Math Problem Statement

B is 844 metres due south of A. C is due west of A. The bearing of C from B is 295 degrees. Find the distance BC, to the nearest metre.

Solution

To solve this problem, we need to use trigonometry and some basic principles of coordinate geometry. Here are the steps:

  1. Understanding the bearings and the problem setup:

    • A is the reference point.
    • B is 844 meters due south of A.
    • C is due west of A.
    • The bearing of C from B is 295 degrees.

    Bearings are measured clockwise from the north. Therefore, a bearing of 295 degrees means that the angle measured clockwise from the north to the line BC is 295 degrees.

  2. Determining the angle for calculations:

    • A bearing of 295 degrees is equivalent to 295270=25295^\circ - 270^\circ = 25^\circ west of the south line from B (since 270 degrees points directly west).
  3. Using trigonometry to find the distance BC:

    • In the triangle formed by points A, B, and C, we can use the right-angled triangle relationships.
    • Let's denote the distance BC as dd.

    Since the angle between the south line from B and the line BC is 2525^\circ:

    tan(25)=oppositeadjacent=distance AC844\tan(25^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{distance AC}}{844}

  4. Calculating AC: distance AC=844tan(25)\text{distance AC} = 844 \cdot \tan(25^\circ)

    Using the tangent value: tan(25)0.4663\tan(25^\circ) \approx 0.4663

    distance AC=8440.4663393.67 meters\text{distance AC} = 844 \cdot 0.4663 \approx 393.67 \text{ meters}

  5. Using the Pythagorean theorem to find BC:

    • Now that we have AC, we can find BC using the Pythagorean theorem in triangle ABC.

    • Since B is due south of A and C is due west of A, triangle ABC is a right triangle with: AB=844 meters (given)AB = 844 \text{ meters (given)} AC=393.67 metersAC = 393.67 \text{ meters} BC=AB2+AC2BC = \sqrt{AB^2 + AC^2}

      Substituting the values: BC=8442+393.672BC = \sqrt{844^2 + 393.67^2} BC=712336+154978.7BC = \sqrt{712336 + 154978.7} BC=867314.7BC = \sqrt{867314.7} BC930.22 metersBC \approx 930.22 \text{ meters}

To the nearest meter, the distance BCBC is 930 meters.

Would you like any further details or explanations?

Further Questions:

  1. What is a bearing, and how is it measured?
  2. How do we determine the angle of a bearing relative to cardinal directions?
  3. What is the significance of using the tangent function in this problem?
  4. How do we derive the tangent of an angle in trigonometry?
  5. Why is the Pythagorean theorem applicable in this problem?
  6. Can you explain the steps for converting the bearing angle into a usable trigonometric angle?
  7. What are some real-life applications of using bearings and distances in navigation?
  8. How can errors in calculating bearings and distances affect real-world navigation?

Tip:

When solving problems involving bearings, always draw a diagram to visualize the relationships between the points and angles. This helps in understanding the problem better and avoiding mistakes.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Coordinate Geometry
Pythagorean Theorem

Formulas

Distance formula in coordinate geometry
Tangent function in trigonometry

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12