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22 A, B, C and D are four towns - OCR - GCSE Maths - Question 21 - 2017 - Paper 1

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22 A, B, C and D are four towns. B is 25 kilometres due East of A. C is 25 kilometres due North of A. D is 45 kilometres due South of A. (a) Work out the bearing of ... show full transcript

Worked Solution & Example Answer:22 A, B, C and D are four towns - OCR - GCSE Maths - Question 21 - 2017 - Paper 1

Step 1

Work out the bearing of B from C.

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Answer

To determine the bearing of B from C, we start by identifying the locations based on the provided details.

  1. Position A: Assume point A is at the origin (0, 0).
  2. Position B: Since B is 25 km due East of A, its coordinates are (25, 0).
  3. Position C: C is located 25 km due North of A, so its coordinates are (0, 25).

Next, we calculate the angle from point C to point B. The bearing is measured clockwise from the North:

  • The direction from C to B forms a right triangle with:
    • The eastward distance = 25 km (from A to B)
    • The northward distance = 25 km (from A to C)

Using trigonometric functions, we need the tangent of the angle θ.

tan(θ)=OppositeAdjacent=2525=1\tan(θ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{25}{25} = 1

Thus, θ = 45°.

Now, to find the bearing from C to B:

  • The bearing from North is calculated as:

Bearing=0°+θ=0°+45°=45°\text{Bearing} = 0° + θ = 0° + 45° = 45°

So, the bearing of B from C is 045°.

Step 2

Calculate the bearing of D from B.

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Answer

To find the bearing of D from B, we first identify the positions:

  1. Position D: D is located 45 km due South of A, making its coordinates (0, -45).
  2. Position B: We've established B's position earlier as (25, 0).

Now, we need to determine the direction from B to D:

  • The vector from B to D can be calculated as follows:
    • The eastward difference = 0 - 25 = -25 km (Westward)
    • The southward difference = -45 - 0 = -45 km (Southward)

We now compute the angle from North:

  • First, we find the angle θ relative to the east-west line using the tangent function:

tan(θ)=OppositeAdjacent=4525\tan(θ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{45}{25}

Using inverse tangent:

  • θ = arctan(1.8) which gives us approximately 61.93°.

Since we are measuring clockwise from North and the direction is in the fourth quadrant (due South and West), we calculate the bearing:

  • The bearing from North is:

Bearing=180°+θ=180°+61.93°241.93°\text{Bearing} = 180° + θ = 180° + 61.93° \approx 241.93°

Thus, we round up the bearing to give the final result as 242°.

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