$Given ellipse \frac{x^2}{16}+\frac{y^2}{12}=1 where a^2=16,b^2=12$

$$\begin{gathered} Given circle x^2+y^2=15 \\ where centre=(0,0),r=\sqrt{15} \end{gathered}$$

let 'm' be slope of common tangent for the ellipse and circle

The equation of tangent is $\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$ 

$\Rightarrow \mathrm{y}=mx\pm \sqrt{16{\mathrm{m}}^2+12} -\left(1\right)$

for circle, equation of tangent is $\mathrm{y}=\mathrm{mx}\pm \sqrt{15} \sqrt{1+{\mathrm{m}}^2} -\left(2\right)$

(1) & (2) represents same tangent then 

$\pm \sqrt{16{\mathrm{m}}^2+12}=\pm \sqrt{15(1+{\mathrm{m}}^2)}$

squaring on both sides 

$\begin{array}{rcl}16{\mathrm{m}}^2+12& =& 15+15{\mathrm{m}}^2\\ & \Rightarrow & {\mathrm{m}}^2=3\\ & \Rightarrow & \mathrm{m}=\sqrt{3}\\ & \Rightarrow & \mathrm{tan\theta }=\tan {60}^0 \Rightarrow \mathrm{\theta }=\frac{\mathrm{\pi }}{3}\end{array}$