Problem 67 - A Graphical Approach to Precalculus with Limits

Question

From shore station A, a ship \(C\) is observed in the direction \(N 22.4^{\circ}\) E. The same ship is observed to be in the direction \(\mathrm{N} 10.6^{\circ} \mathrm{W}\) from shore station \(\mathrm{B}\), located a distance of 25.5 kilometers exactly southeast of A. Find the distance of the ship from station A.

Step-by-Step Solution

Answer

The distance of the ship from station A is 25.5 km.

Step 1: Visualize and Organize the Problem

To understand the problem, first draw a diagram. Place stations A and B on a coordinate system, with station A as the reference point (origin). Since station B is located 25.5 km to the southeast of A, its direction is along the line at 45° to the east and south, resulting in coordinates for B as \[ (x_B, y_B) = (25.5\cos(45^{\circ}), -25.5\sin(45^{\circ})). \]

Step 2: Calculate Coordinates of Station B

Convert the southeast direction of B from station A into Cartesian coordinates. Use \[ x_B = 25.5 \times \cos(45^{\circ}) \quad \text{and} \quad y_B = -25.5 \times \sin(45^{\circ}) \]This gives \[ x_B = y_B = 25.5 \times \frac{1}{\sqrt{2}} \approx 18.02 \text{ km}. \] Thus, the coordinates of B are (18.02, -18.02).

Step 3: Apply the Law of Sines

Construct the angles in the triangle formed by points A, B, and C, using the given directions.From station A, ship C is seen at an angle 22.4° east of north, making a total angle of 67.6° with the vertical downward. From station B, ship C is seen at an angle 10.6° west of north, making a total angle of 79.4° with the vertical downward.The angle at C is the external angle between the lines to A and B, given by sum of provided angles:\[\angle C = 67.6^{\circ} + 79.4^{\circ} = 147^{\circ}. \]

Step 4: Calculate Distance AC Using Law of Sines

Apply the Law of Sines:\[ \frac{AC}{\sin B} = \frac{AB}{\sin C} \,\text{where} \, \angle B = 180^{\circ} - 147^{\circ} = 33^{\circ}.\]Given \(AB = 25.5\, \text{km}\) and \(\angle C = 147^{\circ}\), solve for AC:\[ AC = \frac{25.5 \times \sin(33^{\circ})}{\sin(147^{\circ})}. \]

Step 5: Solve for AC

Compute the sines:- \( \sin(33^{\circ}) \approx 0.5446 \)- \( \sin(147^{\circ}) \approx 0.5446 \)Calculate:\[ AC = \frac{25.5 \times 0.5446}{0.5446} = 25.5 \text{ km}. \]

Key Concepts

Law of Sines

The Law of Sines is a fundamental concept in trigonometry, often used to find unknown lengths or angles in non-right triangles. It's especially handy when dealing with problems where you know two angles and one side, or two sides and a non-included angle. The law states that in any triangle, the ratios of the length of a side to the sine of its opposite angle are equal. This can be expressed mathematically as:

• For any triangle ABC: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

In the given problem with ship C and stations A and B, the Law of Sines helps us find the distance from station A to the ship. We're given the angles seen from both stations, and the distance between the stations A and B. By applying the law, we can effectively solve for the missing side, which in this case is the distance AC. With given angles \(\angle B = 33^{\circ} \) and \(\angle C = 147^{\circ} \), the solution involves calculating the lengths through their respective sine values, ensuring the ratios hold true.

Angles

Angles are a pivotal part of trigonometry and solving geometry problems. In this exercise, we deal with angles formed by directional observations from stations A and B towards the ship. Understanding the references and transformations on these angles is crucial.

• From station A, the angle observed is \(22.4^{\circ}\) east of north. To match this with our axes, consider the full perspective from north downward, giving a complementary angle of \(67.6^{\circ}\).
• From station B, the ship is at \(10.6^{\circ}\) west of north. Again, convert this to the downward angle which gives \(79.4^{\circ}\).

The angle \(C\) at the ship is formed by these two observations and calculated as an external angle summing up to \(147^{\circ}\). These transformations and calculations allow for accurate application of trigonometric identities, like the Law of Sines, to find unknown sides.

Coordinate Systems

Coordinate systems serve as a fundamental framework to map out positions and distances in space using numerical coordinates. For this navigation problem involving stations A and B, and ship C, the coordinate plane simplifies visualization and calculation.

• We start by placing station A at the origin, \((0,0)\), ensuring it acts as a fixed starting reference.
• Station B, being \(25.5\) kilometers southeast of A, integrates basic trigonometry for its placement: along a line \(45^{\circ}\) to the east and south. This results in both x and y values as \(18.02\) km, calculated by multiplying \(25.5\) km by \(\cos(45^{\circ})\) and \(-\sin(45^{\circ})\).

This setup not only aids in forming the visual scope but also in tackling more advanced trigonometric problems, like applying the Law of Sines. Each coordinate value emanates from a clear understanding of direction and distance in the Cartesian framework, paving a way for precise calculations of length and angles in trigonometry exercises.