Higher June 2022 Paper 2 Q7
7.
Diagram not drawn to scale
ABC is a right-angled triangle.
AB = 4·8 cm and \(B\hat{A}C = 57^\circ\)
Calculate the area of triangle ABC. [5]
Area = ……………… cm2
| Scheme | Marks | Comments |
|---|---|---|
| (BC = ) \(4.8\tan 57^\circ\) or \(\dfrac{4.8}{\tan(180 - 90 - 57)}\) | M2 | M1 for \(\tan 57^\circ = \dfrac{BC}{4.8}\) si or \(\tan(180\text{-}90\text{-}57) = \dfrac{4.8}{BC}\) si |
| 7.39 to 7.4 si | A1 | CAO |
| (Area = ) \(\dfrac12 \times 7.39135\ldots \times 4.8\) | M1 | FT ‘\(\dfrac12 \times\) their derived \(BC \times 4.8\)’ provided ‘their derived BC’ is from use of trigonometry |
| 17.7(392...) (cm2) | 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 1
| Scheme | Marks | Comments |
|---|---|---|
| (BC = ) \(\dfrac{4.8 \times \sin 57}{\sin(90 - 57)}\) | M2 | M1 for \(\dfrac{BC}{\sin 57} = \dfrac{4.8}{\sin(90 - 57)}\) |
| 7.39 to 7.4 si | A1 | CAO |
| (Area = ) \(\dfrac12 \times 7.39135\ldots \times 4.8\) | M1 | FT ‘\(\dfrac12 \times\) their derived \(BC \times 4.8\)’ provided ‘their derived BC’ is from use of trigonometry |
| 17.7(392...) (cm2) | 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 = \(\frac12 \times 4.8 \times AC \times \sin 57\) | M1 | Not awarded until \(AC = \dfrac{4.8}{\cos 57}\) oe substituted. |
| \((AC =)\ \dfrac{4.8}{\cos 57}\) | m2 | m1 for \(\cos 57 = \dfrac{4.8}{AC}\) |
| 8.8 to 8.81(3...) si | A1 | CAO |
| Area of ABC = \(\frac12 \times 4.8 \times 8.81.. \times \sin 57\) = 17.7(392...) (cm2) | A1 | FT ‘their derived AC’ If correct, accept answer in range 17.7 to 17.8; accept 18 following correct working |
| (5) |