# convert the vector (-1.0, 1.0) into magnitude/angle form. When using inverse tangent, the angle is -45 degrees. how can the angle be negative 45 degrees?

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We are given the rectangular coordinates (-1,1) and want to convert to polar form `(r,theta)` where r is the magnitude and `theta` is the angle from the initial ray. (The initial ray is taken to be the positive x-axis in the Cartesian coordinate plane.)

Recall that a vector can have more than one description. If the magnitude is positive then given any angle, we can always add/subtract 360 to the angle to get an equivalent expression. E.g. `(5,30^@)` is equivalent to `(5,390^@)` , at least as far as the vector is concerned. If the magnitude is **negative**, then you reflect the vector (or rotate `180^@` ). So for our example, `(5,30^@)` can also be represented as `(-5,210^@)` .

Now given (-1,1): we can find the magnitude since a point given as (x,y) in the rectangular coordinate system has magnitude `sqrt(x^2+y^2)` . This is really just an application of the Pythagorean theorem. Plot the point and drop a perpendicular segment to the x-axis and consider the triangle formed by the origin, the point, and the foot of the altitude. The magnitude of the vector is the hypotenuse of the right triangle.

Thus the magnitude is `r=sqrt((-1)^2+1^2)=sqrt(2)` .

The angle can be found since the x-coordinate is given by `rcostheta` .

We know r and the x-coordinate, so `-1=sqrt(2)costheta` .

Then `costheta=-sqrt(2)/2 ==> theta=cos^(-1)(-sqrt(2)/2)=-45^@` .

Now the angle negative 45 degrees is in the 4th quadrant, while the vector we want is in the second quadrant. So we can change the sign on r.

**One expression for the vector is** `(-sqrt(2),-45^@)` .

We could also use the 45 degrees as a reference angle. We know we want to be in the second quadrant -- with a reference angle of 45 degrees the angle is 135 degrees. **So another expression for the vector is** `(sqrt(2),135^@)`