Higher January 2023 Paper 1R Q21
21 The diagram shows a triangular prism, \(ABCDEF\), with a rectangular base \(ABCD\)
Diagram NOT accurately drawn
\(AB = 6\) cm
\(DE = 2.2\) cm
angle \(DAE = 18°\)
angle \(ADE = 90°\)
Work out the angle that \(BE\) makes with the plane \(ABCD\)
Give your answer correct to one decimal place.
(4)
| Scheme | Marks |
|---|---|
\((AD =)\;\dfrac{2.2}{\tan 18}\;(= 6.77\ldots)\) or \((EA =)\;\dfrac{2.2}{\sin 18}\;(= 7.11\ldots)\) | M1 |
\((DB =)\;\sqrt{(\text{``}{6.77\ldots}\text{''})^2 + 6^2}\;(= 9.04\ldots)\) or \((EB =)\;\sqrt{6^2 + \text{``}{7.11\ldots}\text{''}^2}\;(= 9.31\ldots)\) or \((EB =)\;\sqrt{6^2 + \text{``}{6.77\ldots}\text{''}^2 + 2.2^2}\;(= 9.31\ldots)\) | M1 |
\(\tan DBE = \dfrac{2.2}{\text{``}{9.04\ldots}\text{''}}\) or \(\sin DBE = \dfrac{2.2}{\text{``}{9.31\ldots}\text{''}}\) or \(\sin DBE = \dfrac{2.2 \sin 90}{\text{``}{9.31\ldots}\text{''}}\) \(\cos DBE = \dfrac{\text{``}{9.04\ldots}\text{''}}{\text{``}{9.31\ldots}\text{''}}\) or use of cosine rule | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 13.7 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: a correct method to find \(AD\) or \(AE\)
M1: a correct method to find \(DB\) or \(EB\)
M1: complete method to find one of \(\tan DBE\) or \(\sin DBE\) or \(\cos DBE\) – NB: if using cosine, the student will need to have found \(DB\) and \(EB\) previously
A1: Allow answers in range 13.59 – 13.8