June 2023 Paper 2 Q12
12 \(VABCD\) is a pyramid.
The square horizontal base, \(ABCD\), has side length 15 cm
\(V\) is directly above the centre, \(X\), of the base.
\(VA = 28\) cm
Work out the size of the angle that \(VA\) makes with \(ABCD\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| (\(AC =\)) \(\sqrt{15^2 + 15^2}\) or \(\sqrt{450}\) or \(15\sqrt{2}\) or 21.2(…) or (\(AX =\)) \(\sqrt{7.5^2 + 7.5^2}\) or \(\sqrt{112.5}\) or \(\dfrac{15\sqrt{2}}{2}\) or [10.6, 10.61] | M1 | oe eg (\(DB =\)) \(\sqrt{225 + 225}\) or (\(BX =\)) \(\sqrt{56.25 + 56.25}\) or (\(AX =\)) \(\sqrt{\dfrac{15^2}{2}}\) or (\(AX =\)) \(15\sin 45\) allow any unambiguous indication of side may be seen on diagram |
| \(\cos y = \dfrac{0.5\sqrt{15^2 + 15^2}}{28}\) or \(\cos y = \dfrac{\sqrt{7.5^2 + 7.5^2}}{28}\) or \(\cos y = \dfrac{15\sqrt{2}}{56}\) or \(\cos^{-1}[0.37, 0.38]\) | M1dep | oe eg \(\sin y = \dfrac{\sqrt{28^2 - 7.5^2 - 7.5^2}}{28}\) or \(\tan y = \dfrac{\sqrt{28^2 - 7.5^2 - 7.5^2}}{\sqrt{7.5^2 + 7.5^2}}\) or \(90 - \sin^{-1}\dfrac{\sqrt{7.5^2 + 7.5^2}}{28}\) |
| [66.9, 68.435] | A1 | SC2 [21.565, 23.1] |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| M1dep \(y\) may be \(x\) or \(A\) or \(VAX\) etc | |
| M1dep oe must be of the form \(\cos y =\) (or \(\sin y =\) or \(\tan y =\)) or be a calculation that evaluates to [66.9, 68.435] eg applying cosine rule to triangle \(VAC\) \(\cos y = \dfrac{28^2 + 450 - 28^2}{2 \times 28 \times \sqrt{450}}\) | M2 |
| M1dep allow \(\sin y = [0.92, 0.93]\) or \(\tan y = [2.44, 2.45]\) or \(90 - [21.565, 23.1]\) | M2 |