An isosceles triangle of wood of base 10 feet and height 8 3 feet...

Let, $ABC$ is the triangle and triangle $BCE$ is its shadow on the ground
Height of the triangular shadow $= DE = \dfrac{8}{\sqrt{3}}cot {30}^{o} = 8$ feet
$∴ tan \alpha = \dfrac{5}{8}$ (from $∆ BDE$ )
$\tan \left(2\alpha\right) = \dfrac{2tan \alpha}{1 - \tan^{2} \alpha}$
$= \dfrac{2 \times \dfrac{5}{8}}{1 - \dfrac{25}{64}} = \dfrac{5}{4} \times \dfrac{64}{39} = \dfrac{80}{39}$