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Trigonometry : Trigonometry

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Example Questions

Example Question #571 : Trigonometry

What is the angular velocity of an angle whose measure is $\frac{\pi}{3}$ when it took a point 10 seconds to rotate through the angle.

Possible Answers:

$\omega = \frac{\pi}{30}$ radians/seconds

$\omega = 30\pi$ radians/seconds

$\omega = 10\pi$ radians/seconds

$\omega = \frac{8\pi}{30}$ radians/seconds

Correct answer:

$\omega = \frac{\pi}{30}$ radians/seconds

Explanation:

We have enough information to plug into the angular velocity formula to solve this problem. $\theta = \frac{\pi}{3}$ and t = 10 seconds.

$\omega = \frac{\theta}{t}$

$\omega = (\frac{\pi}{3})(\frac{1}{10})$

$\omega = \frac{\pi}{30}$

And so our answer is $\omega = \frac{\pi}{30}$ radians/seconds.

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Example Question #572 : Trigonometry

If you are walking around a circular track and your angular velocity through the angle $\theta = \frac{\pi}{6}$ is $\omega = \frac{\pi}{12}$ , how long would it take you to complete a full rotation?

Possible Answers:

2 seconds

22 seconds

12 seconds

24 seconds

Correct answer:

24 seconds

Explanation:

This is simply solving for an unknown. Here our unknown is time, . So we know that it takes you have an angular velocity of $\omega = \frac{\pi}{12}$ through the angle $\theta = \frac{\pi}{6}$ . Assuming that your angular velocity is consistent, we will be able to solve for . First we must manipulate our original angular velocity formula to solve for .

$\omega = \frac{\theta}{t}$

$t = \frac{\theta}{\omega}$

Now we will plug in our known value for $\omega$ and the angle of a full rotation. The angle of a full rotation is simply that of a full circle, $2\pi$ .

$t = \frac{\theta}{\omega}$

$t = \frac{2\pi}{\frac{\pi}{12}}$

$t = 2\pi (\frac{12}{\pi})$

t = 24

So it takes you 24 seconds to complete a full rotation

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Example Question #6 : Angular Velocity

You are on a carousel and have completed one full rotation in 15 seconds. What is your angular velocity?

Possible Answers:

$\omega = 15$ radians/seconds

$\omega = \frac{7\pi}{13}$ radians/seconds

$\omega = \frac{2\pi}{15}$ radians/seconds

$\omega = 2\pi$ radians/seconds

Correct answer:

$\omega = \frac{2\pi}{15}$ radians/seconds

Explanation:

Even though we are not directly given an angle measurement, we are told the time it takes to complete one full rotation. One rotation is the same as the angle of an entire circle, $2\pi$ . Now we have enough information to plug into the angular velocity formula to solve this problem.

$\omega = \frac{\theta}{t}$

$\omega = \frac{2\pi}{15}$

So the angular velocity is $\omega = \frac{2\pi}{15}$ radians/seconds

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Example Question #2 : Angle Applications

A wheel is making one full rotation with an angular velocity of $\frac{5\pi }{2}$ radians/seconds. How long will it take the wheel to make 10 full rotations?

Possible Answers:

20 seconds

8 seconds

24 seconds

1 second

Correct answer:

8 seconds

Explanation:

We know that the angular velocity to complete a full rotation (rotate through an angle of $2\pi$ ) is $\frac{5\pi}{2}$ radians/seconds. The angle of 10 full rotations will be 10 rotations through the angle $2\pi$ , so we multiply these two quantities.

$10 *2\pi = 20\pi$

And this will be our $\theta$ . Now we must manipulate our angular velocity formula to solve for .

$\omega = \frac{\theta}{t}$

$t = \frac{\theta}{\omega}$

Now we can plug in our $\theta$ for the 10 rotations and our angular velocity.

$t = \frac{\theta}{\omega}$

$t = \frac{20\pi}{\frac{5\pi}{2}}$

$t = 20\pi \frac{2}{5\pi}$

t = 8

And so it will take 8 seconds to make 10 rotations.

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Example Question #573 : Trigonometry

If it takes you 3 seconds to rotate through an angle with an angular velocity of $\frac{8\pi}{9}$ , what is the measure of the angle you are rotating through?

Possible Answers:

$\theta = \frac{8\pi}{9}$

$\theta = \frac{8\pi}{3}$

$3\pi$

$\theta = \frac{\pi}{3}$

Correct answer:

$\theta = \frac{8\pi}{3}$

Explanation:

We are able to use the angular velocity formula and solve for the unknown $\theta$ .

$\omega = \frac{\theta}{t}$

$\omega t = \theta$

$\frac{8\pi}{9} (3) = \theta$ (making it so we are solving for $\theta$ )

$\frac{24\pi}{9} = \theta$

$\frac{8\pi}{3} = \theta$

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Example Question #1 : Area Of A Sector

Which of the following is the definition for a sector of a circle?

Possible Answers:

A section of a circle which is enclosed by a chord and the chord’s arc

Any section of a circle

A section of a circle which is enclosed by two radii joined at the center of the circle and the arc between the two radii

Any section of a circle that is triangular in shape

Correct answer:

A section of a circle which is enclosed by two radii joined at the center of the circle and the arc between the two radii

Explanation:

Below is an illustration of a sector of a circle. A sector is the area of a circle which has been enclosed by two radii and the arc between them. A sector is not to be confused with a segment of a circle. A segment is when the area enclosed by the chord of a circle and the arc of the chord.

