1. Let the left and right vertices of the hyperbola $x^{2}-y^{2}=6$ be $A_{1}$ and $A_{2}$, respectively, and let $P$ be a point on the right branch of the hyperbola such that $\angle P A_{2} x=3 \angle P A_{1} x+10^{\circ}$. Then the degree measure of $\angle P A_{1} x$ is $\qquad$.
$=、 1.20^{\circ}$.
Let $P\left(x_{0}, y_{0}\right)$, then $x_{0}^{2}-y_{0}^{2}=6$. Since $\tan \angle P A_{1} x=\frac{y_{0}}{x_{0}+\sqrt{6}}, \tan \angle P A_{2} x=\frac{y_{0}}{x_{0}-\sqrt{6}}$, therefore, $\tan \angle P A_{1} x \cdot \tan \angle P A_{2} x=\frac{y_{0}^{2}}{x_{0}^{2}-6}=1$. Hence $\angle P A_{2} x=90^{\circ}-\angle P A_{1} x$.
From $3 \angle P A_{1} x+10^{\circ}=90^{\circ}-\angle P A_{1} x$, we get $\angle P A_{1} x=20^{\circ}$.