Does there exist a function $\tau(\varepsilon)=\tau(\varepsilon,n,K,\mu)$ such that $\lim_{\varepsilon\to +0}\tau(\varepsilon)=0$ and for any $n$dimensional complete Riemannian manifold $M^n$ with sectional curvature $\geq K$ and with the injectivity radius $\geq \mu>0$ the following property is satisfied: if in a geodesic triangle of diameter less than $\mu/100$ one of the angles is at least $\pi\varepsilon$ then the other two angles are less than $\tau(\varepsilon)$?
Yes it is true and the statement follows from Toponogov's comparison.
Assume that triangle $[xyz]$ is small and $\measuredangle [y^x_z]>\pi\varepsilon$. Extend the side $[zy]$ behind $y$ and marke the point $v$ on the extension such that $yv=\tfrac\mu2$. Note that $\measuredangle [y^x_v]<\varepsilon$. By Toponogov's comparison we get an upper bound on $xv$.
Further extend $[zy]$ behind $z$ and mark a point $w$ on the extension such that $yw=\tfrac\mu2$. If $\measuredangle [z^x_y]>\alpha$ then $\measuredangle [z^x_w]<\pi\alpha$. From Toponogov's comparison we get an upper bound on $xw$.
Note that $vw=\mu$ and by triangle inequality $$xv+xw\ge\mu.$$ Assuming that $\alpha$ is big, the later contradicts the estimates above.

$\begingroup$ Following the nice idea you described, I tried to reconstruct the details. I got that the statement is true for triangles of diameter $\leq \varepsilon\mu/100$ rather than $\leq \mu/100$. Did I miss something or this is what you ment? $\endgroup$– maktSep 21 '15 at 8:55

1$\begingroup$ @sva $\mu/100$ should work, do it more carefully better assume $K=0$ for beginning. You should get that $\alpha<\mathrm{const}\cdot\varepsilon$. The estimates are very close the proof of 7.16 in" A.D. Alexandrov spaces with curvature bounded below" by Burago, Gromov and Perelman. $\endgroup$ Sep 21 '15 at 11:04

$\begingroup$ It seems you are right, although I failed to see the relation to the above mentioned reference. Thank you. $\endgroup$– maktSep 23 '15 at 8:14