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Example Question #2 : Area Of A Sector

Which of the following is the formula for the area of a sector?

Possible Answers:

$A = \theta \pi r^2$

$A = \frac{\theta\pi}{2} r^2$

$A = \frac{\theta}{2} r^2$

$A = 2\theta r^2$

Correct answer:

$A = \frac{\theta}{2} r^2$

Explanation:

When thinking about how to derive the formula for a sector, we must consider the angle of an entire circle. The angle of an entire circle, 360 degrees, is $2\pi$ and we know the area of a circle is $\pi r^2$ .

When considering a sector, this is only a portion of the entire circle, so it is a particular $\theta$ out of the entire $2\pi$ . We can plug this into our area for a circle and it will simplify to the area of a sector.

$\frac{\theta}{2\pi}\pi r^2$

$\frac{\theta}{2} r^2$

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Example Question #1 : Area Of A Sector

If a circle has a sector with an angle of $\theta = 60$ and diameter of 4, what is the area of the sector?

Possible Answers:

$\frac{\pi}{6}$

$\frac{\pi}{3}$

$\frac{2\pi}{3}$

$2\pi$

Correct answer:

$\frac{2\pi}{3}$

Explanation:

It is always best to draw a picture in order to visualize the problem you are trying to solve. The figure below shows the sector we are trying to find the area of.

We know that the formula to find the area of a sector is $A = \frac{\theta}{2} r^2$ . From the information given above we know that the diameter is 4. Since we only need the radius for our formula we divide the diameter by 2 to get the radius length. The radius has a length of 2. We also know that we have our angle measure in degrees and must convert it to radians. We use the conversion formula $degrees \frac{\pi}{180} = radians$ .

$60 \frac{\pi}{180} = \frac{\pi}{3}$

Now we can plug everything into our formula and solve.

$A = \frac{\theta}{2}r^2$

$A = \frac{\pi}{3}\frac{1}{2} 2^2$

$A = \frac{\pi}{6}{4}$

$A = \frac{2\pi}{3}$

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Example Question #4 : Area Of A Sector

True or False: The formula to find the area of a sector only works for acute angles.

Possible Answers:

False

True

Correct answer:

True

Explanation:

This is not true. Even obtuse angles are less than $2\pi$ so this formula will still work. We can demonstrate this using the sector below. The radius of the circle is 6 and the obtuse angle is 330 degrees.

Converting 330 degrees to radians:

$330 \frac{\pi}{180} = \frac{11\pi}{6}$

We can now plug this into our formula

$A = \frac{\theta}{2} r^2$

$A = \frac{11\pi}{6}\frac{1}{2} 6^2$

$A = \frac{11\pi}{12} 36$

$A = \frac{396\pi}{12}$

$A = 33\pi$

Now we can confirm this to be true by computing the area of the sector formed by the

area leftover formed by the acute angle, 30 degrees. To do this we will first find the total

area of the circle and then subtract the area of the sector formed by the acute angle. This should be equal to the area of the larger vector if our formula works for all angles because the sum of both sectors should be the total area of the circle.

To find the area of the circle:

$\pi r^2$

$36\pi$

To find the area of the smaller sector (note, 30 degrees in radians is $\frac{\pi}{6}$ :

$A = \frac{\theta}{2} r^2$

$A = \frac{\pi}{6}\frac{1}{2} 6^2$

$A = \frac{\pi}{12} 36$

$A = 3\pi$

Clearly, the total area of the circle minus the area of the small sector is equal to the area

of the larger circle, therefore this formula works for all angles less than $2\pi$

$36\pi - 3\pi = 33\pi$

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Example Question #5 : Area Of A Sector

Knowing that the arc length of a sector $3\pi$ and the angle measure is $\theta = \frac{4\pi}{3}$ , what is the area of the sector?

Possible Answers:

$A = \frac{\pi}{3}$

$A = \frac{81\pi}{127}$

$A = \frac{\pi}{4}$

$A = \frac{27\pi}{8}$

Correct answer:

$A = \frac{27\pi}{8}$

Explanation:

To solve this problem we must know the formula for finding the arc length of a sector. This formula is $l_A = \theta r$ . With the given information we are able to solve for the radius which we can then use to solve for the area of the sector itself.

$l_A = \theta r$

$3\pi = \frac{4\pi}{3} r$

$\frac{9}{4} = r$

Now we can plug this radius into the formula to solve for the area of a sector.

$A = \frac{\theta}{2} r^2$

$A = (\frac{4\pi}{3})(\frac{1}{2})( \frac{9}{4})^2$

$A = (\frac{2\pi}{3})(\frac{81}{16})$

$A = \frac{162\pi}{48}$

$A = \frac{27\pi}{8}$

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #571 : Trigonometry

Example Question #572 : Trigonometry

Example Question #6 : Angular Velocity

Example Question #2 : Angle Applications

Example Question #573 : Trigonometry

Example Question #1 : Area Of A Sector

Example Question #2 : Area Of A Sector

Example Question #1 : Area Of A Sector

Example Question #4 : Area Of A Sector

Example Question #5 : Area Of A Sector

All Trigonometry Resources

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