Comparison of angles in Alexandrov space Let $X$ be a finite dimensional Alexandrov space with curvature bounded below. Let $p\in X$ be a fixed point.
Is it true that for any $\varepsilon >0$ there exists $\delta>0$ such that for any $x,y\in B(p,\delta)$ one has
$$\angle xpy -\tilde \angle xpy<\varepsilon?$$
Here $B(p,\delta)$ denotes the ball of radius $\delta$ centered at $p$, $\angle xpy$ denotes the angle at p in $X$, and $\tilde \angle xpy$ denotes the corresponding angle in the comparison triangle in the fixed model plane.
 A: Yes it is true. 
Assume contrary, then there are two sequences of points $(x_n)$ and $(y_n)$ such that $$|p-x_n|=|p-y_n|=\tfrac1n$$ 
and 
$$\measuredangle [p^{x_n}_{y_n}]-\tilde\measuredangle (p^{x_n}_{y_n})\ge \epsilon$$
for some fixed $\varepsilon>0$.
We can assume that the directions of $[px_n]$ and $[py_n]$ converge correspondingly to $\xi$ and $\upsilon$ in $\Sigma_p$.
Set 
$$\alpha=\measuredangle(\xi,\upsilon)=\lim_{n\to\infty}\measuredangle [p^{x_n}_{y_n}]$$
Choose two geodesics $[pa]$ and $[pb]$ in the directions close to $\xi$ and $\upsilon$ correspondingly. 
Let $a_n\in[pa]$ and $b\in[pb]$ be the points such that 
$|p-a_n|=|p-b_n|=\tfrac1n$.
Note that for large $n$ we have 
$$\lim_{n\to\infty}\tilde\measuredangle (p^{a_n}_{b_n})\approx \alpha.$$
By comparison, the values  $n\cdot |a_n-x_n|$ and $n\cdot|y_n-b_n|$ are small.
It follows that 
$$\tilde\measuredangle (p^{x_n}_{y_n})\approx\tilde\measuredangle (p^{a_n}_{b_n}).$$
Therefore
$$\lim_{n\to\infty}\tilde\measuredangle (p^{x_n}_{y_n})\approx\alpha,$$
a contradiction.
