Polar coordinates give the distance from the origin and the angle from the positive x axis. Cartesian coordinates give the distance from the x and y axes.
You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.
Let's convert (4, 16°) to Cartesian coordinates.
Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.
cosine = adjacent / hypotenuse
cosθ = x/r
cos(16°) = x/4
4cos(16°) = x ≈ 3.84504678375
sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327
So (4, 16°) ⇒ (3.84504678375, 1.10254942327).
Let's convert (-2, 177°) to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)
In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).
Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.
cosine = adjacent / hypotenuse
cosθ = x/r
cos(357°) = x/2
2cos(357°) = x ≈ 1.99725906951
sine = opposite / hypotenuse
sinθ = y/r
sin(357°) = y/2
2sin(357°) = y ≈ -0.10467191248
So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).
Now we must use the distance formula with our two points.
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]d\approx\sqrt{(1.99725906951-3.84504678375)^2+(-0.10467191248-1.10254942327)^2}[/tex]
[tex]d\approx\sqrt{-1.84778771^2+-1.20722134^2}[/tex]
[tex]d\approx\sqrt{3.41431942+1.45738336}[/tex]
[tex]d\approx\sqrt{4.87170278}[/tex]
[tex]\boxed{d\approx2.20719342}[/tex]
