Higher November 2023 Paper 2 Q28
28 \(VABCD\) is a pyramid with a horizontal square base.
\(X\) is the centre of the base.
\(V\) is vertically above \(X\).
\(BD = 18\) cm
Angle \(VBX\) = 72°
Work out the length of \(VB\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\cos 72 = \dfrac{9}{x}\) | M1 | oe eg \(\dfrac{x}{\sin 72} = \dfrac{18}{\sin 36}\) \(x\) can be any letter or \(VB\) or \(VA\) or \(VC\) or \(VD\) |
| \(\dfrac{9}{\cos 72}\) | M1dep | oe eg \(\dfrac{18 \times \sin 72}{\sin 36}\) |
| 29.1(2…) | A1 | accept 29 with M1 scored |
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