Higher November 2020 Paper 2R Q13
13 The diagram shows four congruent right-angled triangles \(ABJ\), \(BCI\), \(CDH\) and \(DEG\).
The diagram also shows the straight line \(ABCDEF\).
Diagram NOT accurately drawn
\(AJ = 15\) cm
Angle \(BAJ = 35^\circ\)
\(AF = 80\) cm
Work out the length of \(EF\).
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
\(\cos 35^\circ = \dfrac{15}{AB}\) or \(\sin 55^\circ = \dfrac{15}{AB}\) or \(\dfrac{15}{\sin 55} = \dfrac{JB}{\sin 35}\) and \(\left(AB^2 =\right) (\text{“}10.50\text{”})^2 + 15^2\) or \(\tan 35^\circ = \dfrac{JB}{15}\) and \(\left(AB^2 =\right) (\text{“}10.50\text{”})^2 + 15^2\) | M1 |
\((AB =)\; \dfrac{15}{\cos 35^\circ}\;(= 18.3\ldots)\) or \((AB =)\; \dfrac{15}{\sin 55^\circ}\;(= 18.3\ldots)\) or \((AB =)\; \sqrt{(\text{“}10.50\text{”})^2 + 15^2}\) or \((AB =)\; \sqrt{(15\tan 35)^2 + 15^2}\) | M1 |
| ‘18.3’ × 4 (= 73.2) | M1 |
| 80 − ‘18.3’ × 4 or 80 – ‘73.2’ | M1 |
| 6.75 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: oe eg \(x\) for \(AB\)
M1: dep 1st M1
M1: dep 1st M1
A1: accept 6.75 – 6.8
Alternative Mark Scheme for Q13 [do not mix and match with above MS]
| Scheme | Marks |
|---|---|
| 15 × 4 (= 60) | M1 |
| \(\cos 35^\circ = \dfrac{\text{“}60\text{”}}{AE}\) or \(\sin 55^\circ = \dfrac{\text{“}60\text{”}}{AE}\) | M1 |
| \((AE =)\; \dfrac{\text{“}60\text{”}}{\cos 35^\circ}\;(= 73.2)\) or \((AE =)\; \dfrac{\text{“}60\text{”}}{\sin 55^\circ}\;(= 73.2)\) | M1 |
| 80 – ‘73.2’ | M1 |
| 6.75 | A1 |
Notes
M1: dep 1st M1
A1: accept 6.75 – 6.8