Consider $\Delta ABD$. Since $AD$ is a vertical pole, $\angle ADB = 90^\circ$. By applying the Pythagoras Theorem, we have
$\begin{array}{rcl}
BD^2 & = & AB^2 – AD^2 \\
BD^2 & = & 25^2 – 15^2 \\
BD & = & 20\text{ m}
\end{array}$
Consider $\Delta BCD$.
$\begin{array}{cl}
& BD^2 + CD^2 \\
= & 20^2 + 21^2 \\
= & 841\text{ m}^2
\end{array}$
Also,
$\begin{array}{cl}
& BC^2 \\
= & 29^2 \\
= & 841\text{ m}^2
\end{array}$
$\because BD^2 + CD^2 = BC^2$,
$\therefore$ by the converse of Pythagoras Theorem, $\Delta BCD$ is a right-angled triangle with $\angle BDC = 90^\circ$.
Since $\angle BDC = 90^\circ$ and $\angle ADB = 90^\circ$, then $D$ is the projection of $B$ on the plane $ACD$. Therefore, the angle between $AB$ and the plane $ACD$ is $\angle BAD$.
In $\Delta ABD$,
$\begin{array}{rcl}
\cos \angle BAD & = & \dfrac{AD}{AB} \\
\cos \angle BAD & = & \dfrac{15}{25} \\
\angle BAD & = & 53.130\ 102\ 35^\circ \\
\angle BAD & \approx & 53^\circ
\end{array}$
