Higher June 2021 Paper 1 Q22
22 The diagram shows a triangular prism \(ABCDEF\) with a horizontal base \(ABEF\).

Diagram NOT accurately drawn
\(AC = BC = FD = ED = 12\) cm \(\qquad AB = 10\) cm \(\qquad BE = 15\) cm
Calculate the size of the angle between \(AD\) and the base \(ABEF\).
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
\([AM =]\ \sqrt{5^2 + 15^2}\ (= \sqrt{250} = 15.8\ldots)\) where \(M\) is midpoint of \(EF\), oe other correct method to find \(AM\) \([AD =]\ \sqrt{12^2 + 15^2}\ (= \sqrt{369} = 19.2\ldots)\) \([DM =]\ \sqrt{12^2 - 5^2}\ (= \sqrt{119} = 10.9\ldots)\) | M2 |
eg \(\tan DAM = \dfrac{\text{“}\sqrt{119}\text{”}}{\text{“}\sqrt{250}\text{”}} \left(= \dfrac{\text{“10.9…”}}{\text{“15.8…”}}\right)\) oe or \(\sin DAM = \dfrac{\text{“}\sqrt{119}\text{”}}{\text{“}\sqrt{369}\text{”}} \left(= \dfrac{\text{“10.9…”}}{\text{“19.2…”}}\right)\) oe or \(\cos DAM = \dfrac{\text{“}\sqrt{250}\text{”}}{\text{“}\sqrt{369}\text{”}} \left(= \dfrac{\text{“15.8…”}}{\text{“19.2…”}}\right)\) oe | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 34.6 | A1 |
| (4) | |
| (4 marks) |
Notes
M2: for a complete method to find two of \(AM\), \(AD\), \(DM\) (where \(M\) is the midpoint of \(EF\))
Other longer ways to find \(AM\), \(AD\), \(DM\) may be used but must be a complete method eg
\(\angle DEM = \cos^{-1}\left(\dfrac{5}{12}\right) (= 65.37\ldots)\) and \(DM = 12\sin 65.37\ldots\)
\(\angle DEM = \cos^{-1}\left(\dfrac{5}{12}\right) (= 65.37\ldots)\) and \(DM = 5\tan 65.37\ldots\)
Use 10 ÷ 2 as 5 throughout
(M1 For a complete method to find one of \(AM\), \(AD\), \(DM\) (where \(M\) is the midpoint of \(EF\)))
M1: a correct method to find the required angle – other longer methods may be used but they must get to the stage of an equation for the required angle
eg \(\sin DAM = \dfrac{\text{“10.9…”}}{\sqrt{\text{“15.8…”}^2 + \text{“10.9…”}^2}}\)
NB: “10.9…” and “15.8…” must come from correct working
A1: any answer which rounds to 34.6