How do you determine the quadrant in which $- \frac{11 π}{9}$ $-(11pi)/9$ lies?

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How do you determine the quadrant in which $- \frac{11 π}{9}$ $-(11pi)/9$ lies?

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Answer 1 · 2018-02-15T23:25:40

The negative means you go clockwise instead of counterclockwise to graph the angle. Then...

Explanation:

Then, since $\frac{11}{9}$ $11/9$ is a little more than one, it means the angle is a little more than $π$ $\pi$ (or 180 degrees). Therefore, when you graph an angle moving clockwise and go past $π$ $\pi$ radians, you will be in Quadrant II

Answer 2 · 2018-02-22T04:31:22

Second quadrant.

Explanation:

$- \frac{11 π}{9} = - 1 (\frac{2 π}{9}) = - π - (\frac{2 π}{9})$ $-(11pi)/9 = -1((2pi)/9) = -pi - ((2pi)/9)$

$\Rightarrow 2 π - π - (\frac{2 π}{9}) = \frac{7 π}{9}$ $=> 2pi - pi - ((2pi)/9) = (7pi)/9$

Since $\frac{7 π}{9} > \frac{π}{2}$ $(7pi)/9 > pi/2$ , it is in second quadrant.

Aliter : -(11pi)/9 = -((11pi)/9) * (360/2pi) = - 220^@#

$\Rightarrow 360 - 220 = 140^{\circ} = {(90 + 50)}^{\circ}$ $=> 360 - 220 = 140^@ = (90 + 50)^@$

It’s in second quadrant, as $140^{\circ}$ $140^@$ is between $90^{\circ}$ $90^@$ and $180^{\circ}$ $180^@$

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How do you determine the quadrant in which $- \frac{11 π}{9}$ $-(11pi)/9$ lies?

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How do you determine the quadrant in which $- \frac{11 π}{9}$ $-(11pi)/9$ lies?