The cartesian coordinates of the point whose polar coordinates are `(13, pi - tan^(-1)((5)/(12)))` is

Options

A. (12, 5)
B. (-12, 5)
C. (-12, -5)
D. (12, -5)

Answer

To convert the polar coordinates \((13, \pi - \tan^{-1}(\frac{5}{12}))\) into Cartesian coordinates, we will follow these steps:

### Step 1: Identify the polar coordinates
The polar coordinates are given as:
- \( r = 13 \)
- \( \theta = \pi - \tan^{-1}(\frac{5}{12}) \)

### Step 2: Determine the Cartesian coordinates formula
The Cartesian coordinates \((x, y)\) can be found using the formulas:
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)

### Step 3: Analyze the angle \(\theta\)
Since \(\theta = \pi - \tan^{-1}(\frac{5}{12})\), we need to understand the position of this angle:
- The angle \(\tan^{-1}(\frac{5}{12})\) is in the first quadrant, which means that \(\pi - \tan^{-1}(\frac{5}{12})\) is in the second quadrant.
- In the second quadrant, the cosine value is negative and the sine value is positive.

### Step 4: Calculate \(\cos(\theta)\) and \(\sin(\theta)\)
Using the right triangle formed by the opposite side (perpendicular) and adjacent side (base):
- Opposite side = 5
- Adjacent side = 12
- Hypotenuse can be calculated using Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

Now we can find:
- \( \cos(\theta) = -\frac{\text{Adjacent}}{\text{Hypotenuse}} = -\frac{12}{13} \)
- \( \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13} \)

### Step 5: Substitute values into the Cartesian formulas
Now substituting the values of \(r\), \(\cos(\theta)\), and \(\sin(\theta)\) into the Cartesian formulas:
- For \(x\):
\[ x = r \cos(\theta) = 13 \cdot \left(-\frac{12}{13}\right) = -12 \]
- For \(y\):
\[ y = r \sin(\theta) = 13 \cdot \frac{5}{13} = 5 \]

### Step 6: Write the final Cartesian coordinates
Thus, the Cartesian coordinates are:
\[ (x, y) = (-12, 5) \]

### Final Answer
The Cartesian coordinates of the point whose polar coordinates are \((13, \pi - \tan^{-1}(\frac{5}{12}))\) is \((-12, 5)\).

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