Higher November 2019 Paper 3 Q23
23 The diagram shows a sector \(OACB\) of a circle with centre \(O\).
The point \(C\) is the midpoint of the arc \(AB\).
The diagram also shows a hollow cone with vertex \(O\).
The cone is formed by joining \(OA\) and \(OB\).
The cone has volume 56.8 cm3 and height 3.6 cm.
Calculate the size of angle \(AOB\) of sector \(OACB\).
Give your answer correct to 3 significant figures.
You must show all your working. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 264 | P1 | correct substitution into the volume formula, eg \(56.8 = \dfrac{1}{3} \times \pi \times r^2 \times 3.6\) |
| P1 | completes process to find base radius or the value of \(r^2\), eg \(r = \sqrt{\dfrac{56.8 \times 3}{\pi \times 3.6}}\ (= 3.88158\ldots)\) or \(r^2 = \dfrac{56.8}{1.2\pi}\ (= 15.066)\) | |
| P1 | Uses Pythagoras to find the sloping length, eg \(\sqrt{\text{``}3.88\ldots\text{''}^2 + 3.6^2}\ (= 5.29\ldots)\) | |
| P1 | process to find an equation in \(AOB\), eg \(\pi \times \text{``}3.88\text{''} \times \text{``}5.29\text{''} = \dfrac{AOB}{360} \times \pi \times \text{``}5.29\text{''}^2\) or \(\dfrac{AOB}{360} \times \pi \times 2 \times \text{``}5.29\text{''} = 2 \times \pi \times \text{``}3.88\text{''}\) or \(\dfrac{AOB}{360} \times \text{``}5.29\text{''} = \text{``}3.88\text{''}\) | |
| A1 | answer in the range 263.9 to 264.1 |
Additional guidance
\(AOB\) does not need to be the subject of the equation