A polar to rectangular calculator converts a point given by distance and angle into ordinary x and y coordinates. Polar form is natural when direction matters, such as in navigation, physics, and engineering, while rectangular form is natural when a point must be plotted on a grid or used in an equation. Being able to move between the two forms is a basic skill in trigonometry and coordinate geometry.
Polar coordinates describe a point by how far it is from the origin, r, and by the angle, theta, measured counterclockwise from the positive x-axis. Rectangular coordinates describe the same point by horizontal and vertical distances from the origin. The two descriptions are equivalent, but each makes different problems easier. A vector given by length and direction is polar; the same vector drawn on graph paper is rectangular.
What this calculator does
The conversion uses right-triangle trigonometry. For a point at distance r and angle theta, the horizontal component is x = r cos theta and the vertical component is y = r sin theta. Together, those two equations project the polar radius onto the x-axis and the y-axis. The calculator below accepts theta in degrees, because most textbook problems use degrees, and converts to radians internally for the trigonometric functions.
A useful check is the distance formula. For the converted point, sqrt(x^2 + y^2) must equal r, except for rounding. If the converted distance is far from r, the angle was probably entered in the wrong unit or the sine and cosine were swapped.
Formula
x = r cos(theta), y = r sin(theta). Theta is measured counterclockwise from the positive x-axis.
How to use it
- Enter r, the distance from the origin. It is normally non-negative.
- Enter theta in degrees, measured counterclockwise from the positive x-axis.
- Compute x = r cos(theta) and y = r sin(theta).
- Round to a sensible number of decimal places for the problem, usually two or three.
- Check with sqrt(x^2 + y^2). It should return r, apart from rounding.
Try the calculator
Worked example
Convert the polar point (r = 5, theta = 53.13 degrees) to rectangular coordinates. Using cos 53.13 degrees = 0.600 and sin 53.13 degrees = 0.800, x = 5 x 0.600 = 3.00 and y = 5 x 0.800 = 4.00. The rectangular point is (3, 4). The check works: sqrt(3^2 + 4^2) = sqrt(25) = 5, which matches r. This is the familiar 3-4-5 triangle, written in polar form.
A second example shows a quadrant change. For r = 10 and theta = 180 degrees, cos 180 = -1 and sin 180 = 0, so x = -10 and y = 0. The point lies on the negative x-axis, which matches an angle pointing directly left.
How to check your answer
Check the distance with sqrt(x^2 + y^2). It must equal r.
Check the quadrant. An angle between 0 and 90 degrees should give positive x and positive y; an angle between 180 and 270 degrees should give both negative.
Convert back if needed: r = sqrt(x^2 + y^2) and theta from the arctangent, adjusted for quadrant.
Test a point you know by heart, such as angle 0 degrees, which must give (r, 0), and angle 90 degrees, which must give (0, r). If either test fails, the angle unit or the sine-cosine assignment is wrong, and that error will affect every other conversion in the same set.
Common mistakes
- Swapping sine and cosine, which swaps x and y and moves the point to the wrong quadrant or angle.
- Entering degrees into a calculator mode set to radians, or the reverse.
- Using a negative r without adjusting the angle, which points in the opposite direction.
- Forgetting that theta is measured from the positive x-axis, not from the y-axis or from north.
Frequently asked questions
What is the difference between polar and rectangular coordinates?
Polar coordinates use distance and angle, (r, theta). Rectangular coordinates use horizontal and vertical distances, (x, y). They describe the same point in different ways.
Can r be negative?
In some conventions, yes, but it means the point lies in the opposite direction from theta. For most introductory problems, keep r non-negative and let theta carry the direction.
How do I convert back to polar?
Use r = sqrt(x^2 + y^2) for distance, and find theta from tan(theta) = y/x, adjusting for the quadrant of (x, y), because arctangent alone does not distinguish opposite quadrants.
