Problem 97 - A Graphical Approach to Precalculus with Limits

Question

Starting at point \(A,\) a ship sails 18.5 kilometers on a bearing of \(189^{\circ}\). then turns and sails 47.8 kilometers on a bearing of \(317^{\circ} .\) Find the distance of the ship from point \(A\).

Step-by-Step Solution

Answer

The ship is approximately 60.98 kilometers from point A.

Step 1: Draw the scenario

First, visualize the problem and draw it. Point \(A\) is the starting point. From \(A\), the ship sails 18.5 kilometers on a bearing of \(189^{\circ}\). Next, the ship turns and sails 47.8 kilometers on a bearing of \(317^{\circ}\). This creates a triangle \( \triangle ABC \), where \(B\) is the point after the first leg of the journey and \(C\) is the final point.

Step 2: Assign coordinates to points

Consider point \(A\) at coordinates \((0, 0)\) since it is the origin. Point \(B\) can be found using the length and bearing: \(B_x = 18.5 \sin(189^\circ)\) and \(B_y = 18.5 \cos(189^\circ)\). Similarly, position \(C\) using relative movement from \(B\) with \(C_x = B_x + 47.8 \sin(317^\circ)\) and \(C_y = B_y + 47.8 \cos(317^\circ)\).

Step 3: Calculate coordinates of B

Calculate the coordinates using bearing and distance for point \(B\). Use the formulas: \(B_x = 18.5 \sin(189^\circ)\approx -18.5 \sin(9^\circ)\) and \(B_y = 18.5 \cos(189^\circ)\approx -18.5 \cos(9^\circ)\). Compute these using a calculator: \(B_x \approx -2.91\) and \(B_y \approx -18.47\).

Step 4: Calculate coordinates of C

Now find \(C_x = B_x + 47.8 \sin(317^\circ)\) and \(C_y = B_y + 47.8 \cos(317^\circ)\). Compute the values: \(\sin(317^\circ) = \sin(43^\circ)\approx 0.682\) and \(\cos(317^\circ) = \cos(43^\circ)\approx -0.731\). Thus, \(C_x = -2.91 + 47.8 \times 0.682 \approx 29.63\), and \(C_y = -18.47 + 47.8 \times -0.731 \approx -53.31\).

Step 5: Calculate the distance from A to C

Using the distance formula to find the distance from \(A\) to \(C\): \(d = \sqrt{(C_x - A_x)^2 + (C_y - A_y)^2}\). Substitute the values: \(d = \sqrt{(29.63 - 0)^2 + (-53.31 - 0)^2}\).Calculate \(d = \sqrt{29.63^2 + 53.31^2}\). Find approximations \(29.63^2 \approx 877.49\) and \(53.31^2 \approx 2842.96\). Thus, \(d \approx \sqrt{3720.45}\approx 60.98\).

Step 6: Conclusion

After calculating, the distance of the ship from point \(A\) is approximately 60.98 kilometers.

Key Concepts

Bearing Calculations

In navigation and surveying, bearing calculations are essential for determining direction relative to a fixed point, like True North. Bearings are typically measured in degrees (°), moving clockwise around the compass. This can help determine the precise path and turn needed between two points.

• A bearing of 0° points directly North.
• A bearing of 90° points directly East.
• If a ship travels on a bearing of 189°, it travels southwards slightly to the west.

In the given exercise, understanding bearings is crucial as the ship changes direction. Transforming these angles into movements helps plot the path on a coordinate plane, crucial for further calculations.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, involves using a coordinate system to analyze geometrical shapes. This study is fundamental when dealing with navigational problems, allowing the projection of real-world paths onto a coordinate plane.
An origin point is typically described as \(0, 0\), from which movements are mapped using \(x, y\) coordinates. For instance, if a ship moves:

• 18.5 km on a specific bearing, geometric functions like sine and cosine can help calculate the new coordinates from the original point, reflecting its new position on this imaginary map.
• Coordinate changes can be calculated using familiar trigonometric relations: \(x = r\sin(\theta)\) and \(y = r\cos(\theta)\).

In our scenario, point \(B\) is calculated from the origin \(A\) by applying these formulas, and the same is done to find \(C\) from \(B\). This approach helps demystify the path, turning complex navigational instructions into simple, interpretable data.

Distance Formula

The distance formula is a powerful tool in geometry for determining the straight-line distance between two points on a coordinate plane. This formula is derived from the Pythagorean theorem and is expressed as:

• \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

By substituting the coordinates of points into this formula, it yields the direct distance, or "displacement," between them. In the context of the exercise, after calculating the coordinates of points \(C\) and \(A\), the formula is directly applied.
Substituting the computed coordinates follows logical arithmetic steps, ensuring accuracy in calculating the exact distance the ship has moved from its original point \(A\) to its final position \(C\). This process is definitive for establishing precise measurements in navigation or when dealing with any travel-based calculation problems.