Higher June 2025 Paper 2R Q23
23 The diagram shows a triangular prism, \(ABCDEF\), with a horizontal rectangular base \(ABCD\)
Diagram NOT accurately drawn
\(M\) is the midpoint of the line \(AC\)
\(AC = 40\) cm angle \(CAE = 35^\circ\) angle \(ABF = 90^\circ\)
Work out the size of angle \(CME\)
Give your answer correct to 3 significant figures.
(3)
| Scheme | Marks |
|---|---|
eg \((CE =)\;40 \times \tan 35\;(= 28(.0\ldots))\) or \((CE =)\;\dfrac{40}{\tan(90 - 35)}\;(= 28(.0\ldots))\) or \((CE =)\;\dfrac{40\sin 35}{\sin(90 - 35)}\;(= 28(.0\ldots))\) | M1 |
eg (angle \(CME\) =) \(\tan^{-1}\left(\dfrac{\text{``}{28}\text{''}}{40 \div 2}\right)\) or (angle \(CME\) =) \(\sin^{-1}\left(\dfrac{\text{``}{28}\text{''}}{\sqrt{(40 \div 2)^2 + \text{``}{28}\text{''}^2}}\right)\) or (angle \(CME\) =) \(\cos^{-1}\left(\dfrac{40 \div 2}{\sqrt{(40 \div 2)^2 + \text{``}{28}\text{''}^2}}\right)\) | M1 |
| Correct answer only scores full marks (unless from obviously incorrect working) Answer: 54.5 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a correct method to find \(CE\)
M1: for a complete method to find angle \(CME\)
A1: awrt 54.5