# How do you determine the quadrant in which the angle -132^circ50' lies?

##### 1 Answer
Nov 29, 2017

−132^@50' is in the ${3}^{r d}$ quadrant

#### Explanation:

If we consider the conventional unit circle, an $\angle \theta$ is measured counter-clockwise from the positive $x -$axis through the four quadrants delimited by ${90}^{\circ} , {180}^{\circ} , {270}^{\circ} , {360}^{\circ}$ respectively. These detemine the ${1}^{s t} , {2}^{n d} , {3}^{r d} \mathmr{and} {4}^{t h}$ quadrants.

However, we can also measure negative $\angle \theta$ clockwise from the positive $x -$axis where the four quadrants are delimited by $- {90}^{\circ} , - {180}^{\circ} , - {270}^{\circ} , - {360}^{\circ}$ as $- \theta$ passes through the ${4}^{t h} , {3}^{r d} , {2}^{n d} \mathmr{and} {1}^{s t}$ quadrants around the circle.

Here we are asked in which quadrant theta = −132^@50'

−132^@50' is between $- {90}^{\circ} \mathmr{and} - {180}^{\circ}$ and hence it is in the ${3}^{r d}$ quadrant.

An altenative approach would be to consider:

−132^@50' = 360^@ −132^@50' =227^@10'

Since ${227}^{\circ} 10 '$ is betwen ${180}^{\circ} \mathmr{and} {270}^{\circ}$ i.e the ${3}^{r d}$ quadrant

-> −132^@50' is in the ${3}^{r d}$ quadrant.