June 2024 Paper 2 Q11
11 \(ABC\) and \(ACD\) are triangles.
Work out the area of triangle \(ACD\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{AC}{\sin 59} = \dfrac{18}{\sin 66}\) | M1 | oe equation in \(AC\) eg \(\dfrac{\sin 59}{AC} = \dfrac{\sin 66}{18}\) accept [0.85, 0.86] for \(\sin 59\) accept [0.91, 0.914] for \(\sin 66\) |
| \(\dfrac{18}{\sin 66} \times \sin 59\) or [16.7, 17.011] | M1dep | oe calculation for \(AC\) accept [0.85, 0.86] for \(\sin 59\) accept [0.91, 0.914] for \(\sin 66\) |
| \(\dfrac{1}{2} \times 15 \times\) their [16.7, 17.011] \(\sin 70\) | M1dep | oe accept [0.93, 0.94] for \(\sin 70\) |
| [119, 119.11] | A1 |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Allow any indication of \(AC\) eg \(x\) or \(b\) | |
| An equivalent method for the first two marks may be from using sine rule and cosine rule eg \(AB = \dfrac{18}{\sin 66} \times \sin(180 - 59 - 66)\) or [16.1, 16.140124] \(AC^2 = 18^2 + 16.14^2 - 2 \times 18 \times 16.14 \times \cos 59\) \(AC = \sqrt{18^2 + 16.14^2 - 2 \times 18 \times 16.14 \times \cos 59}\) | M1 M1dep |
| [119, 119.11] seen with area of \(ABCD\) or \(ABC\) as final answer | M3A0 |