Higher June 2023 Paper 2 Q17
17
Not drawn accurately
Work out the size of angle \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Works out \(AC\) and uses it in triangle \(ABC\) | ||
| \(\cos 37 = \dfrac{AC}{4}\) | M1 | oe eg \(\sin 53 = \dfrac{AC}{4}\) allow [0.798, 0.8] for cos 37 or sin 53 |
| \((AC =)\ 4 \times \cos 37\) or \((AC =)\) [3.19, 3.2] | M1dep | oe eg \((AC =)\ 4 \times \sin 53\) allow [0.798, 0.8] for cos 37 or sin 53 may be seen on diagram |
| \(\sin x = \dfrac{\text{their } [3.19, 3.2]}{9.3}\) or \((x =)\ \sin^{-1}[0.34, 0.3441]\) | M1dep | oe eg \(\cos x = \dfrac{\sqrt{9.3^2 - \text{their } [3.19, 3.2]^2}}{9.3}\) or \((x =)\ 90 - \cos^{-1}[0.34, 0.3441]\) |
| [19.87, 20.13] | A1 | |
| Alternative method 2 Works out angle \(ADC\) and uses it in triangle \(ABD\) | ||
| (angle \(ADC =\)) \(90 - 37\) or (angle \(ADC =\)) 53 | M1 | oe eg (angle \(ADC =\)) \(180 - 90 - 37\) may be seen on diagram |
| \(\dfrac{\sin x}{4} = \dfrac{\sin(90 - 37)}{9.3}\) | M1dep | oe eg \(\dfrac{4}{\sin x} = \dfrac{9.3}{\sin 53}\) |
| \((\sin x =)\ \dfrac{\sin(90 - 37)}{9.3} \times 4\) or \((x =)\ \sin^{-1}[0.34, 0.3441]\) | M1dep | oe |
| [19.87, 20.13] | A1 | |
Additional guidance
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Allow any unambiguous notation for angles eg allow \(B\) for \(x\) | |
| Alt 1 Allow any unambiguous notation for \(AC\) eg \(y\) (condone \(x\) if clearly referring to \(AC\)) | |
| Alt 1 1st M1 must be an equation where \(AC\) is the only variable eg \(AC^2 + (4\sin 37)^2 = 4^2\) | M1 |
| Alt 1 A calculation that leads to \(AC\) scores M1M1 eg \(\sqrt{4^2 - (4\sin 37)^2}\) | M1M1 |
| Alt 1 3rd M1 must have \(\sin x\) (or \(\cos x\)) as the subject or be a calculation that leads to \(x\) | |
| Alt 2 53 only marked at angle \(BAC\) on diagram | M0 |