The angle between one-sided tangents to the curve $y=e^{|x|}$ at x = 0, is

$\frac{\pi}{2}$

We have,

$y=e^{|x|}=\left\{\begin{array}{ll}
e^{-x}, & x<0 \\ e^x, & x \geq 0 \end{array} \Rightarrow \frac{d y}{d x}= \begin{cases}-e^{-x}, & x<0 \\ e^x, & x>0\end{cases}\right.$

The slopes of the tangents at x = 0 to the curves $y=e^{-x}, x<0$ and $y=e^x, x≥0$ are $m_1=-1$ and $m_2=1$ respectively.

Clearly, $m_1m_2=-1$

Hence, the required angle is a right angle.