June 2016 Paper 1 Q20
20 In triangle \(PQR\), \(\cos P = \dfrac{1}{3}\)
Not drawn accurately
Show that triangle \(PQR\) is isosceles. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((3n)^2 + (2n)^2 - 2 \times 3n \times 2n \times \cos P\) | M1 | oe Condone missing brackets for first M1 |
| \(w^2 = 9n^2 + 4n^2 - 12n^2 \times \dfrac{1}{3}\) | A1 | |
| \(w^2 = 9n^2\) | A1 | |
| \(w = 3n\) so the triangle is isosceles or \(QP = QR\) | A1 | |
| Alternative method 2 | ||
| \(\dfrac{(3n)^2 + (2n)^2 - w^2}{2 \times 3n \times 2n}\) | M1 | oe Condone missing brackets for first M1 |
| \(\dfrac{1}{3} = \dfrac{9n^2 + 4n^2 - w^2}{12n^2}\) | A1 | |
| \(w^2 = 9n^2\) | A1 | |
| \(w = 3n\) so the triangle is isosceles or \(QP = QR\) | A1 | |
| Alternative method 3 | ||
| Drop a perpendicular from \(Q\) to \(PR\) | M1 | let this be \(QS\) |
| \(\cos P = \dfrac{PS}{3n}\) or \(\dfrac{1}{3} = \dfrac{PS}{3n}\) | M1 | |
| \(PS = n\) | A1 | |
| \(PR\) is bisected by the perpendicular from \(Q\) hence \(\triangle QPR\) is isosceles | A1 | oe |
Additional guidance
| The final A1 is for a statement saying that two sides have been shown to be equal. | |
| Substituting \(w = 3n\) in either version of the cosine rule and verifying that \(\cos P = \dfrac{1}{3}\) scores SC2 alt 3 ... if they drop a perpendicular from \(Q\) to \(PR\) then assume that \(PS = n\) and then verify that \(\cos P = \frac{1}{3}\) ... eg \(PS = n\), \(QP = 3n\), \(\cos P = \dfrac{PS}{QP} = \dfrac{n}{3n} = \dfrac{1}{3}\) ... they score SC2 | SC2 |