November 2021 Paper 2 Q17
17 \(ABCDEFGH\) is a cube with side length 32 cm
\(M\) and \(N\) are points on \(DH\) and \(CG\) respectively.
Work out the size of the angle that the line \(BM\) makes with the plane \(ABCD\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Works out \(MD\) and \(BD\) and uses \(\tan MBD\) | ||
| \(\tan 28 = \dfrac{GN}{32}\) or \(32\tan 28\) or [17, 17.015] | M1 | oe eg \(\dfrac{32}{\tan(90 - 28)}\) working out \(GN\) or \(HM\) |
| \(32 - 32\tan 28\) or [14.985, 15] | M1dep | oe working out \(NC\) or \(MD\) |
| \(\sqrt{32^2 + 32^2}\) or \(\sqrt{2048}\) or [45.2, 45.3] | M1 | oe eg \(32\sqrt{2}\) working out \(BD\) |
| \(\tan MBD = \dfrac{\text{their } [14.985, 15]}{\text{their } [45.2, 45.3]}\) | M1dep | oe eg \(\tan^{-1}\dfrac{\text{their } [14.985, 15]}{\text{their } [45.2, 45.3]}\) dep on M3 |
| [18.3, 18.4] | A1 | |
| Alternative method 2 Works out \(BD\) and \(MB\) and uses \(\cos MBD\) | ||
| \(\tan 28 = \dfrac{GN}{32}\) or \(32\tan 28\) or [17, 17.015] | M1 | oe eg \(\dfrac{32}{\tan(90 - 28)}\) working out \(GN\) or \(HM\) |
| \(32 - 32\tan 28\) or [14.985, 15] | M1dep | oe working out \(NC\) or \(MD\) |
| \(\sqrt{32^2 + 32^2}\) or \(\sqrt{2048}\) or [45.2, 45.3] or \(\sqrt{32^2 + 32^2 + \text{their } [14.985, 15]^2}\) or [47.67, 47.7] | M1 | oe eg \(32\sqrt{2}\) working out \(BD\) or \(MB\) if awarding this mark for working out \(MB\) it is dependent on M2 |
| \(\cos MBD =\) \(\dfrac{\sqrt{32^2 + 32^2}}{\sqrt{32^2 + 32^2 + \text{their } [14.985, 15]^2}}\) | M1dep | oe eg \(\cos^{-1}\dfrac{\sqrt{32^2 + 32^2}}{\sqrt{32^2 + 32^2 + \text{their } [14.985, 15]^2}}\) dep on M3 |
| [18.3, 18.4] | A1 | |
| Alternative method 3 Works out \(MD\) and \(MB\) and uses \(\sin MBD\) | ||
| \(\tan 28 = \dfrac{GN}{32}\) or \(32\tan 28\) or [17, 17.015] | M1 | oe eg \(\dfrac{32}{\tan(90 - 28)}\) working out \(GN\) or \(HM\) |
| \(32 - 32\tan 28\) or [14.985, 15] | M1dep | oe working out \(NC\) or \(MD\) |
| \(\sqrt{32^2 + 32^2 + \text{their } [14.985, 15]^2}\) or [47.67, 47.7] | M1dep | oe working out \(MB\) |
| \(\sin MBD =\) \(\dfrac{\text{their } [14.985, 15]}{\sqrt{32^2 + 32^2 + \text{their } [14.985, 15]^2}}\) | M1dep | oe eg \(\sin^{-1}\dfrac{\text{their } [14.985, 15]}{\sqrt{32^2 + 32^2 + \text{their } [14.985, 15]^2}}\) |
| [18.3, 18.4] | A1 | |
Additional guidance
| 1st M1 \(GN\) may be seen as a letter, eg \(x\), but do not award if subsequently used as the length of an incorrect side (eg \(MN\)) | |
| 4th M1 \(MBD\) may be seen as a letter, eg \(y\), but do not award if subsequently used as the size of an incorrect angle (eg \(DMB\)) | |
| Alt 1 or Alt 2 \(32\sqrt{1^2 + 1^2}\) | 3rd M1 |
| Alt 1 \(\tan MBD = \dfrac{32(1 - \tan 28)}{32\sqrt{2}}\) or \(\tan MBD = \dfrac{(1 - \tan 28)}{\sqrt{2}}\) | M4 |
Notes
The final answer is printed as “[18.3, 18,4]” in each method (corrected from the printed mark scheme: 18,4 is 18.4).