2012-II-40

Ans: D
Let $x$ be the length of each sides of the tetrahedron.

Add a point $P$ on $BC$ such that $AP\perp BC$ and $DP \perp BC$.

In $\Delta ACB$,

$\begin{array}{rcl}
\sin \angle ACB & = & \dfrac{AP}{AC} \\
AP & = & x \sin 60^\circ \\
& = & \dfrac{\sqrt{3}x}{2}
\end{array}$

Similarly, $DP=\dfrac{\sqrt{3}x}{2}$.

In $\Delta APD$, by cosine law

$\begin{array}{rcl}
\cos \angle APD & = & \dfrac{AP^2+DP^2-AD^2}{2(AP)(DP)} \\
& = & \dfrac{1}{3} \\
\angle APD & = & 70.528~779~37^\circ
\end{array}$