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

According to the tangent-chord theorem, the angle formed between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Therefore: [ \angle ADX = \angle DCY = 30^\circ ] Thus, ( \angle ADX = 30^\circ ).

Using the fact that the angles in a triangle sum to ( 180^\circ ): [ \angle CAB + \angle ZACB + \angle ZBC = 180^\circ ] We already know ( \angle ZACB = 50^\circ ) and can find ( \angle ZBC = \angle ZCYB - \angle ZCB = 100^\circ - 60^\circ = 40^\circ ). Therefore: [ \angle CAB + 50^\circ + 40^\circ = 180^\circ ] This simplifies to: [ \angle CAB = 180^\circ - 90^\circ = 90^\circ ] Thus, ( \angle CAB = 90^\circ ).

To show this, we utilize the property of the focal chord that states if P and Q lie on a parabola with a focal point at (0, a), the product of the slopes of the two lines formed by the focal points is -1: [ PQ: (p + q)yx - 2pq = 0 \ \Rightarrow \text{if } pq = -1, \text{ the geometrical interpretation holds true.} ] Thus, it is shown.

Using the focal chord properties, we set: [\text{Let Q = (2aq, aq^2)}.] Using the information that for focal chords, we have ( pq = -1 ), we can derive: [ q = -\frac{1}{p} = -\frac{1}{\left( \frac{16a}{8a} \right)} = -\frac{1}{2}\text{ or } -4a \text{.} ] Thus, the coordinates of Q are ( (2a(-4), -4^2) = (-8a, 16a) ).

Using the definition of cotangent in the right triangle formed: [ \tan(15^\circ) = \frac{h}{2000}
\Rightarrow OA = \frac{h}{\tan(15^\circ)} = h \cot(15^\circ). ]

Using trigonometric identity, we find h by substituting in the known angle: [ h = 2000 \tan(13^\circ). ] Then compute this value numerically to obtain the specific height h.

This can be derived from the relationship between the chord length and the radius: [ \text{Using the cosine rule: } c^2 = a^2 + b^2 - 2ab \cos(\theta)
\Rightarrow 160^2 = 2r^2(1 - \cos \theta). ]

From previous equation derived, we simplify: [ 160^2 = 2r^2(1 - \cos \theta) = 25 \Rightarrow 80^2 + 25 \cos \theta - 25 = 0. ]

Using Newton's method, the iterative process involves: [ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. ] Where you define your function and its derivative, then compute the iterations based on the approximation until desired accuracy is reached.