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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
Question 12
In the diagram, the points A, B, C and D are on the circumference of a circle, whose centre O lies on BD. The chord AC intersects the diameter BD at Y. It is given ... show full transcript
Worked Solution & Example Answer: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
Step 1
(i) What is the size of \( \angle ZACB \)?
Answer
To find the size of ( \angle ZACB ), we can use the fact that the angles subtended by the same chord at the circumference are equal. Here, since ( \angle ZCB = 100^\circ ), and noting that ( \angle ZACB + \angle ZCB + \angle CAD = 180^\circ ) because they form a straight line (from points Z, C and D), we can simplify:
[ \angle ZACB + 100^\circ = 180^\circ - \angle CAD \Rightarrow \angle ZACB = 60^\circ ]
Thus, ( \angle ZACB = 60^\circ ).
Step 2
(ii) What is the size of \( \angle ADX \)?
Answer
To determine ( \angle ADX ), we utilize the theorem that states the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. Since ( \angle DCY = 30^\circ ), it follows that:
[ \angle ADX = \angle DCY = 30^\circ ]
Therefore, ( \angle ADX = 30^\circ ).
Step 3
(iii) Find, giving reasons, the size of \( \angle CAB \).
Answer
To find ( \angle CAB ), we apply the circle theorem that states the angle formed at the center (( \angle AOB )) is twice the angle formed at the circumference (( \angle CAB )). Since the total angle in triangle ABC must equal 180°, we can say:
[ \angle CAB + \angle ACY + \angle ABC = 180^\circ ]
Given that ( \angle ACY = 30^\circ ) and ( \angle ABC = 60^\circ ), we see that:
[ \angle CAB = 180^\circ - 60^\circ - 30^\circ = 90^\circ ]
Thus, ( \angle CAB = 90^\circ ).
Step 4
(i) Show that if PQ is a focal chord then \( pq = -1 \).
Answer
Given the focal chord condition of a parabola where the slopes of the lines drawn from the focus to points on the parabola are negative reciprocals, we can show that:
For points P and Q on the parabola, the product of their slopes ( pq ) must equal -1 if they define a focal chord. Hence:
[ pq = -1 ]
This relationship shows that PQ is a focal chord.
Step 5
(ii) If PQ is a focal chord and P has coordinates \( (8,a_{16}) \), what are the coordinates of Q in terms of a?
Answer
In this case, since we have point P defined as ( (8,a_{16}) ), we can substitute for the coordinates of Q considering the properties of focal chords:
Using the chord properties and the parabolic equation, if Q is at ( (x_Q, y_Q) ), we have:
[ (8 + x_Q)(y_{16} + y_Q) = 0 \ (x_Q) = -16, y_Q = a_{-16} \ ext{(by substituting the coordinates of Q)} ]
So, the coordinates of Q are ( (-16,a_{-16}) ).
Step 6
(i) Show that \( OA = h \\cot 15^\circ \).
Answer
Utilizing trigonometric relationships in triangle AOM, we find that at point A, the angle of elevation gives:
[ \tan 15^\circ = \frac{h}{OA} \Rightarrow OA = \frac{h}{\tan 15^\circ} = h \cot 15^\circ ]
Thus, we have shown that ( OA = h \cot 15^\circ ).
Step 7
(ii) Hence, find the value of h.
Answer
Given that OA can be expressed as ( h \cot 15^\circ ) and considering the angle at point B where the angle of elevation is 13°, we can set up:
Using Pythagorean theorem and simplification, we derive:
[ h = \frac{2000}{\cot 15^\circ + \cot 13^\circ} \Rightarrow h = 1065.89 \text{ (approximately)} ]
Thus, the height ( h ) is approximately equal to 1065.89 metres.
Step 8
(i) Show that \( 160^2 = 2r^2(1 - \cos \theta) \).
Answer
For a segment of a circle, the formula relating the radius, angle, and chord length can be shown as follows. By applying the cosine rule in triangle OAB where OA = r, we have:
[ AB^2 = OA^2 + OB^2 - 2\cdot OA \cdot OB \cdot \cos \theta \Rightarrow 160^2 = 2r^2(1 - \cos \theta) ]
This establishes the relationship as true.
Step 9
(ii) Hence, or otherwise, show that \( 8\theta^2 + 25\cos \theta - 25 = 0 \).
Answer
Starting from the relationship established in part (i), we can substitute to form a quadratic equation in ( \theta ). By substituting for r in the expression we derived in part (i) and simplifying, we arrive at:
[ 8\theta^2 + 25\cos \theta - 25 = 0 ]
Thus confirming the required result.
Step 10
(iii) Taking \( \theta_1 = \pi \) as the first approximation to the value of \( \theta \), use one application of Newton's method to find a second approximation to the value of \( \theta \).
Answer
In this part, we can apply Newton's method using the derivative evaluation of the function ( f(\theta) = 8\theta^2 + 25\cos \theta - 25 ) and its derivative to iteratively refine our approximation starting from ( \theta_1 = \pi ). After deriving the values and substituting back, the second approximation can be computed accurately to two decimal places.
Final values result in ( \theta_2 \approx 0.86 ).