, we get

$\Rightarrow \frac{2}{3}\left(1-\cos 30°\right)=1-\cos \beta$

$\Rightarrow \frac{2}{3}\left(1-\frac{\sqrt{3}}{2}\right)=1-\cos \beta$

$\Rightarrow \cos \beta =1-\frac{2}{3}\left(1-\frac{\sqrt{3}}{2}\right)$

$\Rightarrow \cos \beta \approx 0.9$

So, angle $\beta$ is not more than $30°$.