Let the distance from A and the break point be $A(t)$ and distance from B and the break point be $B(t)$($t$ is time variable). And the distance between A & B be $l$(which is constant).
Then $A(t)=-l\sin\theta(t)$, $B(t)=l\cos\theta(t)$. ($\theta$ decreases. $90^\circ$~$0^\circ$. That why the A is negative.)
$\frac{\mathrm{d}}{\mathrm{dt}}A(t)=-l\cos\theta(t)\frac{\mathrm{d}\theta}{\mathrm{dt}}$
Where $\mathrm{d}A/dt$ is the speed of A.(1m/s)
Then we know that $\mathrm{d\theta}/\mathrm{d}t=-1/l\cos\theta$
The velocity of B,
$\frac{\mathrm{d}}{\mathrm{dt}}B(t)=-l\sin\theta(t)\frac{\mathrm{d}\theta}{\mathrm{dt}}=l\sin\theta(t)\frac{1}{{l\cos\theta}}=\tan\theta$.
$\tan30^{\circ}=\sqrt{3}/2$ and $\tan45^{\circ}=1$.