Higher January 2022 Paper 2R Q19
19 \(ABCD\) is a horizontal rectangular field.
Diagram NOT accurately drawn
A vertical pole, \(AE\), is placed at the corner \(A\) of the field.
\(AE = 12\) m \(AB = 18\) m \(AD = 8\) m
Calculate the size of the angle between \(EC\) and the plane \(ABCD\)
Give your answer correct to one decimal place.
(3)
| Scheme | Marks |
|---|---|
\((AC =)\ \sqrt{8^2 + 18^2}\left(= \sqrt{388} = 2\sqrt{97} = 19.697\ldots\right)\) or \((CE =)\ \sqrt{8^2 + 18^2 + 12^2}\left(= \sqrt{532} = 2\sqrt{133} = 23.065\ldots\right)\) oe | M1 |
eg \(\tan ECA = \left(\dfrac{12}{\text{‘}{\sqrt{388}}\text{’}}\right)\) or \(\sin ECA = \left(\dfrac{12}{\text{‘}{\sqrt{532}}\text{’}}\right)\) or \(\cos ECA = \left(\dfrac{\text{‘}{\sqrt{388}}\text{’}}{\text{‘}{\sqrt{532}}\text{’}}\right)\) or \(\sin ECA = \dfrac{\sin 90 \times 12}{\text{‘}{\sqrt{532}}\text{’}}\) or \(\cos ECA = \left(\dfrac{(\text{‘}{\sqrt{388}}\text{’})^2 + (\text{‘}{\sqrt{532}}\text{’})^2 - 12^2}{2 \times \text{‘}{\sqrt{388}}\text{’} \times \text{‘}{\sqrt{532}}\text{’}}\right)\) oe | M1 |
| 31.4 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a correct trig statement with \(ECA\) as the only unknown.
NB allow use ‘\(x\)’ or other variable in place of \(ECA\).
A1: allow 31.3 – 31.5