1MA0 Higher November 2016 Paper 2 Q15
15
Diagram NOT accurately drawn
\(AB = 15\) m
\(BC = 24\) m
Angle \(BAD = 62^\circ\)
Work out the size of angle \(BCD\).
Give your answer correct to 1 decimal place. (5)
| Answer | Mark | Notes |
|---|---|---|
| 28.9 | 5 | M1 for \(\sin 62 = \dfrac{BD}{15}\) or \(\dfrac{BD}{\sin 62} = \dfrac{15}{\sin 90}\) oe M1 for \((BD =)\ 15 \times \sin 62\) or \(\dfrac{15}{\sin 90} \times \sin 62\) oe (= 13.24...) M1 for \(\tan BCD = \dfrac{\text{``}13.24\text{''}}{24}\) oe or \(\tan BDC = \dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified M1 for \(BCD = \tan^{-1}\dfrac{\text{``}13.24\text{''}}{24}\) oe or \(BDC = \tan^{-1}\dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified A1 for 28.8 – 28.9 OR M1 for \(\cos(90 - 62) = \dfrac{BD}{15}\) M1 for \((BD =)\ 15 \times \cos(90 - 62)\) (= 13.24...) M1 for \(\tan BCD = \dfrac{\text{``}13.24\text{''}}{24}\) oe or \(\tan BDC = \dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified M1 for \(BCD = \tan^{-1}\dfrac{\text{``}13.24\text{''}}{24}\) oe or \(BDC = \tan^{-1}\dfrac{24}{\text{``}13.24\text{''}}\) with \(BDC\) clearly identified A1 for 28.8 – 28.9 |