Higher November 2020 Paper 2 Q18
18 This shape is made from two right-angled triangles and a rectangle.
Not drawn accurately
Work out the size of angle \(x\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\tan 62 = \dfrac{h}{5}\) | M1 | oe eg \(\tan (90 - 62) = \dfrac{5}{h}\) or \(\dfrac{h}{\sin 62} = \dfrac{5}{\sin 28}\) any letter |
| \(5 \times \tan 62\) or 9.4(0…) | M1dep | oe eg \(\dfrac{5}{\tan 28}\) or \(\dfrac{5}{\sin 28} \times \sin 62\) |
| \(\sin x = \dfrac{\text{their } 9.4(0\ldots)}{12}\) or \(\sin x =\) [0.78, 0.784] | M1dep | oe eg \(\sin x = \dfrac{5 \times \tan 62}{12}\) or \(\cos x = \dfrac{\sqrt{12^2 - \text{their } 9.4^2}}{12}\) |
| [51.536, 51.63] | A1 | accept 52 with M3 seen |
| Alternative method 2 | ||
| \(\left(\dfrac{5}{\cos 62}\right)^2 - 5^2\) or [88.4, 88.43] | M1 | oe |
| \(\sqrt{\left(\dfrac{5}{\cos 62}\right)^2 - 5^2}\) or 9.4(0…) | M1dep | oe |
| \(\sin x = \dfrac{\text{their } 9.4(0\ldots)}{12}\) or \(\sin x =\) [0.78, 0.784] | M1dep | oe eg \(\cos x = \dfrac{\sqrt{12^2 - \text{their } 9.4^2}}{12}\) |
| [51.536, 51.63] | A1 | accept 52 with M3 seen |
Additional guidance
| Answer in range with truncation to 51 | M1M1M1A1 |