Angles

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Trigonometry › Angles

Questions 1 - 10

Which quadrant does $135^\circ$ belong?

III

Explanation

Step 1: Define the quadrants and the angles that go in:

QI:

$0^\circ<x\leq 90^\circ$

QII:

$90^\circ < x \leq 180^\circ$

QIII:

$180^\circ < x \leq 270^\circ$

QIV:

$270^\circ < x \leq 360$

Step 2: Find the quadrant where $135^\circ$ is:

The angle is located in QII (Quadrant II)

Which quadrant does $135^\circ$ belong?

III

Explanation

Step 1: Define the quadrants and the angles that go in:

QI:

$0^\circ<x\leq 90^\circ$

QII:

$90^\circ < x \leq 180^\circ$

QIII:

$180^\circ < x \leq 270^\circ$

QIV:

$270^\circ < x \leq 360$

Step 2: Find the quadrant where $135^\circ$ is:

The angle is located in QII (Quadrant II)

Find all positive values of $\theta$ less than $360^\circ$ for which $\cos\theta=-\frac{1}{2}$ .

$\theta=60^\circ$

$\theta=120^\circ$

$\theta=240^\circ$

$\theta=60^\circ$ and $240^\circ$

$\theta=120^\circ$ and $240^\circ$

Explanation

At first glance, you may think that this problem has infinite answers, since there would be infinitely many negative coterminal angles that could satisfy this; however, notice that the question asks only for positive values of $\theta$ . In other words, this question is simply asking for values of $\theta$ between $0^\circ$ and $360^\circ$ that satisfy this equation.

First, let's think about where the cosine function is negative. Per the chart below, it will be in Quadrants II and III.

Screen shot 2020 07 30 at 10.25.02 am

The reference angle for each angle solution will have its cosine equal to $+\frac{1}{2}$ and is $60^\circ$ . Consult the chart of reference angles below for Quadrants II and III:

Screen shot 2020 07 30 at 11.05.57 am

QII: $\theta=180^\circ-R=180^\circ-60^\circ=120^\circ$

QIII: $\theta=R+180^\circ=60^\circ+180^\circ=240^\circ$

Find all positive values of $\theta$ less than $360^\circ$ for which $\cos\theta=-\frac{1}{2}$ .

$\theta=60^\circ$

$\theta=120^\circ$

$\theta=240^\circ$

$\theta=60^\circ$ and $240^\circ$

$\theta=120^\circ$ and $240^\circ$

Explanation

First, let's think about where the cosine function is negative. Per the chart below, it will be in Quadrants II and III.

Screen shot 2020 07 30 at 10.25.02 am

The reference angle for each angle solution will have its cosine equal to $+\frac{1}{2}$ and is $60^\circ$ . Consult the chart of reference angles below for Quadrants II and III:

Screen shot 2020 07 30 at 11.05.57 am

QII: $\theta=180^\circ-R=180^\circ-60^\circ=120^\circ$

QIII: $\theta=R+180^\circ=60^\circ+180^\circ=240^\circ$

Determine the quadrant that contains the terminal side of an angle measuring $\frac{-7\pi }{6}$ .

Explanation

Each quadrant represents a $\frac{\pi }{2}$ change in radians. Therefore, an angle of $\frac{-7\pi }{6}$ radians would pass through quadrants , , and end in quadrant . The movement of the angle is in the clockwise direction because it is negative.

Determine the quadrant that contains the terminal side of an angle measuring $\frac{-7\pi }{6}$ .

Explanation

Find the least positive coterminal angle to $-1140^{\circ}$ .

$300^{\circ}$

$-780^{\circ}$

$60^{\circ}$

$420^{\circ}$

Explanation

To find a coterminal angle, you must add or subtract $360^{\circ}$ . The question is asking for the least positive coterminal angle, so you must add $360^{\circ}$ until you reach a positive angle.

$-1140^{\circ} + 360^{\circ} = -780^{\circ}$

The angle is still negative, so you must continue.

$-780^{\circ} + 360^{\circ} = -420^{\circ}$

The angle is still negative, so you must continue.

$-420^{\circ} + 360^{\circ} = -60^{\circ}$

The angle is still negative, so you must continue.

$-60^{\circ} + 360^{\circ} = 300^{\circ}$

Find the least positive coterminal angle to $-1140^{\circ}$ .

$300^{\circ}$

$-780^{\circ}$

$60^{\circ}$

$420^{\circ}$

Explanation

To find a coterminal angle, you must add or subtract $360^{\circ}$ . The question is asking for the least positive coterminal angle, so you must add $360^{\circ}$ until you reach a positive angle.

$-1140^{\circ} + 360^{\circ} = -780^{\circ}$

The angle is still negative, so you must continue.

$-780^{\circ} + 360^{\circ} = -420^{\circ}$

The angle is still negative, so you must continue.

$-420^{\circ} + 360^{\circ} = -60^{\circ}$

The angle is still negative, so you must continue.

$-60^{\circ} + 360^{\circ} = 300^{\circ}$

Determine the quadrant that contains the terminal side of an angle $-380^{\circ}$ .

Explanation

Each quadrant represents a $90^{\circ}$ change in degrees. Therefore, an angle of $-380^{\circ}$ radians would pass through quadrants , , , and end in quadrant . The movement of the angle is in the clockwise direction because it is negative.