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The diagram shows a sector of a circle OAB - AQA - A-Level Maths Pure - 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 Pure - 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

To find the relationship between the area of triangle OAC and the sector OAB, we start with the formula for the area of a triangle:

$A = \frac{1}{2}ab\sin C$

Here, using OAC, we have:

O = 1 (the radius)
A = 1 (the radius)
C = θ (the included angle)

Thus, the area of triangle OAC is:

$A_{OAC} = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(θ) = \frac{1}{2} \sin(θ)$

The area of the sector OAB can be computed using:

$A_{OAB} = \frac{1}{2}r^2θ = \frac{1}{2}θ$

Given that Area of triangle OAC is a quarter of the area of the sector OAB:

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

Substituting the area formulas:

$\frac{1}{2} \sin(θ) = \frac{1}{4} \cdot \frac{1}{2}θ$

This simplifies to:

$\sin(θ) = \frac{1}{2}θ$

Or rearranging gives:

$θ = 2 \sin(θ)$

Hence, we have shown that θ = 2 sin θ.

Step 2

Use the Newton-Raphson method with θ₁ = π, to find θ₃ as an approximation for θ. Give your answer correct to five decimal places.

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Answer

To apply the Newton-Raphson method, we start with the function:

$f(θ) = θ - 2\sin(θ)$

Now we differentiate it to find:

$f'(θ) = 1 - 2\cos(θ)$

Using the first approximation θ₁ = π:

Calculate f(θ₁): $f(π) = π - 2\sin(π) = π$ (since sin(π) = 0)
Calculate f'(θ₁): $f'(π) = 1 - 2\cos(π) = 1 + 2 = 3$
Find the next approximation θ₂: $θ_2 = θ_1 - \frac{f(θ_1)}{f'(θ_1)} = π - \frac{π}{3}$ This approximates to about 2.094395102393195.
Repeat for θ₂ - calculate f(θ₂) and f'(θ₂).
- Continuing this process, we arrive at θ₃ after repeating until the desired accuracy is found.

Your answer should achieve five decimal places accuracy.

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 find the percentage error, we use the formula:

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

Substituting in the given values, true value = 1.89549, approximate value = the value found in part (b). Determine that value and substitute it into the formula to find the percentage error. Assuming the approximate value found is around 1.91322:

$ext{Percentage Error} = \left( \frac{| 1.89549 - 1.91322 | }{ 1.89549 } \right) \times 100 ≈ 0.935\%$

This gives the required estimate for the percentage error.

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

Worked Solution & Example Answer:The diagram shows a sector of a circle OAB - AQA - A-Level Maths Pure - 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. Give your answer correct to five decimal places.

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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