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

Since ∠DCY = 30°, by the alternate segment theorem, we find that the angle between the tangent AC at point D and the chord AD (which is ∠ADX) is equal to the angle subtended by the chord AB at point C. Therefore, ∠ADX = ∠DCY = 30°.

To find ∠CAB, we can apply the angle at the center theorem. The angle at the center (which is ∠ZCB) is twice that of the angle at the circumference (which is ∠CAB). Given that ∠ZCB = ∠ZYC + ∠YCB = 100° + 30° = 130°, we can determine that ∠CAB = rac{1}{2} ∠ZCB = rac{1}{2}(130°) = 65°.

In a parabola, a focal chord is a line segment that passes through the focus. Given the equation for the chord PQ, we substitute the coordinates of points P and Q into the parabola's equation. If PQ is a focal chord, then the product of the slopes (p and q) at these points should satisfy the equation pq = -1, hence demonstrating the required relationship.

Given the relationship of focal chords in the parabola, if P has coordinates (8a, 16a) and we denote Q's coordinates as (x, y), we can use the known relationship involving p and q, alongside the coordinates of P, to solve for x and y in terms of a. After plugging values into the equations applicable to the parabola, we can find Q's coordinates as Q(0, 4a).

Using the triangle formed by points A, O, and M, we apply trigonometric ratios. We know that tan(15°) = h/OA, hence OA = h/tan(15°). As cot(15°) is the reciprocal of tan(15°), we derive OA = h cot(15°).

From the previous step, we express OA in terms of h, and since we have a triangle with the total distance of 2000m traveled from A to B. Using the angles of elevation from both A and B to compute the sides, we ultimately solve the equations to derive the precise value of h.

We can approach this by using the chord length relationship where the chord length (160 cm) can be expressed in terms of the radius and central angle θ using the formula l = 2r sin(θ/2). Therefore, squaring both sides yields the required result after simplification and using cosine rule.

Using the result from the previous step, we substitute the expression derived into the problem conditions to create a quadratic equation in terms of θ, which simplifies to the form 8θ² + 25cos θ - 25 = 0.

To apply Newton's method, we first calculate the derivative of the equation derived in the previous part and substitute θ₁ = π into both the function and its derivative to find θ₂. The approximation is given by θ₂ = θ₁ - f(θ₁)/f'(θ₁), and solving gives the second approximation to θ, rounded to two decimal places.