Let a circle of radius 4 be concentric to the ellipse $15{\mathrm{x}}^2+19{\mathrm{y}}^2=285.$ Then the common tangents are inclined to the minor axis of the ellipse at the angle

${\mathrm{a}}^2=19,{\mathrm{b}}^2=15(\mathrm{a}>\mathrm{b})$

$$\begin{gathered} \mathrm{Let}\mathrm{y}=\mathrm{mx}\pm \sqrt{19{\mathrm{m}}^2+15}\mathrm{be}\mathrm{a}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Tangent to the ellipse} \\ \begin{array}{l}\mathrm{mx}-\mathrm{y}\pm \sqrt{19{\mathrm{m}}^2+15}=0\text{\hspace{0.17em}\hspace{0.17em}is also the tangent to }{\mathrm{x}}^2+{\mathrm{y}}^2=16\\ \therefore \mathrm{r}=\mathrm{d}\\ \frac{\left|\sqrt{19{\mathrm{m}}^2+15}\right|}{\sqrt{1+{\mathrm{m}}^2}}=4\end{array} \\ \begin{array}{l}3{\mathrm{m}}^2=1\\ \therefore \mathrm{m}=\pm \frac{1}{\sqrt{3}}\\ ({30}^0,{150}^0\text{\hspace{0.17em}\hspace{0.17em}with major axis})\end{array} \\ \mathrm{Therefore}, \mathrm{angle} \mathrm{made} \mathrm{by} \mathrm{the} \mathrm{common} \mathrm{tangent} \mathrm{with} \mathrm{minor} \mathrm{axis} \mathrm{is} 60° \end{gathered}$$