Higher June 2017 Paper 1 Q22
22 The diagram shows a hexagon \(ABCDEF\).
\(ABEF\) and \(CBED\) are congruent parallelograms where \(AB = BC = x\) cm.
\(P\) is the point on \(AF\) and \(Q\) is the point on \(CD\) such that \(BP = BQ = 10\) cm.
Given that angle \(ABC = 30^\circ\),
prove that \(\cos PBQ = 1 - \dfrac{(2 - \sqrt{3})}{200}x^2\) (5)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| Proof | B1 | (indep) for stating \(\cos 30 = \dfrac{\sqrt{3}}{2}\) | |
| M1 | for \(PQ^2 = 10^2 + 10^2 - 2 \times 10 \times 10 \times \cos PBQ\) or \(AC^2 = x^2 + x^2 - 2 \times x \times x \times \cos 30\ (= x^2(2 - \sqrt{3}))\) oe | ||
| M1 | for \(\cos PBQ = \dfrac{10^2 + 10^2 - PQ^2}{2 \times 10 \times 10}\) (implies previous M1) | ||
| \(\cos PBQ = {}\) \(\dfrac{10^2 + 10^2 - x^2(2 - \sqrt{3})}{200}\) \({= \dfrac{200 - x^2(2 - \sqrt{3})}{200}}\) | M1 | for \(\cos PBQ = \dfrac{10^2 + 10^2 - (x^2 + x^2 - 2 \times x \times x \times \cos 30)}{2 \times 10 \times 10}\) | |
| A1 | conclusion of proof with all working seen |