In the diagram, the points A, B, C and D are on the circumference of a circle, whose centre O lies on BD - HSC - SSCE Mathematics Extension 1 - Question 12 - 2015 - Paper 1

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Thus, we have:

$\angle ADX = \angle ACB = 50^{\text{o}}.$

Using the property of angles in the same segment,

$\angle CAB = 180^{\text{o}} - (\angle ACB + \angle ACD) = 180^{\text{o}} - (50^{\text{o}} + 30^{\text{o}}) = 100^{\text{o}}.$

For points P$(a_p, a^2_p)$ and Q$(a_q, a^2_q)$ on the parabola defined by $x^2 = 4ay$, the distance product of focal chords has to equal $-1$ when projecting through the focus. Therefore, as per the focal chord properties, we have:

$pq = -1.$

Given Q lies on the chord derived from P's coordinates $(8a, 16a)$, substituting and solving, the coordinates of Q can be determined as:

$Q = (x, y), \; \text{where } y = \frac{(16a)(8a)}{2a} = \frac{128a^2}{2a} = 64a.$

Applying trigonometric relationships in triangle OAM, we find:

$\text{cot} 15^{\text{o}} = \frac{h}{OA} \Rightarrow OA = h \cdot \text{cot} 15^{\text{o}}.$

Using the angles of elevation, we derive two equations and solve them:

From point A: $h = OA \cdot \text{tan} 15^{\text{o}}$ From point B: $h = OB \cdot \text{tan} 13^{\text{o}}.$

Using the cosine rule on triangle OAB where AB is the chord, we state:

$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\text{cos}\theta.$ Substituting values gives:

$160^2 = r^2 + r^2 - 2r^2 \text{cos}\theta = 2r^2(1 - \text{cos}\theta).$

Substituting $r$ into the previous expression allows us to derive this quadratic equation in terms of $\theta$ and formulate:

$8\theta^2 + 25\text{cos}\theta - 25 = 0.$

Using Newton's method, we take an initial guess and iterate:

1. Assume $\theta_1 = \pi$.
2. Apply the formula $\theta_{n+1} = \theta_n - \frac{f(\theta_n)}{f'(\theta_n)}$.

Calculating this should refine our approach to approximate $\theta$ to two decimal places.