June 2018 Paper 2 Q20
20 \(PQRSTU\) is a triangular prism.
\(PQRS\) is a rectangle and angle \(QRU = 90^\circ\)
\(PQ = 10\) cm \(\qquad QR = 12\) cm \(\qquad UR = 7\) cm
\(M\) is the midpoint of \(PQ\).
(a) Calculate the size of the angle between the line \(UM\) and the plane \(PQRS\). [4 marks]
(b) Calculate the size of the angle between the planes \(UMR\) and \(UQR\). [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(12^2 + \left(\dfrac{10}{2}\right)^2\) or \(12^2 + 5^2\) or \(144 + 25\) or 169 | M1 | oe \(RM^2\) |
| \(\sqrt{\text{their } 169}\) or 13 | M1dep | oe \(RM\) may be seen on diagram 13 implies M2 |
| \(\tan x = \dfrac{7}{\text{their } 13}\) | M1dep | any letter oe eg \(\tan^{-1} \dfrac{7}{\text{their } 13}\) |
| 28(.3...) | A1 | |
| Alternative method 2 | ||
| \(12^2 + \left(\dfrac{10}{2}\right)^2 + 7^2\) or \(12^2 + 5^2 + 7^2\) or \(144 + 25 + 49\) or 218 | M1 | oe \(UM^2\) |
| \(\sqrt{\text{their } 218}\) or \([14.76,\ 14.8]\) | M1dep | oe \(UM\) may be seen on diagram \([14.76,\ 14.8]\) implies M2 |
| \(\sin x = \dfrac{7}{\text{their } [14.76,\ 14.8]}\) or \(\cos x = \dfrac{\sqrt{12^2 + 5^2}}{\text{their } [14.76,\ 14.8]}\) | M1dep | any letter oe eg \(\sin^{-1} \dfrac{7}{\text{their } [14.76,\ 14.8]}\) or \(\sin x = \dfrac{\sin 90}{\text{their } [14.76,\ 14.8]} \times 7\) or \(\cos x = \dfrac{12^2 + 5^2 + \text{their } [14.76,\ 14.8]^2 - 7^2}{2 \times \sqrt{12^2 + 5^2} \times \text{their } [14.76, 14.8]}\) |
| 28(.3...) | A1 | |
Additional guidance
| Allow \(\tan = \dfrac{7}{\text{their } 13}\) etc | |
| Do not allow \(\tan \dfrac{7}{\text{their } 13}\) etc unless recovered | |
| If using sine or cosine rule, must rearrange to make \(\sin x\) or \(\cos x\) the subject | |
| Allow up to M1M1dep from either alt 1 or alt 2 even if not subsequently used |
| Answer | Mark | Comments |
|---|---|---|
| \(\tan \alpha = \dfrac{10 \div 2}{12}\) or \(\tan \alpha = \dfrac{5}{12}\) or \(\sin \alpha = \dfrac{10 \div 2}{\text{their } 13}\) or \(\sin \alpha = \dfrac{5}{\text{their } 13}\) or \(\cos \alpha = \dfrac{12}{\text{their } 13}\) | M1 | any letter oe eg \(\tan^{-1} \dfrac{5}{12}\) their \(13 = RM\) from (a) may be seen on diagram oe eg \(\sin \alpha = \dfrac{\sin 90}{\sqrt{12^2 + 5^2}} \times 5\) or \(\sin \alpha = \dfrac{\sin 90}{\text{their } 13} \times 5\) or \(\cos \alpha = \dfrac{12^2 + 5^2 + 12^2 - 5^2}{2 \times 12 \times \sqrt{12^2 + 5^2}}\) or \(\cos \alpha = \dfrac{\text{their } 13^2 + 12^2 - 5^2}{2 \times 12 \times \text{their } 13}\) |
| 22.6… | A1ft | allow 23 with working correct or ft ft answers correct to at least 1 dp |
Additional guidance
| Allow \(\tan = \dfrac{5}{12}\) etc | |
| Do not allow \(\tan \dfrac{5}{12}\) etc unless recovered | |
| If using sine or cosine rule, must rearrange to make \(\sin \alpha\) or \(\cos \alpha\) the subject |