4MA0 Higher June 2013 Paper 3R Q24
24 The diagram shows a cube \(ABCDEFGH\).
The sides of the cube are of length 5 cm.
Calculate the size of the angle between the diagonal \(AH\) and the base \(EFGH\).
Give your answer correct to 1 decimal place.
Diagram NOT accurately drawn
(4)
| Scheme | Marks |
|---|---|
| \((FH^2 =)\ 5^2 + 5^2\ (=50)\) | M1 |
| \(\sqrt{50}\) or \(5\sqrt{2}\) (= 7.07..) | A1 |
| \(\tan x = \dfrac{5}{\text{“}\sqrt{50}\text{”}}\) | M1 |
| 35.3 | A1 |
| (4) |
Notes
M1: or correct Pythagoras statement to find any diagonal
A1: for \(\sqrt{50}\) or \(5\sqrt{2}\) or awrt 7.1
M1: dep on previous M1
or \(\sin x = \dfrac{5}{\sqrt{75}}\) or \(\cos x = \dfrac{\text{“}\sqrt{50}\text{”}}{\sqrt{75}}\) or correct statement using Sine or Cosine rule with angle \(AHF\) as the only unknown
(NB. \(\sqrt{75}\) may be \(5\sqrt{3}\) or awrt 8.7 may be used for \(AH\) if any other value used then it must clearly come from correct method to find \(AH\) )
A1: 35.264... awrt 35.3
Alternative scheme
| Scheme | Marks |
|---|---|
| \((AH^2 =)\ 5^2 + 5^2 + 5^2\ (=75)\) | M1 |
| \(\sqrt{75}\) or \(5\sqrt{3}\) (= 8.66..) | A1 |
| \(\sin x = \dfrac{5}{\text{“}\sqrt{75}\text{”}}\) | M1 |
| 35.3 | A1 |
Notes
A1: for \(\sqrt{75}\) or \(5\sqrt{3}\) or awrt 8.7
M1: dep on previous M1
A1: 35.264... awrt 35.3