Specimen 2020 Paper 2 Q20
20 \(VABCD\) is a pyramid with a horizontal rectangular base \(ABCD\).
\(V\) is directly above the centre of the base.
\(VA = VB = VC = VD = 10\) cm
\(AB = 8\) cm \(\quad BC = 6\) cm
\(M\) is the midpoint of \(BC\).
Work out the size of angle \(VMD\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((VM^2 =)\ 10^2 - 3^2\) or \(100 - 9\) or 91 | M1 | oe |
| \((DM^2 =)\ 8^2 + 3^2\) or \(64 + 9\) or 73 | M1 | oe |
| \(10^2 =\) their 91 \(+\) their 73 \(-\ 2 \times \sqrt{\text{their } 91} \times \sqrt{\text{their } 73} \times \cos VMD\) | M1dep | oe dep on M2 may be implied |
| \((\cos VMD =)\ \dfrac{\text{their } 91 + \text{their } 73 - 10^2}{2 \times \sqrt{\text{their } 91} \times \sqrt{\text{their } 73}}\) | M1dep | oe dep on M3 |
| [66.8, 66.9] or 67 | A1 |