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

We need to calculate the slant height (l) of the cone using Pythagoras' theorem. The radius (r) is 3.88 cm and the height (h) is 3.6 cm:

$l = \sqrt{r^2 + h^2} = \sqrt{(3.88)^2 + (3.6)^2}$
$l = \sqrt{15.066 + 12.96} \approx \sqrt{28.026} \approx 5.29 \text{ cm}$

The circumference of the base of the cone formed can be expressed in terms of the sector angle AOB. The relationship between the arc length (l), radius of the circle (r), and angle (( \theta ) in radians) is given by:

$l = r \theta$
In our case, the radius of the sector is equal to the slant height of the cone (5.29 cm) and the arc length is equal to the circumference of the base of the cone:

$C = 2\pi r = 2\pi (3.88)$

We can now substitute the circumference into the equation:
$2\pi (3.88) = 5.29 \theta$
Solving for ( \theta ):
$\theta = \frac{2\pi (3.88)}{5.29} \approx 4.6 \text{ radians}$
To convert this to degrees:
$\theta_{degrees} \approx \frac{4.6}{\pi} \times 180 \approx 263.46 \text{ degrees}$
Thus, the angle AOB is approximately 263 degrees when rounded to 3 significant figures.