Higher June 2019 Paper 2 Q21
21 The diagram shows cuboid \(ABCDEFGH\).
Diagram NOT accurately drawn
For this cuboid
the length of \(AB\) : the length of \(BC\) : the length of \(CF = 4 : 2 : 3\)
Calculate the size of the angle between \(AF\) and the plane \(ABCD\).
Give your answer correct to one decimal place.
(3)
| Scheme | Marks |
|---|---|
e.g. (\(AC\) =) \(\sqrt{(4x)^2 + (2x)^2}\) (= \(\sqrt{20}x\)) or (\(AC\) =) \(\sqrt{(4)^2 + (2)^2}\) (= \(\sqrt{20}\)) or (\(AF\) =) \(\sqrt{(4)^2 + (2)^2 + (3)^2}\) (= \(\sqrt{29}\)) or (\(AF\) =) \(\sqrt{(\sqrt{20})^2 + (3)^2}\) (= \(\sqrt{29}\)) or | M1 |
e.g. (\(CAF\) =) \(\tan^{-1}\left(\dfrac{3x}{\text{``}{\sqrt{20}x}\text{''}}\right)\) (= 33.854…) or (\(CAF\) =) \(\tan^{-1}\left(\dfrac{3}{\text{``}{\sqrt{20}}\text{''}}\right)\) (= 33.854…) or (\(CAF\) =) \(\cos^{-1}\left(\dfrac{\text{``}{\sqrt{20}}\text{''}}{\text{``}{\sqrt{29}}\text{''}}\right)\) (= 33.854…) or (\(CAF\) =) \(\sin^{-1}\left(\dfrac{3}{\text{``}{\sqrt{29}}\text{''}}\right)\) (= 33.854…) | M1 |
| 33.9° | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a method to find an expression for length \(AC\) or length \(AF\) with or without \(x\) or
\(x\) can represent any number
e.g.
\(AB : BC : CF = 2 : 1 : 1.5\)
\(AC = \sqrt{2^2 + 1^2}\;(= \sqrt{5})\) (corrected from the printed mark scheme: it prints \(AC^2 = \sqrt{2^2 + 1^2}\))
M1: for a complete method to find angle \(CAF\) using length \(AC\) or for a complete method to find angle \(CAF\) using length \(AF\) with or without \(x\) or
\(x\) can represent any number
\(AB : BC : CF = 2 : 1 : 1.5\)
(\(CAF\) =) \(\tan^{-1}\left(\dfrac{1.5}{\text{``}{\sqrt{5}}\text{''}}\right)\) (= 33.854…)
A1: answers in the range 33.85 – 33.9