Linear bound on angles in an euclidean triangle. I am trying to understand a proof in the book of Burago "A Course in metric geometry" (Lemma 10.8.13 page 383). 
I have difficulties with a certain inequality for the angle of euclidean triangles: 
Assume we are given a triangle $\triangle paq$ in euclidean space. By the cosine law the angle $\sphericalangle paq$ between the sides $[pa],[aq]$ is given by $$\sphericalangle paq =  \arccos \left( \frac{\vert pa \vert ^2  + \vert aq \vert ^2 - \vert pq \vert^2 }{ 2 \vert pa \vert \vert aq \vert} \right). $$
Where $\vert pa \vert $ is short for $d(p,a)$ (euclidean distance). 
Now assume we have the following inequality 
\begin{equation}\vert pq \vert \leq \frac{\varepsilon}{4} \vert pa\vert. \qquad (*)\end{equation} 
The book says this implies $$\sphericalangle paq < \arcsin (\frac{\varepsilon}{4}) < \varepsilon/2 .$$
I have tried to verify this vor small $\varepsilon$. But have not been successfull so far. 
Here are my (futile) attempts: 
By triangle inequality and (*) we get 
$$ \vert aq \vert \leq \vert pq \vert + \vert pa \vert \leq (1+ \varepsilon/4)\vert pa \vert. $$
by interchanging the roles of $\vert pa \vert, \vert aq \vert$ we get 
$$ (1-\varepsilon/4) \vert pa \vert \leq \vert  aq \vert \leq  (1+ \varepsilon/4)\vert pa \vert. \qquad (**) $$
Now (**) and (*) imply 
\begin{align}
\frac{\vert pa \vert ^2  + \vert aq \vert ^2 - \vert pq \vert^2 }{ 2 \vert pa \vert \vert aq \vert} & \geq \frac{\vert pa \vert ^2 (1 + (1-\varepsilon/4)^2 - (\varepsilon/4)^2) }{ 2 \vert pa \vert \vert aq \vert} \\ 
& \geq \frac{\vert pa \vert ^2 (1 + (1-\varepsilon/4)^2 - (\varepsilon/4)^2)}{ 2 \vert pa \vert^2 (1 + \varepsilon/4)} \\ 
& = \frac{(1-\varepsilon/4)}{(1+\varepsilon/4)} \\ 
\end{align}
so we get by monotonicity of $\arccos$: 
$$ \sphericalangle paq \leq \arccos \left( \frac{(1-\varepsilon/4)}{(1+\varepsilon/4)} \right)$$
But since
$$ \frac{d}{d\varepsilon} \arccos \left( \frac{(1-\varepsilon/4)}{(1+\varepsilon/4)} \right) = \frac{2}{(4+x)\sqrt{x}}$$
The function $$ \arccos \left( \frac{(1-\varepsilon/4)}{(1+\varepsilon/4)} \right) $$ cannot be bounded by a linear function $f(\varepsilon)$. 
I would appreciate any ideas on how to make this inequality work. 
Thanks!  
 A: Here is a geometrical explanation. To find the maximal value for angle  $\sphericalangle paq$  (let me use $\equiv\alpha$ for that angle to save the typing), for given $|ap|$ and $|qp|$, let us draw a circle of radius $\frac{\epsilon}{4}|pa|$ around point $p$. 

The condition $|pq| \le \frac{\epsilon}{4}|pa|$ means that q must lie either inside or on the circumference of the circle. For all such $q$, the largest angle $\alpha$ is realized when $q$ belongs to the circumference and |aq| is tangent to the circle, which means that $\alpha_{max}$ can be found from the right-angled triangle $aqp$, with $\sphericalangle aqp = \pi/2$:
$$
\sin \alpha_{max} = \frac{\frac{\epsilon}{4} |ap|}{|ap|} = \frac{\epsilon}{4}
$$
For all other possible points q that satisfy $|pq| \le \frac{\epsilon}{4}|pa|$ one has
$$
\sin \alpha \le  \frac{\epsilon}{4}
$$
or, equivalently (using that $\arcsin$ is monotonically increasing),
$$
\alpha \le  \arcsin\frac{\epsilon}{4}
$$
The second inequality follows from 
$$
\sin x \ge x/2
$$ for $x\in(0, \pi/2)$, which can be proven, e.g., by inspecting $f(x) = \sin(x) - x/2$. By considering its 1st and 2nd derivatives, it follows that in the interval $(0,\pi/2)$, function $f(x)$ has a single extremum(maximum) at $x_m=\pi/3$ , with $f(x_m)>0$. Given that at the endpoints of the interval the function has values $f(0)= 0$ and $f(\pi/2) = 1-\pi/4 >0$, function $f(x)$ must be positive for all $x$ in $(0,\pi/2)$ by http://en.wikipedia.org/wiki/Extreme_value_theorem. The final step is to notice that $\arcsin$ is a monotonically increasing function, so inequality $x/2 \le \sin(x)$ implies $\arcsin(x/2) \le x$. Your second inequality is obtained by setting  $x=\epsilon/2$.
A: I guess you're not interested in this anymore. However, I struggled with this some time and the following argument was pretty convincing for me: the first inequality it's simpy because of the law of sines applied to $\tilde{\triangle}apq$ in $\mathbb{R}^2$. For the next one, just take a look at the graphs of the functions $f(x) = 2x$ and $g(x) = \arcsin(x)$. It's pretty clear that $f(x) \geq g(x)$ for $x\in [0,1]$. Thus, $\epsilon/2\geq \arcsin(\epsilon/4)$ since $0<\epsilon<<1$. You can even convince yourself of the last inequality by noting that $\lim_{\epsilon \to 0} \frac{\sin(\epsilon/2)}{\epsilon/2} = 1 > 1/2$.   
