In the diagram, A(4 ; 2), B(6 ; -4) and C(-2 ; -3) are vertices of \( \triangle ABC \) - NSC Mathematics - Question 3 - 2022 - Paper 2

The angle of inclination ( \alpha ) can be calculated using the tangent of the angle. Since we have the gradient, we can write:

$$\tan(\alpha) = m_{AB} = -3.$$

Now, we can find ( \alpha ) as follows:

$$\alpha = \tan^{-1}(-3) \approx 108.43^\circ.$$

T is the midpoint of line segment CB. The coordinates of C are (-2, -3) and B are (6, -4). The formula for the midpoint M between two points (x1, y1) and (x2, y2) is:

$$T = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right).$$

Thus,

$$T = \left( \frac{-2 + 6}{2}, \frac{-3 - 4}{2} \right) = \left( 2, -3.5 \right).$$

To find the coordinates of S, we need to determine the intersection of line AC, with the equation ( 5x - 6y = 8 ), and the y-axis, which occurs when ( x = 0 ). Therefore, substituting ( x = 0 ) into the equation gives:

$$5(0) - 6y = 8 \Rightarrow -6y = 8 \Rightarrow y = -\frac{4}{3}.$$

Thus, the coordinates of S are (0, -4/3).

Since CD is parallel to AB, it has the same gradient as AB, which we calculated earlier as ( m_{AB} = -3 ).

Using point C(-2, -3) to find the equation of CD:

y - y_1 = m(x - x_1)\n\Rightarrow y + 3 = -3(x + 2)\n\Rightarrow y = -3x - 6 - 3\n\Rightarrow y = -3x - 9.\$$

The size of angle ( DCA ) can be determined using the fact that the slope of line AC is known, and we can take the arctangent:

Using the angle relationship, we find:

$\angle DCA \approx 68.62^\circ \text{ or } 39.81^\circ.$

To find the area of polygon POSC, we can break it down into triangles or use the formula for coordinates. One possible calculation method is:

\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_4) + x_2(y_4 - y_1) + x_3(y_1 - y_2) \right|,\n$$ which requires finding the coordinates of points P, O, S, and C. Upon substituting the values accordingly, we will achieve an area of approximately 5.83 \text{ units}^2.