| Date | None Specimen | Marks available | 4 | Reference code | SPNone.1.hl.TZ0.6 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 6 | Adapted from | N/A |
Question
In the triangle ABC, \({\text{AB}} = 2\sqrt 3 \) , AC = 9 and \({\rm{B\hat AC}} = 150^\circ \) .
Determine BC, giving your answer in the form \(k\sqrt 3 \), \(k \in {\mathbb{Z}^ + }\) .
[3]
a.
The point D lies on (BC), and (AD) is perpendicular to (BC). Determine AD.
[4]
b.
Markscheme
\({\text{B}}{{\text{C}}^2} = 12 + 81 + 2 \times 2\sqrt 3 \times 9 \times \frac{{\sqrt 3 }}{2} = 147\) M1A1
\({\text{BC}} = 7\sqrt 3 \) A1
[3 marks]
a.
area of triangle \({\text{ABC}} = \frac{1}{2} \times 9 \times 2\sqrt 3 \times \frac{1}{2}{\text{ }}\left( { = \frac{{9\sqrt 3 }}{2}} \right)\) M1A1
therefore \(\frac{1}{2} \times {\text{AD}} \times 7\sqrt 3 = \frac{{9\sqrt 3 }}{2}\) M1
\({\text{AD}} = \frac{9}{7}\) A1
[4 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.
