The diagram shows the positions of three towns, Acton (A), Barston (B) and Chorlton (C) - Edexcel - GCSE Maths - Question 4 - 2019 - Paper 3

Next, we can use the cosine rule to find the length of side AC:

$$AC^2 = AB^2 + BC^2 - 2(AB)(BC) ext{cos(ABC)}$$

Substituting the known values:

• AB = 8 km
• BC = 9 km,
• Angle ABC = 113°:

$$AC^2 = 8^2 + 9^2 - 2(8)(9) ext{cos(113°)}$$

Simplifying gives:

$$AC^2 = 64 + 81 - 144 imes -0.9135 = 145.15$$

Thus,

$$AC = ext{√}145.15 ≈ 12.1 ext{ km}.$$

Now we need to find the angle ACB: Using the sine rule:

rac{AC}{ ext{sin(Angle ACB)}} = rac{BC}{ ext{sin(Angle ABC)}}

Substituting the values:

rac{12.1}{ ext{sin(ACB)}} = rac{9}{ ext{sin(113°)}}

From this, we can solve for angle ACB and hence the bearing from A to C.

To find the bearing of Chorlton from Acton, we add angle ACB to the bearing of AB:

• Let’s assume angle ACB is calculated to be θ. Thus,

$$ext{Bearing from A to C} = 037° + θ$$

Calculating gives: The correct answer will be adjusted to ensure it's correct to 1 decimal place.