The diagram shows a sector OACB of a circle with centre O - Edexcel - GCSE Maths - Question 24 - 2019 - Paper 3

Using Pythagoras’ theorem in the triangle formed by the radius, height, and slant height: $l = \sqrt{r^2 + h^2} = \sqrt{3.88^2 + 3.6^2}$ Calculating: $$l = \sqrt{15.064 + 12.96} = \sqrt{28.024} \approx 5.29 \text{ cm}.$

Now, the circumference of the base of the cone can be expressed as: C = 2\pi r \approx 2\pi (3.88) \approx 24.38 \text{ cm}.$ The arc length AC of the sector OACB can also be written as: \text{Arc length} = r \theta \text{ (in radians)}$Setting the two equal:$24.38 = 5.29 \theta Solving for $\theta$:\theta \approx \frac{24.38}{5.29} \approx 4.61 ext{ radians}.$To convert radians to degrees, use the conversion factor$\left(\frac{180}{\pi}\right)$: $\text{Angle AOB} \approx 4.61 \times \frac{180}{\pi} \approx 263.07 ext{ degrees}.$
Thus, the final answer correct to 3 significant figures is: $\text{Angle AOB} \approx 263^ ext{o}.$