Question

How do you sketch the angle in standard position `(2pi)/9`?

Answer 1

See graph and and explanation.

Explanation:

The direction angle `theta = 2/9pi = (2/9(180)^o = 40^o`.

The Cartesian `P ( 1, tan 40^o ) = ( 1, 0.8391 )`.

In polar coordinates, it is `P ( OP, 40^o )`

`= P ( sqrt ( 1^2 + (0.8391)^2), 40^o ) = P ( 1.2054, 40^o )`. See grapk.

graph{(arctan(y/x) - 2/9pi)((x-1)^2 + (y- tan(2/9pi))^2-0.0005)=0[0 2 0 1]}