June 2022 Paper 2 Q18
18 \(ABCDEFGH\) is a cuboid.
Not drawn accurately
\(AB = 40\text{ cm} \qquad BC = 9\text{ cm} \qquad CG = 20\text{ cm}\)
\(P\) is a point on \(HG\) such that \(\quad HP : PG = 3 : 7\)
\(AP = 25\text{ cm}\)
Work out the size of angle \(APC\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{40}{3 + 7} \times 7\) or 28 | M1 | oe eg \(40 - \dfrac{40}{3 + 7} \times 3\) or \(40 - 12\) may be seen on diagram may be implied |
| \(20^2 + \text{their } 28^2\) or \(400 + 784\) or 1184 or \(4\sqrt{74}\) or [34.4, 34.41] | M1 | oe eg \(\sqrt{20^2 + \text{their } 28^2}\) or \(\sqrt{1184}\) their 28 must be \(\lt 40\) may be seen on diagram |
| \(40^2 + 9^2\) or \(1600 + 81\) or 1681 or 41 | M1 | oe eg \(\sqrt{40^2 + 9^2}\) or \(\sqrt{1681}\) may be seen on diagram |
| their 1681 \(= 25^2 +\) their 1184 \(- 2 \times 25 \times \sqrt{\text{their } 1184} \times \cos x\) | M1dep | oe eg \(\cos^{-1}\dfrac{25^2 + \text{their } 1184 - \text{their } 1681}{2 \times 25 \times \sqrt{\text{their } 1184}}\) or \(\cos^{-1}[0.07, 0.07442]\) dep on 2nd and 3rd M1 \(x\) may be \(APC\) or \(A\) etc |
| [85.7, 86] | A1 |
Additional guidance
| Up to M4 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts | |
| If their \(PG\) is 28 do not allow use of a value other than 28 in subsequent working | |
| \(\dfrac{40}{3 + 7} \times 3 = 12\) \(20^2 + 12^2 = 544\) \(40^2 + 9^2 = 1681\) \(\cos^{-1}\dfrac{25^2 + 544 - 1681}{2 \times 25 \times \sqrt{544}}\) | M0 M1 M1 M1A0 |
| 4th M1 Condone \(\cos^{-1} = 0.07\) for \(\cos^{-1} 0.07\) etc | |
| 4th M1 oes must be a fully correct method eg Uses cosine rule to work out angle \(PCA\) then uses sine rule to work out angle \(APC\) Must get to correct sine rule equation with no errors in method | |
| Missing brackets must be recovered eg 4th M1 Do not allow \(4\sqrt{74}^2\) unless recovered in subsequent working | |
| When \(AP\) is used it must be 25 |