Geometric Area Equation Setup

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\includegraphics[max width=\textwidth, alt={}, center]{1a4ddaa9-1ec2-4138-bfcb-a482fe6c942f-2_358_618_1082_762} The diagram shows a sector of a circle with radius \(r \mathrm {~cm}\) and centre \(O\). The chord \(A B\) divides the sector into a triangle \(A O B\) and a segment \(A X B\). Angle \(A O B\) is \(\theta\) radians.

1. In the case where the areas of the triangle \(A O B\) and the segment \(A X B\) are equal, find the value of the constant \(p\) for which \(\theta = p \sin \theta\).
2. In the case where \(r = 8\) and \(\theta = 2.4\), find the perimeter of the segment \(A X B\).

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\includegraphics[max width=\textwidth, alt={}, center]{8c533469-393c-4e4c-a6ec-eab1303741e7-2_385_476_1653_836} The diagram shows a sector \(O A B\) of a circle with centre \(O\) and radius \(r\). The angle \(A O B\) is \(\alpha\) radians, where \(0 < \alpha < \frac { 1 } { 2 } \pi\). The point \(N\) on \(O A\) is such that \(B N\) is perpendicular to \(O A\). The area of the triangle \(O N B\) is half the area of the sector \(O A B\).

1. Show that \(\alpha\) satisfies the equation \(\sin 2 x = x\).
2. By sketching a suitable pair of graphs, show that this equation has exactly one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\).
3. Use the iterative formula $$x _ { n + 1 } = \sin \left( 2 x _ { n } \right)$$ with initial value \(x _ { 1 } = 1\), to find \(\alpha\) correct to 2 decimal places, showing the result of each iteration.

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\includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-2_551_567_1416_788} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(O A B\) is \(\theta\) radians. The shaded region is bounded by the circumference of the circle and the arc with centre \(A\) joining \(B\) and \(C\). The area of the shaded region is equal to half the area of the circle.

1. Show that \(\cos 2 \theta = \frac { 2 \sin 2 \theta - \pi } { 4 \theta }\).
2. Use the iterative formula $$\theta _ { n + 1 } = \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 2 \sin 2 \theta _ { n } - \pi } { 4 \theta _ { n } } \right)$$ with initial value \(\theta _ { 1 } = 1\), to determine \(\theta\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.
   \(7 \quad\) Let \(\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 7 x - 1 } { ( x - 2 ) \left( x ^ { 2 } + 3 \right) }\).
3. Express \(\mathrm { f } ( x )\) in partial fractions.
4. Hence obtain the expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\).

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\includegraphics[max width=\textwidth, alt={}, center]{a5f8f007-5176-4686-93d2-4caabeea182e-2_551_569_1416_788} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(O A B\) is \(\theta\) radians. The shaded region is bounded by the circumference of the circle and the arc with centre \(A\) joining \(B\) and \(C\). The area of the shaded region is equal to half the area of the circle.

1. Show that \(\cos 2 \theta = \frac { 2 \sin 2 \theta - \pi } { 4 \theta }\).
2. Use the iterative formula $$\theta _ { n + 1 } = \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 2 \sin 2 \theta _ { n } - \pi } { 4 \theta _ { n } } \right) ,$$ with initial value \(\theta _ { 1 } = 1\), to determine \(\theta\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

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\includegraphics[max width=\textwidth, alt={}, center]{9e1bd528-e7c4-4936-a05a-dde1d1ace7c2-2_512_775_1318_683} The diagram shows the curve \(y = \cos x\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). A rectangle \(O A B C\) is drawn, where \(B\) is the point on the curve with \(x\)-coordinate \(\theta\), and \(A\) and \(C\) are on the axes, as shown. The shaded region \(R\) is bounded by the curve and by the lines \(x = \theta\) and \(y = 0\).

1. Find the area of \(R\) in terms of \(\theta\).
2. The area of the rectangle \(O A B C\) is equal to the area of \(R\). Show that $$\theta = \frac { 1 - \sin \theta } { \cos \theta }$$
3. Use the iterative formula \(\theta _ { n + 1 } = \frac { 1 - \sin \theta _ { n } } { \cos \theta _ { n } }\), with initial value \(\theta _ { 1 } = 0.5\), to determine the value of \(\theta\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

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\includegraphics[max width=\textwidth, alt={}, center]{96a4df57-b3c7-4dbf-9bea-bb00ed6a4a16-2_512_775_1318_683} The diagram shows the curve \(y = \cos x\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). A rectangle \(O A B C\) is drawn, where \(B\) is the point on the curve with \(x\)-coordinate \(\theta\), and \(A\) and \(C\) are on the axes, as shown. The shaded region \(R\) is bounded by the curve and by the lines \(x = \theta\) and \(y = 0\).

1. Find the area of \(R\) in terms of \(\theta\).
2. The area of the rectangle \(O A B C\) is equal to the area of \(R\). Show that $$\theta = \frac { 1 - \sin \theta } { \cos \theta }$$
3. Use the iterative formula \(\theta _ { n + 1 } = \frac { 1 - \sin \theta _ { n } } { \cos \theta _ { n } }\), with initial value \(\theta _ { 1 } = 0.5\), to determine the value of \(\theta\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

2. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the vertical cross-section of the entrance to a tunnel. The width at the base of the tunnel entrance is 2 metres and its maximum height is 3 metres. The shape of the cross-section can be modelled by the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = 3 \cos \left( \frac { \pi } { 2 } x ^ { 2 } \right) \quad x \in [ - 1,1 ]$$ A wooden door of uniform thickness 85 mm is to be made to seal the tunnel entrance.
Use Simpson's rule with 6 intervals to estimate the volume of wood required for this door, giving your answer in \(\mathrm { m } ^ { 3 }\) to 4 significant figures.