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]

Mark scheme (a)

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

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

Mark scheme (b)

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

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