1MA0 Higher June 2012 Paper 2 Q24
24
Diagram NOT accurately drawn
\(ABC\) is a triangle.
\(AB = 8.7\) cm.
Angle \(ABC = 49^\circ\).
Angle \(ACB = 64^\circ\).
Calculate the area of triangle \(ABC\).
Give your answer correct to 3 significant figures. (5)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\dfrac{AC}{\sin 49} = \dfrac{8.7}{\sin 64}\) \(AC = \dfrac{8.7}{\sin 64} \times \sin 49\) (= 7.305…) \(\frac{1}{2} \times 8.7 \times 7.305... \times \sin(180 - 64 - 49)\) | 29.3 | 5 | M1 for \(\dfrac{AC}{\sin 49} = \dfrac{8.7}{\sin 64}\) oe M1 for (\(AC =\)) \(\dfrac{8.7}{\sin 64} \times \sin 49\) A1 for 7.3(05…) M1 for \(\frac{1}{2} \times 8.7 \times\) ‘7.305’ \(\times \sin(180 - 64 - 49)\) A1 for 29.19 – 29.3 OR M1 for \(\dfrac{BC}{\sin(180 - 64 - 49)} = \dfrac{8.7}{\sin 64}\) oe M1 for (\(BC =\)) \(\dfrac{8.7}{\sin 64} \times \sin\)‘67’ A1 for 8.9(10...) M1 for \(\frac{1}{2} \times 8.7 \times\) ‘8.910’ \(\times \sin 49\) A1 for 29.19 – 29.3 OR (\(X\) is point such that \(AX\) is perpendicular to \(BC\)) M1 for \(AX = 8.7 \times \sin 49\) (= 6.565…) or \(XB = 8.7 \times \cos 49\) (= 5.707...) M1 for \(XB = 8.7 \times \cos 49\) (= 5.707...) and \(CX =\) ‘6.565’ ÷ tan 64 oe (= 3.202…) A1 for 8.9(10...) or 5.7(07...) and 3.2(02...) M1 for ½ × ‘6.565…’ × (‘5.707’ + ‘3.202’) oe A1 for 29.19 – 29.3 |