The diagram shows a sector AOB of a circle with centre O and radius r cm - AQA - A-Level Maths Pure - Question 5 - 2017 - Paper 1

Solve the Quadratic Equation: We start with the equation derived earlier:

$2r^2 - 15r + 18 = 0$

Using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where a = 2, b = -15, c = 18, gives:

$r = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot 18}}{2 \cdot 2}$

Simplifying this:

$r = \frac{15 \pm \sqrt{225 - 144}}{4} = \frac{15 \pm \sqrt{81}}{4} = \frac{15 \pm 9}{4}$

This yields two potential solutions:

$r = \frac{24}{4} = 6 \quad \text{and} \quad r = \frac{6}{4} = 1.5$

Calculate θ for Both Values:

• For r = 6:

  Substituting back into equation (1):

  $\theta = \frac{18}{6^2} = \frac{18}{36} = \frac{1}{2} \text{ radians}$

• For r = 1.5:

  $\theta = \frac{18}{(1.5)^2} = \frac{18}{2.25} = 8 \text{ radians}$

Determine the Valid Value of θ: Since the perimeter formula requires that θ must be less than or equal to 2π radians (approximately 6.28 radians), the only suitable value is:

$\theta = \frac{1}{2} \text{ radians}.$

Conclusion: Therefore, the only possible value for θ is (\theta = \frac{1}{2}) radians.