The diagram shows the positions of three towns, Acton (A), Barston (B) and Chorlton (C) - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 1

We will use the Sine Rule to find the length AC:

$\frac{AC}{\sin(150°)} = \frac{8}{\sin(67°)}$

Cross multiplying gives:

$AC \cdot \sin(67°) = 8 \cdot \sin(150°)$

Calculating the sine values, we have:

$\sin(150°) = 0.5 \quad \text{and} \quad \sin(67°) \approx 0.920 \implies$

$AC = \frac{8 \cdot 0.5}{0.920} \approx 4.35 \text{ km}$

Now to find the bearing of Chorlton from Acton. The bearing is calculated by taking the north direction as 0°, moving clockwise:

Since we have angle ABC as 67° and the angle at B extending from A (037°) towards C is 150°, we combine these as follows:

$\text{Bearing of Chorlton from Acton} = 037° + 67° = 104°$

Thus, the correct bearing of Chorlton from Acton rounded to one decimal place is:

104.0°