Problem 90 - Algebra and Trigonometry

Question

Radar station \(A\) uses a coordinate system where \(A\) is located at the pole and due east is the polar axis. On this system, two other radar stations, \(B\) and \(C,\) are located at coordinates \(\left(150,-24^{\circ}\right)\) and \(\left(100,32^{\circ}\right)\) respectively. If radar station \(B\) uses a coordinate system where \(B\) is located at the pole and due east is the polar axis, then what are the coordinates of radar stations \(A\) and \(C\) on this second system? Round answers to one decimal place.

Step-by-Step Solution

Answer

A: (150, 155.2°), C: (125.1, 115.6°).

Step 1 - Convert Polar Coordinates to Cartesian Coordinates

Start by converting the polar coordinates of stations B and C to Cartesian coordinates. Use the formulas: \( x = r \cos(\theta) \)\( y = r \sin(\theta) \)For station B, \( r_B = 150 \) and \( \theta_B = -24^{\circ} \): \( x_B = 150 \cos(-24^{\circ}) \)\( y_B = 150 \sin(-24^{\circ}) \). For station C, \( r_C = 100 \) and \( \theta_C = 32^{\circ} \): \( x_C = 100 \cos(32^{\circ}) \) and \( y_C = 100 \sin(32^{\circ}) \).

Step 2 - Compute Cartesian Coordinates

Calculate the Cartesian coordinates:For station B: \( x_B = 150 \cos(-24^{\circ}) = 150 \cdot 0.9135 = 137 \)\( y_B = 150 \sin(-24^{\circ}) = 150 \cdot -0.4067 = -61 \)For station C: \( x_C = 100 \cos(32^{\circ}) = 100 \cdot 0.848 \)\( y_C = 100 \sin(32^{\circ}) = 100 \cdot 0.5299 \)So, the Cartesian coordinates of stations B and C are \( (137, -61) \) and \( (84.8, 52.99) \), respectively.

Step 3 - Translate the Axes

Translate the origin of the Cartesian coordinate system to the location of station B. Subtract the coordinates of B from the coordinates of A and C. For station A: \( x_A' = 0 - 137 = -137 \)\( y_A' = 0 - (-61) = 61 \) For station C:\( x_C' = 84.8 - 137 = -52.2 \)\( y_C' = 52.99 - (-61) = 113.99 \)So, the coordinates of A and C in the new system are \( (-137, 61) \) and \( (-52.2, 113.99) \).

Step 4 - Convert Back to Polar Coordinates

Convert the translated Cartesian coordinates back to polar coordinates using the formulas: \( r = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1}(\frac{y}{x}) \).For station A: \( r_A' = \sqrt{(-137)^2 + 61^2} = 150 \)\( \theta_A' = \tan^{-1}(\frac{61}{-137}) \approx 155.2^{\circ} \)For station C: \( r_C' = \sqrt{(-52.2)^2 + 113.99^2} \approx 125.1 \)\( \theta_C' = \tan^{-1}(\frac{113.99}{-52.2}) \approx 115.6^{\circ} \)So, the polar coordinates of A and C in the new system are \( (150, 155.2^{\circ}) \) and \( (125.1, 115.6^{\circ}) \).

Key Concepts

polar to cartesian coordinates

When we need to convert from polar coordinates to Cartesian coordinates, we use the formulas involving trigonometry. Polar coordinates are represented as \( (r, \theta) \), where \( r \) is the radius (distance from the origin) and \( \theta \) is the angle from the positive x-axis in radians or degrees. To convert these to Cartesian coordinates (x, y), you use these formulas:
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
For example, to convert \( (150, -24\degree) \) to Cartesian, you'd calculate:
\( x = 150 \cos(-24\degree) \) and \( y = 150 \sin(-24\degree) \).
This helps in visualizing the position in a regular coordinate system.

cartesian to polar coordinates

Converting back from Cartesian coordinates to polar coordinates is also crucial. When you have \( (x, y) \) coordinates, you can find the polar coordinates \( (r, \theta) \) using these relationships:
- \( r = \sqrt{x^2 + y^2} \) (Pythagorean theorem)
- \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \).
Ensure your angle \( \theta \) is measured correctly with respect to the coordinate axis. For instance, if the Cartesian coordinates are \( (-137, 61) \), then:
\( r = \sqrt{(-137)^2 + 61^2} \) and
\( \theta = \tan^{-1}\left(\frac{61}{-137}\right) \).
This gives you the distance from the origin and the angle from the positive x-axis.

coordinate translation

To translate coordinates effectively, move the origin of the current coordinate system. Translation involves shifting all coordinates by subtracting the coordinates of the new origin. Suppose radar station B needs to be the new origin. Original coordinates of B are:
\( x_B = 137 \)
\( y_B = -61 \).
The translation for any point originally at \( (x, y) \) becomes:
- \( x' = x - x_B \)
- \( y' = y - y_B \).
Therefore, the new coordinates of A and C relative to B are recalculated by subtracting B’s values. For instance, coordinates of C shift from
\( (84.8, 52.99) \) to
\( (84.8 - 137, 52.99 - (-61)) = (-52.2, 113.99) \).
This step ensures all relevant positions are accurately mapped to the new origin.

trigonometry in coordinate systems

Trigonometry is paramount in translating and converting coordinates. It involves the relationships between the angles and sides of triangles.
In the context of our coordinate transformations:
- Sine (\( \sin(\theta) \)) relates to the opposite side over the hypotenuse.
- Cosine (\( \cos(\theta) \)) relates to the adjacent side over the hypotenuse.
- Tangent (\( \tan(\theta) \)) relates the opposite side over the adjacent side.
These relationships help convert polar to Cartesian coordinates and vice versa. Also, translating the coordinates involves understanding vector addition and subtraction, underpinned by trigonometric concepts. These principles simplify navigating multi-step transformations and ensure precise calculations.