Problem 20 - Algebra and Trigonometry with Analytic Geometry

Question

Flight distance An airplane flies 165 miles from point \(A\) in the direction \(130^{\circ}\) and then travels in the direction \(245^{\circ}\) for 80 miles. Approximately how far is the airplane from \(A\) ?

Step-by-Step Solution

Answer

Approximately 139 miles.

Step 1: Understand the Problem

The airplane starts at point \( A \), flies 165 miles at an angle of \( 130^{\circ} \), and then flies another 80 miles at an angle of \( 245^{\circ} \). We need to find the direct distance from point \( A \) to the airplane's final position.

Step 2: Represent the Path with Vectors

Consider the airplane's path as two vectors. The first leg of the journey can be represented by a vector \( \overrightarrow{v_1} \) which, using polar to Cartesian conversion, has components \( 165 \cos (130^{\circ}) \) and \( 165 \sin (130^{\circ}) \). The second leg is represented by vector \( \overrightarrow{v_2} \), with components \( 80 \cos (245^{\circ}) \) and \( 80 \sin (245^{\circ}) \).

Step 3: Calculate the Cartesian Components

Using trigonometric identities, calculate the components of the vectors:\[ \overrightarrow{v_1} = 165 (\cos 130^{\circ}, \sin 130^{\circ}) \approx (\ -106.36, \ 126.14) \]\[ \overrightarrow{v_2} = 80 (\cos 245^{\circ}, \sin 245^{\circ}) \approx (\ -32.58, \ -72.72) \]

Step 4: Find the Resultant Vector

Add the two vectors to find the resultant vector \( \overrightarrow{v} \):\[\overrightarrow{v} = \overrightarrow{v_1} + \overrightarrow{v_2} = (-106.36 + (-32.58), 126.14 + (-72.72)) = (-138.94, 53.42)\]

Step 5: Calculate the Distance from Point A

The distance from point \( A \) to the airplane is the magnitude of the resultant vector \( \overrightarrow{v} \):\[\text{Distance} = \sqrt{(-138.94)^2 + (53.42)^2} \approx \sqrt{19303.68}\approx 138.92\]

Step 6: Conclude the Solution

The approximate distance from point \( A \) to the airplane's final position is 138.92 miles.

Key Concepts

Polar to Cartesian Conversion

To understand the flight path of the airplane, we first need to convert polar coordinates into Cartesian coordinates.
This is essential when dealing with directions specified in degrees. In the airplane scenario, vectors are used to depict the direction and magnitude of its travel. These vectors are initially in polar form, expressed as

• Magnitude: the distance travelled
• Angle: the direction relative to a reference line, such as north (zero degrees)

To convert them into Cartesian form, where vectors are expressed with

• X (horizontal)
• Y (vertical) coordinates

, we use the following conversions:

• For the X-component: Multiply the magnitude by the cosine of the angle: \( X = r\cos(\theta) \).
• For the Y-component: Multiply the magnitude by the sine of the angle: \( Y = r\sin(\theta) \).

For example, for the first vector, using the conversion, we find: \( 165\cos(130^{\circ}) \approx -106.36 \) and \( 165\sin(130^{\circ}) \approx 126.14 \). This effectively translates the vector from polar to Cartesian coordinates.

Trigonometric Identities

While converting angles and solving vector problems, trigonometric identities come into play. These are mathematical equations that relate the various trigonometric functions: sine, cosine, and tangent. In the context of vector addition:

• **Cosine** helps find the horizontal component.
• **Sine** assists in calculating the vertical component.

Using functions like \( \cos(\theta) \) and \( \sin(\theta) \), you can break down any angle into x and y components, critical for solving many physics and engineering problems.Trigonometric identities also ensure the correctness of conversions from polar to Cartesian forms and verify the calculations' integrity, providing exact results needed for plotting or assessing an object's position.

Resultant Vector

When an object, such as an airplane, moves through a multi-leg journey involving different directions, the concept of the resultant vector becomes crucial.
The resultant vector is essentially the sum of all individual vectors. To determine the resultant:

• Add up all the corresponding x-components of the vectors.
• Sum up all the corresponding y-components of the vectors.

In the airplane problem, after converting both vectors into Cartesian form:

• First vector: \( \overrightarrow{v_1} \approx (-106.36, 126.14) \)
• Second vector: \( \overrightarrow{v_2} \approx (-32.58, -72.72) \)

Adding these gives the resultant vector \( \overrightarrow{v} = (-138.94, 53.42) \). This resultant represents the airplane's net vector position from the starting point, factoring in both the distance and direction of travel.

Distance Calculation

After finding the resultant vector of the airplane's path, the next step is to determine the direct distance from the starting point to its new position.
This is achieved by calculating the magnitude of the resultant vector. The formula for the magnitude \( \sqrt{x^2 + y^2} \) helps find the direct "as-the-crow-flies" distance, which works in two dimensions using Pythagoras' theorem.For instance, using the resultant vector components:

• X-component: \( -138.94 \)
• Y-component: \( 53.42 \)

The formula becomes \( \sqrt{(-138.94)^2 + (53.42)^2} \) yielding approximately \( 138.92 \) miles. This gives the student a practical understanding of how these mathematical concepts translate a series of movements into a singular, straightforward measurement from their origin.