According to a Theorem of Herman ([@Hershort], see also Theorem 4.2 in [@widths] for a simpler version) for all $s\in \R$, $$\trans(F_{\alpha+sa, a}) = \alpha + a\cdot (s+\cal M_\p)+ o(a).$$ As a consequence, if $s>-\cal M_\p$ and $a$ is sufficiently close to $0$, then $\trans(F_{\alpha+sa, a}) >\alpha$ . If $s<-\cal M_\p$ and $a$ is sufficiently close to $0$, then $\trans(F_{\alpha+sa, a}) <\alpha$. Hence $$\gamma_{\alpha}(a)=\alpha-\cal M(\p)a+o(a).$$

A small calculation gives the following corollary.

The angle between the left and right bounding curves of $\cal T_{p/q}$ is $$\arctan \frac {(M_{\cal A}-m_{\cal A})(1+m_{\cal A}M_{\cal A})}{(m_{\cal A}M_{\cal A})^2}.$$

For a fixed $q$ the angle between the two bounding curves of $\cal T_{p/q}$ remains same for all $p$ coprime to $q$.

If $\cal A_q(\p) $ is non constant for some well chosen $\p$ then we have a non zero angle between the boundaries of the $p/q$ tongue. Checking the Fourier expansion of $\p$ we note that if all the Fourier coefficients of $\p$ are non zero then we would have angle between the boundaries of each rational tongue. Such an example is $$\p(x)=e^{\cos (2\pi x)}\sin(\sin(2\pi x))=\displaystyle {\sum_{n\ge0} \frac{\sin(2\pi nx)}{n!}}.$$ For this choice of $\p$ we have $$\cal A_q(\p) (x)=\displaystyle{\sum_{m\ge0} \frac{\sin(2\pi mqx)}{m!}};$$ which is non constant 