## Part (a)

| Fix A: $v_{\min} = 15\sin 50° = 11.5 \text{ km h}^{-1}$ (3 s.f.) | M1A1 |
| or: triangle without the right angle identified and $\frac{15}{\sin\theta} = \frac{v_B}{\sin 50°}$ | A1 |
| $\Rightarrow v_B = \frac{15\sin 50°}{\sin\theta}$ | |
| minimum value $\Rightarrow \theta = 90°$ for M1 | |
| As above for A1A1 | |
| **Total: 3 marks** | |

## Part (b)

| Ambiguous Sine Rule: 2 possible solutions for $\alpha$ | B1B1 |
| **Total: 2 marks** | |

## Part (c)

| $\frac{\sin\alpha}{\sin 50°} = \frac{15}{13}$ | M1A1 |
| $\alpha = 62.1°$ (or $118°$) (smaller value gives larger relative velocity) | A1 |
| $\Rightarrow$ either $v = 13\cos 62.1° + 15\cos 50° = 15.72 km h^{-1}$ | M1A1 |
| Or $v^2 = 15^2 + 13^2 - 390\cos 67.9 = 247.27$ | M1 |
| $v = 15.7 km h^{-1}$ | A1 |
| $\text{Time} = \frac{20}{\text{their } 15.72...} = 1.272... \text{ hrs}$ | M1A1 |
| Earliest time is $13.16 \text{ hrs}$ or $13.17 \text{ hrs}$ accept $1.16$ (pm) or $1.17$ (pm) | A1 |
| **Total: 8 marks** | |

### Guidance:

**a)** M1: Velocity of B relative to A is in the direction of the line joining AB. Minimum V requires a right angled triangle. Convincing attempt to find the correct side. A1: $15 \times \sin(\text{their } 50°)$. A1: Q specifies 3sf, so 11.5 only.

**b)** B1B1: Convincing argument. B1B0: Argument with some merit.

**c)** M1: Use of Sine Rule. A1: Correct expression. A1: (2 possible values.) pick the correct value. M1: Use trig. to form an equation in $v$. A1: correct equation. M1: $\text{time} = \frac{\text{distance}}{\text{speed}}$. A1ft: correct expression with their $v$ (not necessarily evaluated). A1: correct time in hours & minutes. Or: M1: Use of cosine rule. A1: $13^2 = 15^2 + v^2 - 2 \times 15 \times v \times \cos 50°$. A1: (Award after the next two marks) $15.72$ or awrt $15.72$. M1: Attempt to solve the equation for $v$. A1: $\frac{30\cos 50 \pm \sqrt{(30\cos 50)^2 - 4 \times 56}}{2}$ (15.72 or 3.562). Finish as above.

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