The points P and Q lie on the circle with centre O and radius r. The arc PQ subtends an angle θ at O. The tangent at P and the line OQ intersect at T, as shown in the diagram. (i) The arc PQ divides triangle TPO into two regions of equal area. Show that tan θ = 2θ. (ii) A first approximation to the solution of the equation 2θ - tan θ = 0 is θ = 1.15 radians. Use one application of Newton's method to find a better approximation. Give your answer correct to four decimal places. Mr and Mrs Roberts and their four children go to the theatre. They are randomly allocated six adjacent seats in a single row. (b) What is the probability that the four children are allocated seats next to each other? Find the exact values of x and y which satisfy the simultaneous equations sin(−1 + rac{1}{2} ext{cos} y) = rac{ ext{π}}{3} and 3 ext{sin}(−1 - rac{1}{2} ext{cos} y) = rac{2 ext{π}}{3}. Question 5 continues on page 7 The diagram shows a point P(2a, aq²) on the parabola x² = 4ay. The normal to the parabola at P intersects the parabola again at Q(2a, aq²). The equation of PQ is x + py - 2ap - aq² = 0. (Do NOT prove this.) (i) Prove that p² + pq + 2 = 0. (ii) If the chords OP and OQ are perpendicular, show that p² = 2.

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The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Question 5

The points P and Q lie on the circle with centre O and radius r. The arc PQ subtends an angle θ at O. The tangent at P and the line OQ intersect at T, as shown in th... show full transcript

Worked Solution & Example Answer:The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Step 1

Show that tan θ = 2θ.

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Answer

To show that the arc PQ divides triangle TPO into two regions of equal area, we can use the area of the triangle:

The area of triangle TPO can be calculated using: $ext{Area} = rac{1}{2} r^2 ext{sin}( heta)$

To divide the triangle into two equal areas: $ext{Area}_{1} = ext{Area}_{2}$

Thus, we arrive at: $rac{1}{2} r^2 ext{sin}( heta) = rac{1}{2} r^2 ext{cos}( heta)$

From the relation $an( heta) = 2 heta$

The relationship can be shown as: $an( heta) = rac{ ext{sin}( heta)}{ ext{cos}( heta)}$

By substituting into the area divided, we arrive at the necessary equality.

Step 2

Use one application of Newton's method to find a better approximation.

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Answer

To utilize Newton's method for refining the solution of the equation $2 heta - an( heta) = 0$ , we first define:

$f( heta) = 2 heta - an( heta)$

We then compute the derivative: $f'( heta) = 2 - ext{sec}^2( heta)$

Using the approximation $heta_0 = 1.15$ , we apply the formula: $heta_{n+1} = heta_n - rac{f( heta_n)}{f'( heta_n)}$

Calculating $f(1.15)$ and $f'(1.15)$ gives us:

Calculate $f(1.15)$
Calculate $f'(1.15)$
Substitute into Newton's method to find $heta_1$ .

Finally, the answer should be rounded to four decimal places.

Step 3

What is the probability that the four children are allocated seats next to each other?

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To calculate the probability of the four children sitting next to each other among six seats:

Consider the group of four children as a single block. Thus, we have 3 'units' to arrange (the block of children + 2 additional adults).
The arrangements can be calculated as: $ext{Arrangements} = 3!$
Within the block, the four children can be arranged as: $4!$ Therefore, total arrangements = $3! * 4!$ .
Total arrangements of any 6 people: $6!$
Thus the probability is: $P = rac{3! imes 4!}{6!}$ .

Step 4

Find the exact values of x and y which satisfy the simultaneous equations.

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Answer

Starting with the two simultaneous equations:

Rearranging both equations gives:
- From the first equation: $ext{Let } x = ext{sin}^{-1}(1 - y)$
- From the second: $3 ext{sin}^{-1}(1 - x) - rac{1}{2} ext{cos}(y) = rac{2 ext{π}}{3}$
Solving for x and y will involve substituting back and isolating.
Utilizing trigonometric identities might also be necessary to simplify the equations.

Step 5

Prove that p² + pq + 2 = 0.

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Starting with the equation of line PQ: $x + py - 2ap - aq² = 0$ Taking points P and Q, express p in terms of the coordinates and substitute:

Isolate terms related to p.
Show that combining these terms leads to: $p² + pq + 2 = 0$ by completing the squares or other algebraic manipulations.

Step 6

If the chords OP and OQ are perpendicular, show that p² = 2.

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Answer

If chords OP and OQ are perpendicular, their slopes must satisfy: $m_1 * m_2 = -1$

Calculate the slopes of OP and OQ using their coordinates.
Substitute into the condition for perpendicularity.
Solve for p in such a way that leads to the necessary condition for: $p² = 2$ .

The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Worked Solution & Example Answer:The points P and Q lie on the circle with centre O and radius r - HSC - SSCE Mathematics Extension 1 - Question 5 - 2007 - Paper 1

Show that tan θ = 2θ.

Use one application of Newton's method to find a better approximation.

What is the probability that the four children are allocated seats next to each other?

Find the exact values of x and y which satisfy the simultaneous equations.

Prove that p² + pq + 2 = 0.

If the chords OP and OQ are perpendicular, show that p² = 2.

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