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

To find the value of θ, we first need to solve the quadratic equation derived in part (a):

1. Solving the Quadratic Equation: Using the quadratic formula: $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where a = 2, b = -15, and c = 18: $r = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot 18}}{2 \cdot 2}$ Calculate the discriminant: $r = \frac{15 \pm \sqrt{225 - 144}}{4}$ $r = \frac{15 \pm \sqrt{81}}{4}$ $r = \frac{15 \pm 9}{4}$ This results in:

   • $r = \frac{24}{4} = 6$
   • $r = \frac{6}{4} = 1.5$

2. Finding θ: Using the value of r = 6: From Equation 1: $r^2 \theta = 18\implies\ 6^2 \theta = 18\implies\ 36θ = 18\implies\ θ = \frac{1}{2}$

   Using the value of r = 1.5: $r^2 \theta = 18\implies\ (1.5)^2 \theta = 18\implies\ 2.25θ = 18\implies\ θ = 8$ Since θ must be less than 2π for it to make sense geometrically, only θ = \frac{1}{2} is valid. Hence the only possible value for θ is ( \frac{1}{2} ) radians, because the other value results from a less practical radius for the context.