1MA0 Higher June 2016 Paper 2 Q25
25
Diagram NOT accurately drawn
\(ABC\) is an acute-angled triangle.
\(BA = 7\) cm
\(BC = 8\) cm
The area of triangle \(ABC\) is 18 cm\(^2\).
Work out the size of angle \(BAC\).
Give your answer correct to 3 significant figures.
You must show all your working. (6)
| Answer | Mark | Notes |
|---|---|---|
| 80.4 | 6 | M1 for \(0.5 \times 7 \times 8 \times \sin x = 18\) M1 (dep) for \((x =)\ \sin^{-1}\left(\dfrac{18}{0.5 \times 7 \times 8}\right)\) oe (= 40) M1 (dep on at least M1) for \((AC^2 =)\ 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos \text{``}40\text{''}\) M1 (dep on previous M1) for correct order of evaluation or 27.2(03…) or 5.2(15…) M1 (dep) for \(\sin A = \dfrac{8 \times \sin \text{``}40\text{''}}{\text{``}5.2(15..)\text{''}}\) or \(\sin A = \dfrac{18}{0.5 \times 7 \times \text{``}5.2(15..)\text{''}}\) or \(\cos A = \dfrac{\text{``}5.2(15..)\text{''}^2 + 7^2 - 8^2}{2 \times \text{``}5.2(15..)\text{''} \times 7}\) A1 for answer in the range 80.3 to 80.4 from correct working OR (with perpendicular from \(A\) meeting \(BC\) at a point \(X\)) M1 for \(0.5 \times 8 \times h = 18\) M1 (dep) for \((h =)\ \dfrac{18}{0.5 \times 8}\) (= 4.5) M1 (dep on at least M1) for \(BAX = \cos^{-1}\dfrac{\text{``}4.5\text{''}}{7}\) (= 49.99 …) M1 (dep) for \((BX =)\ \sqrt{7^2 - \text{``}4.5\text{''}^2}\) (= 5.3619…) M1 (dep) for \(CAX = \tan^{-1}\dfrac{\text{``}4.5\text{''}}{8 - \text{``}5.3619\text{''}}\) (= 30.38 …) A1 for answer in the range 80.3 to 80.4 from correct working NB Similar method applies for use of perpendicular from \(C\) to \(AB\) |