a\ge 0,\\
                                    \gamma_{p/q}^r(a)\;\text{for}\; a<0. \end{array}\right.$

    2.  $\gamma_{p/q}^+(a)=p/q-m_{\cal A}a+o(a)$, where $\gamma_{p/q}^+(a)
        =\left\{ \begin{array}{ll} \gamma_{p/q}^r(a)\; \text{for}\; a\ge 0,\\
                                    \gamma_{p/q}^l(a)\;\text{for}\; a<0. \end{array} \right.$

The proof depends on how $F_{t,a}^{\circ q}$ behaves near $(p/q,0)$.

\[lemma\_angle\] For small values of $\e$ and $a$, $$F_{p/q+\e,a}^{\circ q}(x)=x+p+q\e + qa \cal A_q(\p) (x)+a\Psi_{p/q}(x,a,\e)$$ where $\Psi_{p/q}$ is a uniformly continuous function which is $0$ for all $x$ if $a=\e=0$.

Note that $$\begin{aligned}
F_{p/q+\e,a}^{\circ q}(x) & = x+p + q\e \\
& \quad + a\p\bigl(F_{p/q+\e,a}^{\circ 0}(x)\bigr)+ a\p\bigl(F_{p/q+\e,a}^{\circ 1}(x)\bigl)
+\ldots+ a\p\bigl(F_{p/q+\e,a}^{\circ q-1}(x)\bigr).\end{aligned}$$ We write $\p(F_{p/q+\e,a}^{\circ k}(x))=\p(x+kp/q)+\psi_k(x,a,\e)$ for each $0 \le k \le q-1$, where $\psi_k$ is defined as $\psi_k(x,a,\e)=\p(F_{p/q+\e,a}^{\circ k}(x))-\p(x+kp/q)$. By definition $\psi_k$ is periodic and continuous, thus it is uniformly continuous and $\psi_k(x,0,0)=0$ for all $x$.

Therefore $$\begin{aligned}
F_{p/q+\e,a}^{\circ q}(x)& =x+p+q\e + a\sum_{k=0}^ {q-1}\p(x+ kp/q) +
   a\sum_{k=0}^{q-1}\psi_k(x,a,\e) \\
& = x+p+q\e + aq\cal A_q(\p) (x)+ a\Psi_{p/q}(x,a,\e);\end{aligned}$$ where $\Psi_{p/q}(x,a,\e)=\ds \sum_{ k=0}^{q-1}\psi_k(x,a,\e)$ and $\Psi_{p/q}(x,0,0)=
\ds \sum_{k=0}^{q-1}\psi_k(x,0,0)=0$.

\[Proof of Proposition \[prop\_angle\]\] First we look at the boundaries of the rational tongue $\cal T_{p/q}$ assuming their modified definition. We consider a particular case here, when we are approaching the left boundary curve $\gamma_{p/q}^-$ from above i.e. $a>0$. Define $$\lambda(a)=\displaystyle\frac{\gamma_{p/q}^-(a)-p/q}{a}.$$ It is enough to show that $\lambda(a) \rightarrow
-M_{\cal A}$ as $a \rightarrow 0$ with 