How do you sketch the angle in standard position $\frac{2 π}{9}$ $(2pi)/9$ ?

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Answer 1 · 2018-08-09T08:29:06

See graph and and explanation.

The direction angle $θ = \frac{2}{9} π = (\frac{2}{9} {(180)}^{o} = 40^{o}$ $theta = 2/9pi = (2/9(180)^o = 40^o$ .

The Cartesian $P (1, {tan 40}^{o}) = (1, 0.8391)$ $P ( 1, tan 40^o ) = ( 1, 0.8391 )$ .

In polar coordinates, it is $P (O P, 40^{o})$ $P ( OP, 40^o )$

$= P (\sqrt{1^{2} + {(0.8391)}^{2}}, 40^{o}) = P (1.2054, 40^{o})$ $= P ( sqrt ( 1^2 + (0.8391)^2), 40^o ) = P ( 1.2054, 40^o )$ . See grapk.

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

How do you sketch the angle in standard position 2π9(2pi)/9?