Chapter 11.3, Problem 39E

To determine

To Find: The polar coordinates of the image of point $A$ after ${90}^{°},{180}^{°},{270}^{°}$ rotation.

Answer to Problem 39E

  $\begin{array}{l}{90}^{°}:A(2,30)=A\text{'}(2,120)\\ {180}^{°}:A(2,30)=A\text{'}(2,210)\\ {270}^{°}:A(2,30)=A\text{'}(2,300)\end{array}$

Explanation of Solution

Given: Point $A(2,30)$ is $2units$ from the origin and $\angle XOA={30}^{°}.$

Rotating the point $A(2,30)$

  $\begin{array}{l}{90}^{°}:A(2,30)=A\text{'}(2,120)\\ {180}^{°}:A(2,30)=A\text{'}(2,210)\\ {270}^{°}:A(2,30)=A\text{'}(2,300)\end{array}$

Radius remains the same.