Higher January 2019 Paper 1 Q10
10 Here is triangle \(ABD\).
Diagram NOT accurately drawn
The point \(C\) lies on \(BD\).
\(AD = 13\) cm \(BC = 8\) cm angle \(ADB = 90°\) angle \(CAD = 20°\)
Calculate the size of angle \(BAC\).
Give your answer correct to 1 decimal place.
(5)
| Scheme | Marks |
|---|---|
Working with \(CD\) and then triangle \(ABD\) E.g. \(\tan 20 = \dfrac{CD}{13}\) | M1 |
| E.g. (\(CD\) =) \(13\tan 20\) or 4.7(316...) | M1 |
| E.g. \(\tan(BAD) = \dfrac{8 + \text{``}{4.73}\text{''}}{13}\) or \(\tan(BAD) = 0.97(93...)\) | M1 |
| E.g. (\(BAD\) =) \(\tan^{-1}(\text{``}{0.979}\text{''})\) or 44.4(024...) | M1 |
| 24.4 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a correct statement or equation including angle \(BAD\) as the only variable
M1: for a correct method to find angle \(BAD\)
A1: for 24.3 - 24.41
Award M1A1M1M1A0 for an answer in the range 44.3 – 44.41
| Scheme | Marks |
|---|---|
Alternative mark scheme – working with \(AC\) and then triangle \(ABC\) E.g. \(\cos 20 = \dfrac{13}{AC}\) | M1 |
| E.g. (\(AC\) =) \(\dfrac{13}{\cos 20}\) or 13.8(3…) | M1 |
| E.g. (\(AB\) =) \(\sqrt{\text{``}{13.8}\text{''}^2 + 8^2 - 2 \times 13.8 \times 8 \times \cos(110)}\) (=18.1(9..) or 18.2 | M1 |
E.g. \(\dfrac{\sin BAC}{8} = \dfrac{\sin 110}{\text{``}{18.1}\text{''}}\) or \(8^2 = \text{``}{13.8}\text{''}^2 + \text{``}{18.1}\text{''}^2 - 2 \times \text{``}{13.8}\text{''} \times \text{``}{18.1}\text{''} \times \cos BAC\) | M1 |
| 24.4 | A1 |
Notes
M1: for a correct statement or equation including \(AC\) as the only variable
E.g. \(AC^2 = 13^2 + (13\tan 20)^2\)
M1: for a correct method to find \(AB\)
M1: for a correct statement or equation including angle \(BAC\) as the only variable
A1: for ans in range 24.3 - 24.41
Award M4A0 for an answer in the range 44.3 – 44.41