Higher November 2018 Paper 6 Q19
19 In this triangle:
- AB = 9 cm
- AC = 10 cm
- BC > 5 cm
- angle BCA = 60°
- angle ABC < 90°.
Not to scale
Calculate the area of triangle ABC. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 32.2 to 32.3 | 6 | M2 for \(x^2 - 10x + 19 = 0\) oe or M1 for \(9^2 = 10^2 + x^2 - 2 \times 10 \times x \times \cos 60\) AND M1FT for \(\dfrac{10 \pm \sqrt{10^2 - 4 \times 1 \times 19}}{2}\) A1 for \(x = 7.45\) or \(5 + \sqrt{6}\) AND M1 for \(\frac{1}{2} \times 10 \times \textit{their}\ 7.45 \times \sin 60\) oe Alternative M1 for \(\dfrac{\sin 60}{9} = \dfrac{\sin B}{10}\) oe M1 for \(\sin B = \dfrac{10}{9}\sin 60\) or better A1 for B = 74.2(…) AND M1 for A = 180 – 60 – their 74.2 soi by 45.8 AND M1 for \(\frac{1}{2} \times 9 \times 10 \times \sin(\textit{their}\ 45.8)\) | Accept 32 after full correct method Use of cosine rule FT their quadratic = 0 Alternative: M1 for \((x - 5)^2 - 6 = 0\) Ignore 2.55 or \(5 - \sqrt{6}\) Their 7.45 should be from cosine rule followed by quadratic (not from measuring etc.) Use of sine rule Isolates sinB Their 45.8 should be from sine rule followed by 180 – their sine rule answer (not from measuring etc.) |