June 2016 Paper 2 Q22
22 Pyramid \(VABCD\) has a horizontal rectangular base.
\(X\) is the centre of the base.
\(V\) is vertically above \(X\).
\[VB = VC = 17\text{ cm} \qquad AB = 22\text{ cm} \qquad BC = 16\text{ cm}\]
Work out the angle between the planes \(VBC\) and \(ABCD\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Triangles \(VMB\) and \(VXM\) (\(M\) is the midpoint of \(BC\)) | ||
| \(17^2 - (16 \div 2)^2\) or 225 or \(17^2 = VM^2 + (16 \div 2)^2\) | M1 | oe |
| \((VM =)\ \sqrt{17^2 - (16 \div 2)^2}\) or \(\sqrt{225}\) or 15 | M1 | Implies M1 M1 May be seen on diagram |
| \(\cos x = \dfrac{22 \div 2}{\text{their } 15}\) or \(\sin x = \dfrac{\sqrt{15^2 - (22 \div 2)^2}}{\text{their } 15}\) or \(\tan x = \dfrac{\sqrt{15^2 - (22 \div 2)^2}}{22 \div 2}\) | M1dep | \(x\) is required angle dep on M1 M1 oe eg correct method using cosine rule or sine rule simplified to \(\cos x =\) or \(\sin x =\) or \(90 - \sin^{-1}\dfrac{22 \div 2}{\text{their } 15}\) |
| 42.8… | A1 | Allow 43 with correct working SC2 Answer 36.8... or 36.9 |
| Alternative method 2 Triangles \(BXM\) and \(VXB\) and \(VXM\) (\(M\) is the midpoint of \(BC\)) | ||
| \(BX^2 = (16 \div 2)^2 + (22 \div 2)^2\) or 185 and \(17^2\) − their \(BX^2\) or \(17^2 = VX^2\) + their \(BX^2\) | M1 | oe eg for \(BX^2\) \(BX^2 = \left(\dfrac{1}{2}BD\right)^2 = \dfrac{1}{4}(16^2 + 22^2)\) |
| \((VX =)\ \sqrt{17^2 - (\text{their } BX)^2}\) or \(\sqrt{104}\) or \(2\sqrt{26}\) or [10.19, 10.2] | M1 | Implies M1 M1 May be seen on diagram |
| \(\tan x = \dfrac{\text{their } [10.19, 10.2]}{22 \div 2}\) or \(\sin x = \dfrac{\text{their } [10.19, 10.2]}{\text{their } VM}\) or \(\cos x = \dfrac{22 \div 2}{\text{their } VM}\) | M1dep | \(x\) is required angle dep on M1 M1 oe eg correct method using cosine rule or sine rule simplified to \(\cos x =\) or \(\sin x =\) or \(90 - \tan^{-1}\dfrac{22 \div 2}{\text{their } [10.19, 10.2]}\) |
| 42.8… | A1 | Allow 43 with correct working SC2 Answer 36.8... or 36.9 |
| Alternative method 3 Triangles \(VMB\) and \(VMN\) (\(M\) is the midpoint of \(BC\), \(N\) is the midpoint of \(AD\)) | ||
| \(17^2 - (16 \div 2)^2\) or 225 or \(17^2 = VM^2 + (16 \div 2)^2\) | M1 | oe |
| \((VM =)\ \sqrt{17^2 - (16 \div 2)^2}\) or \(\sqrt{225}\) or 15 | M1 | Implies M1 M1 May be seen on diagram |
| \(\dfrac{1}{2} \times \left(180 - \cos^{-1}\dfrac{\text{their } 15^2 + \text{their } 15^2 - 22^2}{2 \times \text{their } 15 \times \text{their } 15}\right)\) | M1dep | dep on M1 M1 |
| 42.8… | A1 | Allow 43 with correct working SC2 Answer 36.8... or 36.9 |
Additional guidance
Alt 2 3rd M1 their \(VM\) must be from correct method