The diagram represents a tower, AB, built on horizontal ground. The height of the tower is 12 m. Point D is 11 m from C and DCB is a straight line. It is given that $\sin \angle ACB = \dfrac{12}{13}$. Calculate the angle of depression in degrees of point D from the top of the tower. Answer ___ °

1. Use the given sine ratio to identify the 5-12-13 triangle and find $BC$.

In right-angled triangle $ABC$,
$\sin \angle ACB=\frac{AB}{AC}=\frac{12}{13}.$
Since $AB=12$, $AC=13$.

By Pythagoras,
$BC=\sqrt{13^2-12^2}=\sqrt{169-144}=5\text{ m}.$

2. Find the horizontal distance from the foot of the tower $B$ to point $D$.

Since $D,C,B$ are collinear and $DC=11$ m,
$DB=DC+CB=11+5=16\text{ m}.$

3. Set up the trigonometric ratio for the required angle.

Let the angle of depression of $D$ from $A$ be $\theta$.

The horizontal through $A$ is parallel to the ground, so this angle equals $\angle ADB$.

In right-angled triangle $ABD$,
$\tan \theta=\frac{AB}{DB}=\frac{12}{16}.$

4. Calculate the angle and round appropriately.

$\theta=\tan^{-1}\left(\frac{12}{16}\right)=36.869\ldots^\circ$

Hence the angle of depression is
$\theta\approx 36.9^\circ.$