Higher January 2023 Paper 2 Q23
23 The diagram shows a solid prism \(ABCDEFGHIJ\)
Diagram NOT accurately drawn
The prism is such that each cross section is a pentagon where
\(AE = BC = x\) cm \(\qquad AB = 2x\) cm \(\qquad ED = CD = 8\) cm
angle \(EAB\) = angle \(CBA = 90^\circ \qquad\) angle \(AED\) = angle \(BCD = 120^\circ\)
Given that \(\;AG = BH = EF = DJ = CI = 12\) cm
calculate the angle that \(AJ\) makes with the base \(ABHG\) of the prism.
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
\([DN =]\;8\sin 30\) or \(8\cos 60\) (= 4) oe [where \(N\) is the midpoint of \(EC\)] or \([x =]\;8\cos 30\) or \(8\sin 60\;(= 4\sqrt{3} = 6.928\ldots)\) or \(2x = \sqrt{8^2 + 8^2 - 2 \times 8 \times 8 \times \cos 120}\;(= \sqrt{192} = 8\sqrt{3} = 13.85\ldots)\) | M1 |
\([DN =]\;8\sin 30\) or \(8\cos 60\) (= 4) oe eg \(\sqrt{8^2 - (4\sqrt{3})^2}\;(= 4)\) And 1 of \([x =]\;8\cos 30\) or \(8\sin 60\;(= 4\sqrt{3} = 6.928\ldots)\) oe or \(\sqrt{8^2 - \text{``}{4}\text{''}^2}\;(= 4\sqrt{3} = 6.928\ldots)\) or \(2x = \sqrt{8^2 + 8^2 - 2 \times 8 \times 8 \times \cos 120}\;(= 8\sqrt{3} = 13.85\ldots)\) | M1 |
\([AM =]\;\sqrt{12^2 + \left(\text{``}{4\sqrt{3}}\text{''}\right)^2}\;(= \sqrt{192} = 8\sqrt{3} = 13.856\ldots)\) oe [where \(M\) is the midpoint of \(GH\)] | M1 |
\(\tan MAJ = \left(\dfrac{\text{``}{4}\text{''} + \text{``}{4\sqrt{3}}\text{''}}{\text{``}{8\sqrt{3}}\text{''}}\right)\) oe eg \(\tan MAJ = \left(\dfrac{10.928\ldots}{13.856\ldots}\right)\) if student uses sin or cos, then \(AJ = 17.647\ldots\) to award marks this must come from a correct method or be correct | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 38.3 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: [\((JM =)\;4 + 4\sqrt{3}\) implies M2]
M1: Clear intention to be \(AM\) (not \(EC\))
A1: Accept 38.1 – 38.4
If no marks scored then award
SCB2 for \(\tan MAJ = \dfrac{4 + x}{\sqrt{x^2 + 12^2}}\)
or SCB1 for \(AM = \sqrt{x^2 + 12^2}\)