4400 Higher June 2009 Paper 3 Q21
21
Diagram NOT accurately drawn
The diagram shows the length, in centimetres, of each side of triangle \(ABC\).
Angle \(BAC = 60^\circ\).
Find the value of \(x\).
(5)
| Scheme | Marks |
|---|---|
\((x + 4)^2 = x^2 + (x + 6)^2 - 2x(x + 6)\cos 60^\circ\) or \(\cos 60^\circ = \dfrac{(x + 6)^2 + x^2 - (x + 4)^2}{2x(x + 6)}\) | M1 |
| \(x^2 + 4x + 4x + 16\) or \(x^2 + 8x + 16\) and \(x^2 + 6x + 6x + 36\) or \(x^2 + 12x + 36\) | B1 |
| \(x^2 + 8x + 16 = x^2 + x^2 + 12x + 36 - x^2 - 6x\) or \(x^2 + 6x = x^2 + 12x + 36 + x^2 - x^2 - 8x - 16\) oe | B1 |
| \(2x = 20\) oe | B1 |
| 10 | A1 |
| (5) | |
| (5 marks) |
Notes
B1: dep on M1 for correct expansion of \((x + 4)^2\) and \((x + 6)^2\) in correct statement of Cosine Rule
Omitted brackets may be implied by correct subsequent working.
B1: for correctly dealing with \(\cos 60^\circ\) and obtaining a correct equation with no fractions and no brackets
B1: for correct linear equation e.g. \(2x = 20\), \(-2x = -20\), \(4x = 40\), \(2x - 20 = 0\)
A1: cao dep on all preceding marks