Finding Missing Angles Using the Unit Circle

Quiz summary

0 of 3 questions completed

Questions:

–

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading...

You must sign in or sign up to start the quiz.

You have to finish following quiz, to start this quiz:

Lets see your results!

Points

Score

Time

Retry Quiz

Question 1 of 3

Given that

$0≤theta≤2pi$ , solve for

$theta$

$tan^2theta=3$

1. $pi/6,(5pi)/6,(7pi)/6,(11pi)/6$
2. $pi/4,(7pi)/6$
3. $pi/3,(2pi)/3,(4pi)/3,(5pi)/3$
4. $pi/3,(4pi)/3$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Trigonometric Functions

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

First, simplify the given equation

$tan^2theta$	$=$	$3$
$sqrt(tan^2theta)$	$=$	$sqrt3$	Get the square root of both sides
$tan theta$	$=$	$+-sqrt3$

Remember that

$tan theta=(sin theta)/(cos theta)$ . Knowing that

$tan theta$ can be either positive or negative, this means that

$theta$ lies on all

$4$ quadrants

Next, find the reference angle to work with by using special triangles

$tan theta$	$=$	$+-sqrt3$	$=$	$\frac{\text{opposite}}{\text{adjacent}}$

$tan theta$	$=$	$+-sqrt3$	$=$	$(sqrt3)/1$	Using $pi/3$ as $theta$

$tan theta$	$=$	$+-sqrt3$	$=$	$sqrt3$

Hence, the reference angle and the value of

$theta$ on the First Quadrant is $pi/3$

Finally, use the reference angle to find the value of

$theta$ on the other

$3$ quadrants

Remember that the reference angle is always relative to the

$x$ -axis

Second Quadrant:

$theta$	$=$	$pi-pi/3$

	$=$	$(3pi)/3-pi/3$

	$=$	$(2pi)/3$

Third Quadrant:

$theta$	$=$	$pi+pi/3$

	$=$	$(3pi)/3+pi/3$

	$=$	$(4pi)/3$

Fourth Quadrant:

$theta$	$=$	$2pi-pi/3$

	$=$	$(6pi)/3-pi/3$

	$=$	$(5pi)/3$

$theta=pi/3,(2pi)/3,(4pi)/3,(5pi)/3$

Question 2 of 3

2. Question
Given that $0≤theta≤2pi$ , solve for $theta$

$sin^2theta=1$
- 1. $pi/4,(3pi)/4,(5pi)/4,(7pi)/4$
- 2. $pi/4,(3pi)/4$
- 3. $pi/2,pi,(3pi)/2,2pi$
- 4. $pi/2,(3pi)/2$
Hint

Help Video
Correct

Correct!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

Trigonometric Functions

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

First, simplify the given equation

$sin^2theta$ $=$ $1$

$sqrt(sin^2theta)$ $=$ $sqrt1$ Get the square root of both sides

$sin theta$ $=$ $+-1$

In the unit circle. mark out the coordinates on the horizontal and vertical axis

Keep in mind that the radius of a unit circle is $1$

Now, recall that in a coordinate, $(x,y)=(cos theta, sin theta)$

$sin theta=+-1$

The $y$ value in these coordinates match the value of $sin theta$ . Therefore, $theta=pi/2,(3pi)/2$

$theta=pi/2,(3pi)/2$
Question 3 of 3

3. Question
Given that $0≤theta≤2pi$ , solve for $theta$

$costheta=-1$
- 1. $theta=pi/2$
- 2. $theta=pi$
- 3. $theta=(3pi)/2$
- 4. $theta=2pi$
Hint

Help Video
Correct

Exceptional!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

Trigonometric Functions

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

In the unit circle. mark out the coordinates on the horizontal and vertical axis

Keep in mind that the radius of a unit circle is $1$

Now, recall that in a coordinate, $(x,y)=(cos theta, sin theta)$

$cos theta=-1$

The $x$ value in $pi$ the value of $cos theta$ . Therefore, $theta=pi$

$theta=pi$

Finding Missing Angles Using the Unit Circle

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Excellent!

Trigonometric Functions

2. Question

Hint

Correct!

Trigonometric Functions

3. Question

Hint

Exceptional!

Trigonometric Functions

Quizzes

$sin^2theta$	$=$	$1$
$sqrt(sin^2theta)$	$=$	$sqrt1$	Get the square root of both sides
$sin theta$	$=$	$+-1$