Higher November 2017 Paper 2 Q9
9 Work out the size of angle \(x\). [2 marks]
Not drawn accurately
| Answer | Mark | Comments |
|---|---|---|
| \(\tan x = \dfrac{3}{7}\) or \(\tan^{-1} \dfrac{3}{7}\) or \(\sin x = \dfrac{3(\sin 90)}{\sqrt{3^2 + 7^2}}\) or \(\sin x = \dfrac{3(\sin 90)}{\sqrt{58}}\) or \(\cos x = \dfrac{7}{\sqrt{3^2 + 7^2}}\) or \(\cos x = \dfrac{7}{\sqrt{58}}\) or \(90 - \tan^{-1} \dfrac{7}{3}\) or 90 − [66.7, 66.81] or 90 − 67 | M1 | oe eg \(\cos x = \dfrac{7^2 + \left(\sqrt{7^2 + 3^2}\right)^2 - 3^2}{2 \times \sqrt{3^2 + 7^2} \times 7}\) Any letter |
| [23, 23.3] | A1 |
Additional guidance
| \(\tan = \dfrac{3}{7}\) or \(\tan \dfrac{3}{7}\) or \(\tan^{-1} = \dfrac{3}{7}\) (unless recovered) | M0 |
| Answer [23, 23.3] (possibly coming from scale drawing) | M1A1 |
| If using sine rule must rearrange to \(\sin x =\) for M1 | |
| If using cosine rule must rearrange to \(\cos x =\) for M1 | |
| Allow [0.42, 0.43] for \(\dfrac{3}{7}\) | |
| Allow 2.33… for \(\dfrac{7}{3}\) | |
| Allow [7.6, 7.62] for \(\sqrt{3^2 + 7^2}\) |