June 2019 Paper 2 Q21
21 \(ABCDEFGH\) is a cuboid.
\(BC = 15\) cm \(\qquad CD = 12\) cm \(\qquad DH = 8\) cm
Work out the size of the angle between the line \(CE\) and the plane \(CDHG\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(8^2 + 12^2\) or \(64 + 144\) or 208 or \(8^2 + 12^2 + 15^2\) or \(64 + 144 + 225\) or 433 | M1 | \(HC^2\) or \(CE^2\) implied by 2nd M1 |
| \(\sqrt{8^2 + 12^2}\) or \(\sqrt{208}\) or \(4\sqrt{13}\) or 14.4… or \(\sqrt{8^2 + 12^2 + 15^2}\) or \(\sqrt{433}\) or 20.8… | M1dep | oe may be on diagram fully correct trigonometry method leading to 14.4… or 20.8… can score M2 eg \(8 \div \sin\left(\tan^{-1} \dfrac{8}{12}\right)\) or \(8 \div \sin\left(\tan^{-1} \dfrac{8}{\sqrt{12^2 + 15^2}}\right)\) |
| \(\tan x = \dfrac{15}{\sqrt{8^2 + 12^2}}\) or \(\cos x = \dfrac{\sqrt{8^2 + 12^2}}{\sqrt{8^2 + 12^2 + 15^2}}\) or \(\sin x = \dfrac{15}{\sqrt{8^2 + 12^2 + 15^2}}\) | M1dep | oe eg \(\tan x = [1.04,\ 1.042]\) or or \(\cos x = [0.69,\ 0.6934]\) or \(\sin x = [0.72,\ 0.7212]\) or \(90 - \tan^{-1} \dfrac{\sqrt{8^2 + 12^2}}{15}\) dep on M2 any letter |
| 46(.1…) | A1 |
Additional guidance
| 3rd M1 If using sine rule or cosine rule, must be in the form \(\cos x =\) or \(\sin x =\) eg \(\cos x = \dfrac{20.8^2 + 14.4^2 - 15^2}{2 \times 20.8 \times 14.4}\) (oe eg \(\cos^{-1} [0.69,\ 0.6934]\)) | |
| 3rd M1 Condone \(\ \tan = \dfrac{15}{\sqrt{8^2 + 12^2}}\) etc | M3 |
| Allow the first 2 M marks even if not subsequently used | |
| Allow recovery of missing brackets |