4400 Higher November 2009 Paper 4 Q18
18
Diagram NOT accurately drawn
\(ABC\) is a triangle.
\(D\) is a point on \(AB\).
\(CD\) is perpendicular to \(AB\).
\(AD = 7.2\) cm, \(DB = 3.9\) cm, \(AC = 8.3\) cm.
Calculate the size of angle \(DBC\).
Give your answer correct to 1 decimal place.
(5)
| Scheme | Marks |
|---|---|
| \(8.3^2 - 7.2^2 = 68.89 - 51.84 = 17.05\) | M1 |
| \(\sqrt{8.3^2 - 7.2^2} = 4.129...\) | M1 |
| tan and \(\dfrac{\text{“}4.129...\text{”}}{3.9}\) | M2 |
| 46.6 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for \(8.3^2 - 7.2^2\)
M1: for \(\sqrt{8.3^2 - 7.2^2}\)
M2: M1 for tan and \(\frac{3.9}{\text{“}4.129...\text{”}}\)
Accept \(CD\) rounded or truncated to at least 1 dp (4.12916...)
A1: Accept answer rounding to 46.6 (4.1 → 46.43... 4.12 → 46.57... 4.13 → 46.64...)
Alternative methods (each M1 M1 M2 A1, answer 46.6):
1. M1 M1 as above, then \(\sqrt{4.129^2 + 3.9^2} = 5.679...\); M2 for cos and \(\frac{3.9}{\text{“}5.679\text{”}}\) (M1 for cos and \(\frac{\text{“}5.679\text{”}}{3.9}\)). Accept \(BC\) rounded or truncated to at least 1 dp (5.67978...)
2. M1 M1 as above, then \(BC = 5.679...\); M2 for sin and \(\frac{\text{“}4.129\text{”}}{\text{“}5.679\text{”}}\) (M1 for sin and \(\frac{\text{“}5.679\text{”}}{\text{“}4.129\text{”}}\)). Accept \(CD\) and \(BC\) rounded or truncated to at least 1 dp
3. M1 for a correct method for finding \(\angle A\), eg \(\cos\angle A = \frac{7.2}{8.3}\) (\(\angle A = 29.83...°\)); M1 for correct Cosine Rule expression for \(BC\), \(\sqrt{11.1^2 + 8.3^2 - 2 \times 11.1 \times 8.3\cos\text{“}29.8\text{”}}\); M2 for cos and \(\frac{3.9}{\text{“}5.679\text{”}}\) (M1 for cos and \(\frac{\text{“}5.679\text{”}}{3.9}\))
4. M1 for \(\angle A\) and M1 for the Cosine Rule for \(BC\) as in 3; M2 for \(\sin B = \frac{8.3\sin\text{“}29.8\text{”}}{\text{“}5.68\text{”}}\) (M1 for correct statement of Sine Rule eg \(\frac{\sin B}{8.3} = \frac{\sin\text{“}29.8\text{”}}{\text{“}5.68\text{”}}\))