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

Using the tangent chord theorem, we know that the angle formed between a line drawn from the point of tangency to the circle and the chord through that point is equal to the angle subtended by the chord at the opposite arc. Thus, we have:

$\angle ADX = \angle DCY = 30^\circ$

To find $\angle CAB$, we can use the fact that angles subtended by the same chord are equal. Therefore:

$\angle CAB = \angle ZACB = 80^\circ$

For the points P and Q to be focal chords in the parabola, their coordinates must satisfy the relationship between focal length and the directrix. By substituting $pq = -1$, we utilize the properties of the chord from the definitions of focal points in parabolas.

Since P has coordinates determined under the context of the parabolic geometry, the coordinates of Q can be derived as follows:

If $P(8a, 16a)$ satisfies the parabolic condition, we find the corresponding $Q(x, y)$ by ensuring it symmetrically aligns with respect to the focal point. So, for this curve, we can derive point Q based on the reflective property.

In triangle OMA, using trigonometric definitions, we can express:

$\tan(15^\circ) = \frac{h}{OA} \implies OA = \frac{h}{\tan(15^\circ)} = h \cot(15^\circ)$

Using the angles and lengths M and O to form two right triangles with known angles, we can use the law of sines or other relations to find:

$h = OA \tan(15^\circ)$, where OA can be calculated based on the road distance and angle.

Using the relationship of the arc and chord in the circle, we apply the law of cosines in triangle OAC, establishing the radius r in those terms which leads to:

$160^2 = 2r^2(1 - \cos(\theta))$

After deriving the previous equation, we can adjust the terms and manipulate to form this quadratic equation through rearrangement and substitution, ultimately yielding:

$80^2 + 25\cos(\theta) - 25 = 0$

Applying Newton's method involves setting $f(\theta) = 80^2 + 25\cos(\theta) - 25$ and iteratively refining the estimate through the derivative calculated at the surrounding quadrant ensuring convergence towards an accurate solution to two decimal places.