4MA0 Higher January 2017 Paper 4 Q9
9 The diagram shows a ladder, \(EF\), leaning against a vertical wall.
The foot, \(E\), of the ladder is on horizontal ground.
Diagram NOT accurately drawn
\(EG = 2.1\) m \(\qquad FG = 3.5\) m \(\qquad\) angle \(EGF = 90^\circ\)
(a) Work out the length of the ladder.
Give your answer correct to 1 decimal place. (3)
Give your answer correct to 1 decimal place. (3)
(b) Work out the size of angle \(EFG\).
Give your answer correct to the nearest degree. (3)
Give your answer correct to the nearest degree. (3)
| Scheme | Marks |
|---|---|
| (\(EF^2 =\)) \(2.1^2 + 3.5^2\) (\(= 4.41 + 12.25 = 16.66\)) | M1 |
| (\(EF =\)) \(\sqrt{2.1^2 + 3.5^2}\) or \(\sqrt{16.66}\) | M1 |
| 4.1 | A1 |
| (3) |
Notes
M1: dep
A1: allow 4.08(166…) rounded or truncated to at least 2DP
| Scheme | Marks |
|---|---|
\(\tan F = \dfrac{2.1}{3.5}\) or \(\tan F = 0.6\) \(\sin F = \dfrac{2.1}{4.1}\) or \(\sin F\) = 0.512(195...) \(\cos F = \dfrac{3.5}{4.1}\) or \(\cos F\) = 0.853(658...) | M1 |
\(\tan^{-1}\left(\dfrac{2.1}{3.5}\right)\) or \(\tan^{-1}0.6\) or \(\sin^{-1}\left(\dfrac{2.1}{4.1}\right)\) or \(\sin^{-1}\)0.512(195…) or \(\cos^{-1}\left(\dfrac{3.5}{4.1}\right)\) or \(\cos^{-1}\)0.853(658…) | M1 |
| 31 | A1 |
| (3) | |
| (6 marks) |
Notes
M1: ft 4.1 from (a)
M1: ft 4.1 from (a)
A1: ft 4.1 from (a)
Accept 30.8 – 31.4