Higher June 2022 Paper 6 Q21
21 The diagram shows triangle ABC.
X lies on BC such that angle AXC = 90°.
Not to scale
BC = 7.5 cm, angle ABC = 32° and angle ACB = 43°.
Work out length AX.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 2.8 to 2.81 with correct working | 6 | First M1 may be on diagram or within other M marks M1 for BAC = 180 − (32 + 43) or 58 + 47 or 105 | All methods: “correct working” requires at least M1ANDM1ANDM1 First M1 may be on diagram or within other M marks M1 for BAC = 180 − (32 + 43) or 58 + 47 or 105 |
| AND M2 for AC = \(\dfrac{7.5 \times \sin 32}{\sin(\textit{their } 105)}\) or M1 for \(\dfrac{\text{AC}}{\sin 32} = \dfrac{7.5}{\sin(\textit{their } 105)}\) oe | AND M2 for AB = \(\dfrac{7.5 \times \sin 43}{\sin(\textit{their } 105)}\) or M1 \(\dfrac{\text{AB}}{\sin 43} = \dfrac{7.5}{\sin(\textit{their } 105)}\) oe | ||
| AND A1 for AC = 4.11[4…] or 4.115 (M1A1 implies M2A1) | AND A1 for AB = 5.29[5…] or 5.3[0] (M1A1 implies M2A1) | ||
| AND M1 for \(\dfrac{\text{AX}}{\textit{their } 4.11} = \sin 43\) or their 4.11 × sin 43 oe | AND M1 for \(\dfrac{\text{AX}}{\textit{their } 5.3[0]} = \sin 32\) or their 5.3[0] × sin 32 oe | ||
| If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working If 0 or 1 scored, instead award SC2 for 4.11[4…], 4.115 with no or insufficient working | If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working If 0 or 1 scored, instead award SC2 for 5.29[5…] or 5.3[0] with no or insufficient working | ||
| Alternative Method splitting BC: First M1 may be on diagram or within other M marks. Accept other terms for \(d\). Do not accept numerical partition of BC. M1 for BX = \(d\) and XC = \(7.5 - d\) | First M1 may be on diagram or within other M marks. Accept other terms for \(d\). Do not accept numerical partition of BC. M1 BX = \(7.5 - d\) and XC = \(d\) | ||
| AND M2 for \(d = \dfrac{7.5\tan 43}{\tan 32 + \tan 43}\) oe or M1 for \(d\tan 32 = (7.5 - d)\tan 43\) oe | AND M2 for \(d = \dfrac{7.5\tan 32}{\tan 43 + \tan 32}\) oe or M1 for \(d\tan 43 = (7.5 - d)\tan 32\) oe | ||
| AND A1 for \(d\) = 4.49 to 4.5 (M1A1 implies M2A1) | AND A1 for \(d\) = 3 to 3.01 (M1A1 implies M2A1) | ||
| AND M1 for \(\dfrac{\text{AX}}{\textit{their } 4.49} = \tan 32\) or their 4.49 × tan 32 or for \(\dfrac{\text{AX}}{7.5 - \textit{their } 4.49} = \tan 43\) or (7.5 − their 4.49) × tan 43 | AND M1 for \(\dfrac{\text{AX}}{\textit{their } 3.01} = \tan 43\) or their 3.01 × tan 43 or \(\dfrac{\text{AX}}{7.5 - \textit{their } 3.01} = \tan 32\) or (7.5 − their 3.01) × tan 32 | ||
| If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working. If 0 or 1 scored, instead award SC2 for 4.49[…] or 4.5[0] with no or insufficient working. | If 0, 1 or 2 scored, instead award SC3 for 2.8 to 2.81 with no or insufficient working. If 0 or 1 scored, instead award SC2 for 4.49[…] or 4.5[0] with no or insufficient working. | ||
| Alternative Method using Areas The main scheme still applies. The first four marks will be identical and the area method eventually simplifies to the final M1 expression. This is just a much more complex method. | |||