**Triangle BAC with:**
- $BC = 700$ m, $AC = 500$ m, angle $BAC = 15°$

Using cosine rule:
$$BC^2 = 700^2 + 500^2 - 2 \times 500 \times 700\cos 15°$$ | M1 A1 |
$(= 63851.92...)$
$BC = 253$ (awrt) | A1 (3) |

**(a)** $$\frac{\sin B}{700} = \frac{\sin 15°}{\text{candidate's } BC}$$ | M1 |

$\sin B = \sin 15° \times 700/253_c = 0.716...$ and giving an **obtuse** $B$ $(134.2°)$ | M1 (dep)

**(b)** $\theta = 180° - \text{candidate's angle } B$ (Dep. on first M only, $B$ can be acute) | M1 |

$\theta = 180 - 134.2 = 45.8°$ (allow 46 or awrt 45.7, 45.8, 45.9) | M1; A1 (4) [7] |

[46 needs to be from correct working]

**Notes:**
- (a) If use $\cos 15° = ...$, then A1 not scored until written as $BC^2 = ...$ correctly

**Splitting into 2 triangles BAX and CAX, where X is foot of perp. from B to AC**
- Finding value for $BX$ and $CX$ and using Pythagoras | M1
- $BC^2 = (500\sin 15°)^2 + (700 - 500\cos 15°)^2$ | A1
- $BC = 253$ (awrt) | A1

**(b) Several alternative methods:** (Showing the M marks, 3rd M dep. on first M)

(i) $\cos B = \frac{500^2 + \text{candidate's } sBC^2 - 700^2}{2 \times 500 \times \text{candidate's } BC}$ or $700^2 = 500^2 + BC_c^2 - 2 \times 500 \times BC_c$ | M1
Finding angle $B$ | M1, then M1 as above

(ii) 2 triangle approach, as defined in notes for (a)
$$\tan CBX = \frac{700 - \text{value for } AX}{\text{value for } BX}$$ | M1
Finding value for $\angle CBX$ (≈ 59°) | M1
$\theta = [180° - (75° + \text{candidate's } \angle CBX)]$ | M1

(iii) Using sine rule (or cos rule) to find $C$ first:
Correct use of sine or cos rule for $C$ | M1, Finding value for $C$ | M1
Either $B = 180° - (15° + \text{candidate's } C)$ or $\theta = (15° + \text{candidate's } C)$ | M1

(iv) $700\cos 15° = 500 + BC\cos\theta$ | M2 (first two Ms earned in this case)
Solving for $\theta$; $\theta = 45.8$ (allow 46 or 5.7, 45.8, 45.9) | M1; A1

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