1MA0 Higher June 2017 Paper 2 Q25
25 \(ABC\) is an acute angled triangle.
Diagram NOT accurately drawn
The area of triangle \(ABC\) is 19 cm\(^2\).
Work out the size of angle \(ACB\).
Give your answer correct to 3 significant figures. (6)
| Answer | Mark | Notes |
|---|---|---|
| 41.6 | 6 | M1 for \(0.5 \times 5.4 \times 7.3 \times \sin B = 19\) M1 for \((B =) \sin^{-1}\left(\dfrac{19}{0.5 \times 5.4 \times 7.3}\right)\) oe (= 74.6) M1 for \((AC^2 =) 5.4^2 + 7.3^2 - 2 \times 5.4 \times 7.3 \times \cos \text{``}74.6\text{''}\) M1 (dep) for correct order of evaluation or 61.(479…) M1 for \(\dfrac{\sin C}{5.4} = \dfrac{\sin \text{``}74.6\text{''}}{\text{``}7.84\text{''}}\) oe or \(0.5 \times 7.3 \times \text{``}7.84\text{''} \times \sin C = 19\) A1 for answer in the range 41.55 to 41.65 from correct working OR (with perpendicular from \(C\) meeting \(AB\) at a point \(X\)) M1 for \(0.5 \times 5.4 \times CX = 19\) M1 for \((CX =) \dfrac{19}{0.5 \times 5.4}\) (= 7.037...) M1 for \(BCX = \cos^{-1} \dfrac{\text{``}7.04\text{''}}{7.3}\) (= 15.425…) M1 for \((BX =) \sqrt{7.3^2 - \text{``}7.04\text{''}^2}\) (= 1.94…) M1 for \(ACX = \tan^{-1} \dfrac{5.4 - \text{``}1.94\text{''}}{\text{``}7.04\text{''}}\) (= 26.17…) A1 for answer in the range 41.55 to 41.65 from correct working OR (with perpendicular from \(A\) meeting \(BC\) at a point \(Y\)) M1 for \(AY = \dfrac{19}{7.3 \div 2} = 5.20547..\) M1 for \(BY = \sqrt{5.4^2 - \text{``}5.20547\text{''}^2}\) (=1.43630..) M1 for \(CY = 7.3 - \text{``}1.43630..\text{''}\) (= 5.86369 ...) M1 for \(\tan C = \dfrac{\text{``}5.20547..\text{''}}{\text{``}5.86369..\text{''}}\) (= 0.8877..) M1 for \(C = \tan^{-1} \text{``}0.8877..\text{''}\) A1 for answer in the range 41.55 to 41.65 from correct working |