June 2019 Paper 2 Q14
14 Here is a triangle.
Use the cosine rule to work out the ratio \(\quad b^2 : a^2\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(a^2 + (3a)^2 - 2 \times a \times 3a \times \cos 120\) or \(\cos 120 = \dfrac{a^2 + (3a)^2 - b^2}{2 \times a \times 3a}\) | M1 | oe eg may substitute \(\cos 120 = -0.5\) may be seen in a square root |
| \(b^2 = a^2 + 9a^2 + 3a^2\) or \(b^2 = 13a^2\) or \(-b^2 = -13a^2\) or \(b = \sqrt{13}\,a\) | A1 | oe equation of the form \(b^2 =\) or \(b =\) with brackets expanded and terms fully simplified \(\cos 120 = -0.5\) substituted |
| 13 : 1 | A1 | SC1 7 : 1 |
Additional guidance
| Allow recovery of missing brackets | |
| \(a^2 = a^2 + (3a)^2 - 2 \times a \times 3a \times \cos 120\) not recovered | M1M0A0 |
| \(b^2 = 10a^2 - -3a^2\) | M1A0A0 |
| \(b^2 = 10a^2 + 3a^2\) | M1A1A0 |