June 2023 Paper 2 Q8
8 \(ABC\) is a triangle with perpendicular height \(AD\).
Not drawn accurately
Area of \(ABC = 25\text{ cm}^2\)
\(BD : DC = 2 : 3\)
Work out the size of angle \(w\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{1}{2} \times BC \times 4 = 25\) or (\(BC =\)) \(25 \times 2 \div 4\) or (\(BC =\)) 12.5 | M1 | oe eg \(\dfrac{1}{2} \times BD \times 4 + \dfrac{1}{2} \times DC \times 4 = 25\) allow any unambiguous indication of \(BC\) or \(BD\) or \(DC\) may be seen on diagram |
| (\(DC =\)) \(\dfrac{3}{3 + 2} \times \text{their } 12.5\) or 7.5 | M1dep | oe eg (\(DC =\)) their \(12.5 \div 5 \times 3\) may be seen on diagram |
| \(\tan w = \dfrac{4}{\text{their } DC}\) or \(\tan w = \dfrac{8}{15}\) or \(\tan^{-1} 0.53(\ldots)\) | M1 | oe eg \(\sin w = \dfrac{4}{\sqrt{4^2 + (\text{their } DC)^2}}\) or \(\cos w = \dfrac{\text{their } DC}{\sqrt{4^2 + (\text{their } DC)^2}}\) or \(90 - \tan^{-1}\dfrac{\text{their } DC}{4}\) |
| [27.9, 28.4] | A1 | SC3 [46.6, 46.94] or 47 |
| Alternative method 2 | ||
| Expression for area = 25 with \(BD : DC = 2 : 3\) | M1 | eg \(\dfrac{1}{2} \times 2x \times 4 + \dfrac{1}{2} \times 3x \times 4 = 25\) or \(\dfrac{1}{2} \times y \times 4 + \dfrac{1}{2} \times \dfrac{3}{2}y \times 4 = 25\) implied by \(BD = 5\) may be seen on diagram |
| Method to work out \(DC\) or (\(DC =\)) 7.5 | M1dep | eg (\(DC =\)) \(2.5 \times 3\) or (\(DC =\)) \(5 \times \dfrac{3}{2}\) may be seen on diagram |
| \(\tan w = \dfrac{4}{\text{their } DC}\) or \(\tan w = \dfrac{8}{15}\) or \(\tan^{-1} 0.53(\ldots)\) | M1 | oe eg \(\sin w = \dfrac{4}{\sqrt{4^2 + (\text{their } DC)^2}}\) or \(\cos w = \dfrac{\text{their } DC}{\sqrt{4^2 + (\text{their } DC)^2}}\) or \(90 - \tan^{-1}\dfrac{\text{their } DC}{4}\) |
| [27.9, 28.4] | A1 | SC3 [46.6, 46.94] or 47 |
Additional guidance
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| \(BC = 12\) (no working seen) \(\qquad \dfrac{3}{5} \times 12 = 7.2\) \(\tan w = \dfrac{4}{7.2} \qquad w = 29.1\) | M0M0 M1A0 |
| SC3 is from omitting the \(\dfrac{1}{2}\) from area of triangle formula | |
| Alt 1 Using \(\dfrac{2}{3 + 2}\) instead of \(\dfrac{3}{3 + 2}\) is not a misread | 2nd M0 |
| 3rd M1 oe must be of the form \(\tan w =\) (or \(\sin w =\) or \(\cos w =\)) or be a calculation that evaluates to [27.9, 28.4] eg applying cosine rule to triangle \(BAC\) \(\cos w = \dfrac{12.5^2 + 72.25 - 41}{2 \times 12.5 \times \sqrt{72.25}}\) | M3 |
| 3rd M1 Allow \(\sin w = 0.47(\ldots)\) or \(\cos w = 0.88(\ldots)\) or \(90 - [61.6, 62.1]\) | M3 |
| 3rd M1 Condone their \(DC = 3\) if not contradicted |