Foundation June 2022 Paper 2 Q24
24.
Diagram not drawn to scale
ABC is a right-angled triangle.
\(AB = 4{\cdot}8\) cm and \(B\hat{A}C = 57^\circ\).
Calculate the area of triangle ABC. [5]
| Scheme | Marks | Comments |
|---|---|---|
| (BC = ) \(4.8\tan 57^\circ\) or \(\frac{4.8}{\tan(180-90-57)}\) | M2 | M1 for \(\tan 57^\circ = \frac{BC}{4.8}\) si or \(\tan(180\text{-}90\text{-}57) = \frac{4.8}{BC}\) si |
| 7.39 to 7.4 si | A1 | CAO |
| (Area = ) \(\frac{1}{2} \times 7.39135\ldots \times 4.8\) | M1 | FT ‘\(\frac{1}{2} \times\) their derived BC \(\times\ 4.8\)’ provided ‘their derived BC’ is from use of trigonometry |
| 17.7(392...) (cm\(^2\)) | A1 | FT ‘their derived BC’ If correct, accept answer in range 17.7 to 17.8; accept 18 following correct working |
| (5) |
Alternative
Alternative (Higher tier) method 1
| Scheme | Marks | Comments |
|---|---|---|
| (BC = ) \(\frac{4.8 \times \sin 57}{\sin(90-57)}\) | M2 | M1 for \(\frac{BC}{\sin 57} = \frac{4.8}{\sin(90-57)}\) |
| 7.39 to 7.4 si | A1 | CAO |
| (Area = ) \(\frac{1}{2} \times 7.39135\ldots \times 4.8\) | M1 | FT ‘\(\frac{1}{2} \times\) their derived BC \(\times\ 4.8\)’ provided ‘their derived BC’ is from use of trigonometry |
| 17.7(392...) (cm\(^2\)) | A1 | FT ‘their derived BC’ If correct, accept answer in range 17.7 to 17.8; accept 18 following correct working |
Alternative (Higher tier) method 2
| Scheme | Marks | Comments |
|---|---|---|
| Area of ABC = \(\frac{1}{2} \times 4.8 \times AC \times \sin 57\) | M1 | Not awarded until \(AC = \frac{4.8}{\cos 57}\) oe substituted. |
| \((AC =)\ \frac{4.8}{\cos 57}\) | m2 | m1 for \(\cos 57 = \frac{4.8}{AC}\) |
| 8.8 to 8.81(3...) si | A1 | CAO |
| Area of ABC = \(\frac{1}{2} \times 4.8 \times 8.81\ldots \times \sin 57\) = 17.7(392...) (cm\(^2\)) | A1 | FT ‘their derived AC’ If correct, accept answer in range 17.7 to 17.8; accept 18 following correct working |