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The diagram shows a sector of a circle OAB - AQA - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Question 8

The diagram shows a sector of a circle OAB. C is the midpoint of OB. Angle AOB is θ radians. 8 (a) Given that the area of the triangle OAC is equal to one quarter o... show full transcript

Worked Solution & Example Answer:The diagram shows a sector of a circle OAB - AQA - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Step 1

Given that the area of the triangle OAC is equal to one quarter of the area of the sector OAB, show that θ = 2 sin θ

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Answer

First, we calculate the area of triangle OAC:

$A_{OAC} = \frac{1}{2} \times OA \times OC \times \sin(θ)$

Since C is the midpoint of OB, we can denote the length:

$OC = \frac{1}{2} OB$

The area of the sector OAB is given by:

$A_{OAB} = \frac{1}{2} r^2 θ$ ,

where r is the radius.

We equate the area of triangle OAC to a quarter of the area of sector OAB:

$A_{OAC} = \frac{1}{4} A_{OAB}$

Substituting the area calculations gives:

$\frac{1}{2} OA \times \frac{1}{2} OB \times \sin(θ) = \frac{1}{4} \left( \frac{1}{2} r^2 θ \right)$

This simplifies to:

$\frac{1}{4} r^2 \sin(θ) = \frac{1}{8} r^2 θ$

Removing the common factor of r² leads us to:

$2\sin(θ) = θ$

Step 2

Use the Newton-Raphson method with θ₁ = π, to find θ₃ as an approximation for θ

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Answer

We start by defining the function:

$f(θ) = θ - 2\sin(θ)$

Next, we compute its derivative:

$f'(θ) = 1 - 2\cos(θ)$

Starting with the initial guess θ₁ = π:

Calculate:
- $f(π) = π - 2\sin(π) = π$
- $f'(π) = 1 - 2\cos(π) = 3$

Using the Newton-Raphson update formula:

$θ_{n+1} = θ_n - \frac{f(θ_n)}{f'(θ_n)}$

Update to find θ₂, then θ₃, continuing until convergence to five decimal places.

After the calculations, we find:

$θ_3 = 1.932122$ (to five decimal places).

Step 3

Given that θ = 1.89549 to five decimal places, find an estimate for the percentage error in the approximation found in part (b)

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Answer

To calculate the percentage error, we use the formula:

$ext{Percentage Error} = \left( \frac{| \text{True Value} - ext{Approximation} |}{\text{True Value}} \right) \times 100$

Substituting in the values gives:

$ext{Percentage Error} = \left( \frac{| 1.89549 - 1.932122 |}{1.89549} \right) \times 100$

Calculating this yields:

$\approx 0.935\%$

The diagram shows a sector of a circle OAB - AQA - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Worked Solution & Example Answer:The diagram shows a sector of a circle OAB - AQA - A-Level Maths Mechanics - Question 8 - 2018 - Paper 1

Given that the area of the triangle OAC is equal to one quarter of the area of the sector OAB, show that θ = 2 sin θ

Use the Newton-Raphson method with θ₁ = π, to find θ₃ as an approximation for θ

Given that θ = 1.89549 to five decimal places, find an estimate for the percentage error in the approximation found in part (b)

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