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 $\tau(\varepsilon)$?



A property of geodesic triangles in manifolds with lower bounds on curvature and injectivity radius
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 $|y-v|=\tfrac\mu2$.
Note that $\measuredangle [y^x\_v]<\varepsilon$.
By Toponogov's comparison we get an upper bound on $|x-v|$.


Further extend $[zy]$ behind $z$ and mark a point $w$ on the extension such that $|y-w|=\tfrac\mu2$.
If $\measuredangle [z^x\_y]>\alpha$ then $\measuredangle [z^x\_w]<\pi-\alpha$.
From Toponogov's comparison we get an upper bound on $|x-w|$.


Note that $|v-w|=\mu$ and by triangle inequality
$$|x-v|+|x-w|\ge\mu.$$
Assuming that $\alpha$ is big, the later contradicts the estimates above. 


