At what angle do the curves $x^{2}+y^{2}=16$ and $y^{2}=6 x$ intersect?
The sought angle is the angle between the tangents at the intersection point of the two curves. We determine the intersection point of the two curves. Substituting the value of $y^{2}=6 x$ into the equation of the circle, we obtain the quadratic equation

$$
x^{2}+6 x-16=0
$$

whose roots are $x_{1}=2$ and $x_{2}=-8$. Consequently, $y_{1,2}=\sqrt{6 \cdot 2}= \pm 2 \sqrt{3}$. The root $x_{2}=-8$ does not correspond to a real intersection point. Since both curves are symmetric with respect to the $x$-axis, it is sufficient to calculate the angle between the tangents at the point $(2,2 \sqrt{3})$. The equation of the tangent to the parabola and the circle at a point $\left(x_{1}, y_{1}\right)$ on the curve is $y_{1} y=p\left(x+x_{1}\right)$ and $x_{1} x+y_{1} y=r^{2}$, respectively, so the slopes of the tangents are $\frac{p}{y_{1}}$ and $-\frac{x_{1}}{y_{1}}$.

Thus, in this case, the slope of the tangent to the parabola is $m_{1}=\frac{p}{y_{1}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$, and the slope of the tangent to the circle is $m_{2}=-\frac{x_{1}}{y_{1}}=-\frac{2}{2 \sqrt{3}}=-\frac{\sqrt{3}}{3}$, and thus the tangent of the angle $\omega$ between the two tangents is

$$
\operatorname{tg} \omega=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}=\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{3}}{1-\frac{1}{2}}=\frac{5 \sqrt{3}}{3}=\frac{5}{\sqrt{3}}
$$

from which $\omega=70^{\circ} 54^{\prime}$.