Higher June 2018 Paper 2R Q10
10 \(ABCD\) is a trapezium.
Diagram NOT accurately drawn
Work out the size of angle \(x\).
Give your answer correct to 1 decimal place.
(4)
| Scheme | Marks |
|---|---|
| 24.3 – 16 (= 8.3) | M1 |
\(\tan y = \dfrac{12.5}{\text{``}{8.3}\text{''}}\) or \(\tan z = \dfrac{\text{``}{8.3}\text{''}}{12.5}\) OR \(\sqrt{\text{``}{8.3}\text{''}^2 + 12.5^2}\) (= 15.004…) and \(\sin y = \dfrac{12.5}{\text{``}{15.0}\text{''}}\) or \(\sin z = \dfrac{\text{``}{8.3}\text{''}}{\text{``}{15.0}\text{''}}\) or \(\cos y = \dfrac{\text{``}{8.3}\text{''}}{\text{``}{15.0}\text{''}}\) or \(\cos z = \dfrac{12.5}{\text{``}{15.0}\text{''}}\) | M1 |
\(\tan^{-1}\left(\dfrac{12.5}{\text{``}{8.3}\text{''}}\right)\) (= 56.415…) or \(\tan^{-1}\left(\dfrac{\text{``}{8.3}\text{''}}{12.5}\right)\) (= 33.584…) or \(\sin^{-1}\left(\dfrac{12.5}{\text{``}{15.0}\text{''}}\right)\) (= 56.415…) or \(\sin^{-1}\left(\dfrac{\text{``}{8.3}\text{''}}{\text{``}{15.0}\text{''}}\right)\) (= 33.584…) or \(\cos^{-1}\left(\dfrac{\text{``}{8.3}\text{''}}{\text{``}{15.0}\text{''}}\right)\) (= 56.415…) or \(\cos^{-1}\left(\dfrac{12.5}{\text{``}{15.0}\text{''}}\right)\) (= 33.584…) | M1 |
| 123.6 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: Forming a right angled triangle with 24.3 – 16 on one side, 8.3 may be seen on diagram
M1: for a correct trig statement involving angle \(CDE\) or \(DCE\) where \(E\) is on the line \(AD\) and \(CE\) is perpendicular to \(AD\)
M1: complete method to find angle \(CDE\) or \(DCE\)
A1: 123.5 – 123.6
(In the working, \(y\) is angle \(CDE\) and \(z\) is angle \(DCE\).)