A, B and C are points (6, 10), (6,6) and (q, 2) respectively. The area of triangle ABC is 16 square units.

1. Find the opposite and adjacent lengths for $\angle CAB$ and apply the tangent ratio.

From part (a), $q=-2$, so $C=(-2,2)$.

From $A(6,10)$ to $C(-2,2)$,
$\text{horizontal change}=6-(-2)=8,$
$\text{vertical change}=10-2=8.$

Since $AB$ is vertical, for $\angle CAB$,
$\tan \angle CAB=\frac{\text{opposite}}{\text{adjacent}}=\frac{8}{8}=1.$

1. Find the distance between $A$ and $C$ using their coordinates.

From part (a), $q=-2$, so $C=(-2,2)$.

$AC=\sqrt{(6-(-2))^2+(10-2)^2}$
$=\sqrt{8^2+8^2}$
$=\sqrt{128}=8\sqrt{2}$

1. Relate the given area to base $AC$ and the required perpendicular distance.

Let the perpendicular distance from $B$ to $AC$ be $h$.

Using $AC=4\sqrt{5}$ and area of $\triangle ABC=16$,
$\frac12 \times AC \times h=16$
$\frac12 \times 4\sqrt{5}\times h=16$

2. Solve for the perpendicular distance and simplify the exact surd form.

$2\sqrt{5}h=16$
$h=\frac{16}{2\sqrt{5}}=\frac{8}{\sqrt{5}}$
Rationalising the denominator,
$h=\frac{8\sqrt{5}}{5}$