The triangle with vertices A (2, 5), B (−2, −3) and C (2, −2) is shown in the diagram below.

1. Identify a suitable base and perpendicular height from the coordinates

Since $A(2,5)$ and $C(2,-2)$ have the same $x$-coordinate, $AC$ is vertical.

$AC=5-(-2)=7$

The perpendicular distance from $B(-2,-3)$ to the vertical line $x=2$ is

$2-(-2)=4$

2. Calculate the area of the triangle

$\text{Area of }\triangle ABC=\frac{1}{2}\times 7\times 4=14$

1. Find the translation from $B$ to $C$ and apply it to $A$ to locate $D$.

For parallelogram $ABCD$, $\overrightarrow{AD}=\overrightarrow{BC}$.

$B(-2,-3)$ to $C(2,-2)$ gives
$\overrightarrow{BC}=\begin{pmatrix}2-(-2)\\-2-(-3)\end{pmatrix}=\begin{pmatrix}4\\1\end{pmatrix}.$

So
$D=A+\overrightarrow{BC}=(2,5)+(4,1)=(6,6).$

1. Use the triangle area from part (a) to find the area of the parallelogram

Area of parallelogram $ABCD = 2 \times$ area of triangle $ABC$

$= 2 \times 14$

$= 28\text{ unit}^2$

1. Identify the relevant side lengths for angle $BAC$

Since $A(2,5)$ and $C(2,-2)$ have the same $x$-coordinate, $AC$ is vertical.

$AC=5-(-2)=7$

The horizontal distance from $B(-2,-3)$ to the line $AC$ is

$2-(-2)=4$.

So, for angle $BAC$, the opposite side is $4$ and the adjacent side is $7$.

2. Use the tangent ratio to find angle $BAC$

$\tan \angle BAC=\frac{4}{7}$

$\angle BAC=\tan^{-1}\left(\frac{4}{7}\right)=29.7^\circ$

1. Determine the gradient of line $AC$ from the coordinates of $A$ and $C$

For $A(2,5)$ and $C(2,-2)$,
$\text{gradient of } AC=\frac{5-(-2)}{2-2}=\frac{7}{0}.$
Since division by zero is undefined, the gradient of $AC$ is undefined.

1. Identify the adjacent side and hypotenuse for angle $BCA$, then apply the cosine ratio

From the coordinates, $A(2,5)$ and $C(2,-2)$, so

$AC=5-(-2)=7$.

Given $p=BC$, for angle $BCA$,

$\cos \angle BCA=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AC}{BC}=\frac{7}{p}.$