Question

Explain why the inequalities $r>0$and $0<\theta <\frac{\pi }{2}$ together describe the points in the first quadrant. Use similar inequalities to describe the points in the third quadrant.

Short answer

The answer is $r>0$and $\pi <\theta <\frac{3\pi }{2}$

Step-by-step solution

Given information

The inequalities $r>0$and $0<\theta <\frac{\pi }{2}$

Simplification

Consider the inequalities $r>0$and $0<\theta <\frac{\pi }{2}$.

The objective is to write the inequality which is in the third quadrant.

In the first quadrant, all the values are more than $0°$and less than $\frac{\pi }{2}$.

All the angles are positive in the first quadrant.

Every point $(r,\theta )$lies on the ray since $r>0$.and the $\theta$values are in between role="math" localid="1652103013215" $0^{\text{ο}}to\frac{\pi }{2}$.

In the third quadrant $r>0$and all the $\theta$values are in between $\pi$to $\frac{3\pi }{2}$.

Every point $(r,\theta )$lies on the ray and $\theta$values are in between $\pi$to $\frac{3\pi }{2}$.

Thus, representation of angles by using inequality is as follows,

$r>0$and $\pi <\theta <\frac{3\pi }{2}$

Therefore, the answer is $r>0$and $\pi <\theta <\frac{3\pi }{2}$