Math Problem Statement

The bearing of a point A from P is 310o and the bearing of another point B from P is 292o. Both A and B are due north of a point C which is 5600 m due west of P.

Find the distance between A and B.

Solution

To solve the problem, we'll break it down step by step.

Step 1: Understand the Bearing Information

  • The bearing of a point is the direction measured in degrees clockwise from the north.
  • The bearing of point A from point P is 310310^\circ.
  • The bearing of point B from point P is 292292^\circ.

Since bearings are measured clockwise from the north:

  • 310310^\circ means A is slightly west of north.
  • 292292^\circ means B is even further west of north.

Step 2: Visualize and Set Up the Triangle

We are given that both points A and B are due north of a point C, which is 5600 meters due west of P. This setup implies that the distance between A and B is simply the horizontal (west-east) distance between them, as both points lie on the same vertical line (due north).

Given that bearings are measured clockwise from the north:

  • The angle θ1\theta_1 from the north to PA (from P to A) is 360310=50360^\circ - 310^\circ = 50^\circ.
  • The angle θ2\theta_2 from the north to PB (from P to B) is 360292=68360^\circ - 292^\circ = 68^\circ.

Step 3: Calculate the Horizontal Distances to Points A and B

Using trigonometry, we can calculate the horizontal distances (westward components) to points A and B.

For point A: xA=Distance from P to C×tan(θ1)=5600×tan(50)x_A = \text{Distance from P to C} \times \tan(\theta_1) = 5600 \times \tan(50^\circ)

For point B: xB=Distance from P to C×tan(θ2)=5600×tan(68)x_B = \text{Distance from P to C} \times \tan(\theta_2) = 5600 \times \tan(68^\circ)

Step 4: Calculate the Distance Between A and B

The distance between A and B, dABd_{AB}, is the difference between the horizontal distances to A and B: dAB=xBxAd_{AB} = |x_B - x_A|

Let’s calculate it.

xA=5600×tan(50)5600×1.19186674.08 metersx_A = 5600 \times \tan(50^\circ) \approx 5600 \times 1.1918 \approx 6674.08 \text{ meters} xB=5600×tan(68)5600×2.475113860.56 metersx_B = 5600 \times \tan(68^\circ) \approx 5600 \times 2.4751 \approx 13860.56 \text{ meters}

Thus, the distance between A and B is: dAB=13860.566674.087186.48 metersd_{AB} = |13860.56 - 6674.08| \approx 7186.48 \text{ meters}

Final Answer

The distance between points A and B is approximately 7186.48 meters.

Would you like further details on any part of the solution or have any questions?

Follow-up Questions

  1. How would the solution change if point C was east of P instead of west?
  2. What would the bearing of A and B be if they were southeast of P?
  3. How would you calculate the distance between A and B if they were not due north of point C?
  4. What trigonometric principles are involved in solving this problem?
  5. How would the problem change if P was at a different location, say, due south of C?

Tip

When dealing with bearings, always convert them to angles measured from the north in the counterclockwise direction to simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Trigonometry

Formulas

Horizontal distance calculation using tangent

Theorems

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Suitable Grade Level

Grades 10-12