Higher November 2018 Paper 2 Q25
25 \(ABC\) and \(ACD\) are triangles.
Not drawn accurately
Work out the size of angle \(x\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\tan 49 = \dfrac{AC}{16}\) | M1 | oe eg \(\tan (90 - 49) = \dfrac{16}{AC}\) or \(AC^2 + 16^2 = \left(\dfrac{16}{\cos 49}\right)^2\) |
| \(\tan 49 \times 16\) or [18.4, 18.41] | M1dep | oe eg \(\dfrac{16}{\tan (90 - 49)}\) or \(\sqrt{\left(\dfrac{16}{\cos 49}\right)^2 - 16^2}\) |
| \(\dfrac{\sin x}{\text{their } [18.4, 18.41]} = \dfrac{\sin 35}{20}\) or \(\dfrac{\text{their } [18.4, 18.41]}{\sin x} = \dfrac{20}{\sin 35}\) | M1dep | oe eg \(\dfrac{\sin x}{16 \tan 49} = \dfrac{\sin 35}{20}\) dep on 1st M1 |
| \(\sin x = \dfrac{\sin 35}{20} \times \text{their } [18.4, 18.41]\) | M1dep | oe eg \(\sin x = \dfrac{16 \tan 49 \sin 35}{20}\) or \(\sin^{-1}\left(\dfrac{\sin 35}{20} \times \text{their } [18.4, 18.41]\right)\) or \(\sin^{-1}\) [0.527, 0.528] dep on 1st and 3rd M1 |
| [31.8, 31.9] | A1 | allow 32 with full method seen |
Additional guidance
| Answer [31.8, 31.9] possibly from scale drawing | 5 marks |
| Answer 32 possibly from scale drawing | Zero |