Higher June 2024 Paper 2 Q23
23.
Diagram not drawn to scale
WXY is a straight line.
Calculate the length of YZ. [6]
| Scheme | Marks | Comments |
|---|---|---|
| \((W\hat{X}Z =)\ \sin^{-1}\!\left(\dfrac{\sin(61)\times11.8}{13.1}\right)\) | M2 | Award marks for similar complete methods M1 for \(\dfrac{\sin(W\hat{X}Z)}{11.8} = \dfrac{\sin(61)}{13.1}\) oe |
| \(= 51.9(82\ldots)(^\circ)\) or \(52(^\circ)\) | A1 | Accept answers in the range 51.9 to 52 |
| (YZ =) \(\sqrt{13.1^2 + 7.8^2 - 2\times13.1\times7.8\times\cos(180 - 51.9(82..))}\) | M2 | FT ‘their derived 51.9(82…)’ provided at least M1 previously awarded. M1 for \((YZ^2 =)\ 13.1^2 + 7.8^2 - 2\times13.1\times7.8\times\cos(180 - 51.9(82..))\) |
| = 18.9(2…) (cm) or 19 (cm) | A1 | FT Accept answers in the range 18.9 to 19 inclusive. |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| (Height of triangle = \(h\) =) \(11.8\sin(61)\) AND \((d =)\ \sqrt{13.1^2 - h^2}\) | M2 | \(h = 10.32\ldots\) |
| \((d =)\ 8.068(2\ldots.)\) or 8.07 | A1 | |
| \((YZ =)\ \sqrt{10.32^2 + (8.07 + 7.8)^2}\) | M2 | FT ‘their 8.07’ provided M2 previously awarded M1 for \((YZ^2 =)\ 10.32^2 + (8.07 + 7.8)^2\) |
| = 18.9(2…) (cm) or 19 (cm) | A1 | FT Accept answers in the range 18.9 to 19 inclusive |
| (6) |
Corrected by Westie's Workshop
- The board's mark scheme printed “√(10.32² + (8.07 + 7.8²))”. Changed to “√(10.32² + (8.07 + 7.8)²)” because the square should apply to the whole bracket (8.07 + 7.8); as printed the working gives 13.2, not the correct 18.9.
- The board's mark scheme printed “(YZ² =) 10.32² + (8.07 + 7.8²)”. Changed to “(YZ² =) 10.32² + (8.07 + 7.8)²” because the square should apply to the whole bracket (8.07 + 7.8), as in the corrected line above.
Original: see the board's mark scheme PDF (link above).