3. Given that $AB$ is a chord of the circle $\odot O$ with radius 1, and the length of $AB$ is the positive root of the equation $x^{2}+x-1=0$. Then the degree of $\angle AOB$ is $\qquad$ .
3. $36^{\circ}$.

From $x^{2}+x-1=0$, we know $x=1-x^{2}<0$, so $AB < OB$. Therefore, on $OB$, we take $OC = AB = x$. Also, from $x^{2}+x-1=0$, we can get $\frac{x}{1-x}=\frac{1}{x}$. As shown in the figure, $\frac{AB}{BC} = \frac{OA}{AB}$, thus $\triangle OAB \sim \triangle ABC$. Since $\triangle OAB$ is an isosceles triangle, we have $AC = AB = OC$. Given $\angle 1 = \angle 2 = \angle 3$, we have $5 \angle AOB = 180^{\circ}$.