Higher November 2018 Paper 5 Q13
13 In the diagram, ABC is a right-angled triangle.
P is a point on AB.
BC = 40 m, AP = 20 m and angle ABC = 30°.
Not to scale
(a) Show that AC = 20 m. [3]
(b) Find the length of PB.
Give your answer in the form \(a(\sqrt{3} - b)\), where \(a\) and \(b\) are integers. [5]
Give your answer in the form \(a(\sqrt{3} - b)\), where \(a\) and \(b\) are integers. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| AC = 40sin30 | M2 | M1 for \(\dfrac{\text{AC}}{40} = \sin 30\) oe | |
| 20 and evidence that sin 30 = 0.5 | A1 | If 0 scored, B1 for sin 30 = 0.5 oe | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(20(\sqrt{3} - 1)\) | 5 | B4 for \(20\sqrt{3} - 20\) or \(\sqrt{1200} - 20\) or B3 for \(\sqrt{1200}\) or \(\dfrac{40\sqrt{3}}{2}\) or M2 for 40cos30 oe or M1 for \(\cos 30 = \dfrac{\text{AB}}{40}\) oe | Other methods are possible e.g. Pythag e.g. M2 for [AB =] \(\sqrt{40^2 - 20^2}\) e.g. M1 for AB² + 20² = 40² |