In the diagram, BDCE is a straight line. BD = 4 cm, AC = 10 cm and AB = AD.

1. Apply the triangle area formula and solve for the vertical height.

Let the vertical height of triangle $ABD$ be $h$ cm.

$\text{Area of }\triangle ABD=\frac12\times BD\times h$

Given $\text{Area}=16$ and $BD=4$,

$16=\frac12\times4\times h$

$16=2h$

$h=8$

So the vertical height is $8\text{ cm}$.

1. Use the vertical height as the opposite side and AC as the hypotenuse.

The vertical height from A to the straight line is $8\text{ cm}$ and $AC=10\text{ cm}$.

$\sin \angle ACE=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{8}{10}=\frac{4}{5}$

1. Find the horizontal distance from C to the foot of the perpendicular from A.

Let the foot of the perpendicular from $A$ to $BDCE$ be $F$.

From part (a), $AF = 8\text{ cm}$ and $AC = 10\text{ cm}$.

In right-angled triangle $ACF$,
$CF^2 = AC^2 - AF^2$
$CF^2 = 10^2 - 8^2 = 100 - 64 = 36$
$CF = 6\text{ cm}$

2. Use the tangent ratio for angle ACD.

Since $D$, $C$ and $F$ lie on the same straight line, $\angle ACD = \angle ACF$.

$\tan \angle ACD = \frac{\text{opposite}}{\text{adjacent}} = \frac{AF}{CF}$
$\tan \angle ACD = \frac{8}{6} = \frac{4}{3}$